SCHOOL Mathematics Book 10 - kedcnepal.com

AuthorsHansha Raj Dhungana

Narayan Prasad Dahal

Nava Raj Baral

Kantipur Education Development CouncilMaharajgunj, Chakrapath, Kathmandu, NepalPhone: 01-4720202/4720517Email: [email protected]: www.kedcnepal.com

KED

C

MathematicsMathematicsSCHOOL

Book 10

Approved by Government of Nepal, Ministry of Education, Curriculum Development Centre, Sanothimi, Bhaktapur

Kantipur Education Development Council

Maharajgunj, Chakrapath, Kathmandu, Nepal

Phone: 4720202/4720517

Email: [email protected]

Website: www.kedcnepal.com

Kantipur Education Development Council 2058 ALL

RIGHTS RESERVED. No part of this book may be

reproduced or transmitted in any form by any means,

electronic or mechanical, including photocopying and

recording, or by any information storage and retrieval

system, except as may be expressly permitted in writing

by the publisher.

Edited by:

Mahendra Raj Ghimire

Yubaraj Bhandari

Written by:

Hansha Raj Dhungana

Narayan Prasad Dahal

Nava Raj Baral

Edition:

First, 2019, Revised edition, 2020

Layout/Design

Jaydev Koirala

KED

C

Preface

This book hopes to develop and deep-root interest in learners in mathematics. The fun in learning mathematics is possible only when there is right co-ordination between the learners and the facilitators. It can helps us raise the curiosity to emancipate the bundle of mysteries of the world.

Mathematics is the most essential discipline among so many subjects. So, the strong foundation is necessary to be built up. The learner will find the content in this book interesting to practice and certainly find it useful in day to day life.

The exercises at design as per the curriculum of CDC of Nepal in a very simple language and in clear an concise form. Complete theory in each chapter, figures, notes, tables are the key component of this book. Simple but different techniques, sufficient project works and practice sets bind both the learners and facilitators for easy preparation of National Examination Board or any other examination.

KEDC will like to extent it heartfelt thanks to Mr. Ghanashyam Dhakal for his emmance help and contribution in producing such wonderful books.

Constructive suggestion for further improvement are heartily apperciated.

Contents

Unit 1: Set 1-18Unit 2: Arithmetic 19-642.1. Discount and VAT 202.2 Money and Exchange 332.3. Compound Interest 382.4. Population Growth 532.5 Depreciation 59Unit 3: Mensuration 65-1413.1. Area of Triangle 663.2. Prism 813.3. Cylinder 923.4. Sphere 1003.5. Pyramid 1113.6. Cone 1203.7. Combined Solids 1293.8. Application of Mensuration 135Unit 4: Algebra 142-2194.1 HCF 1434.2. LCM 1484.3. Simplification 1534.4. Indices 1644.5. Exponential Equation 1764.6. Rational Number 180

4.7. Rationalization of Surds 1854.8. Equations Involving Surds 1904.9. Simultaneous Equations 1954.10. Quadratic Equation 207Unit 5: Geometry 220-2825.1. Area of Triangle and

Quadrilaterals 221

5.2. Circle 2415.3. Tangent 2675.4. Constructions 275Unit 6: Trigonometry 283-3056.1. Area of triangle 2846.2. Height and Distance 295Unit 7: Statistics 306-3287.1. Mean 3077.2. Median 3147.3. Quartiles 3187.4. Ogive 320Unit 8 : Probability 329-3538.1. Addition Law 3328.2. Multiplication law 3378.3. Tree Diagram 345Model Question 354

School Mathematics Book - 10 1

SetEstimated period : 8

1 Unit

Specific objectivesAt the end of this unit, students will be able to

• find cardinal number of the given set.• perform set operation (union, intersection, difference and complement).• represent set operation in a Venn-diagram.• solve the word problems based on two or three sets by using the Venn- diagram. • relate the set operations with the daily life situation and solve the problems

using mathematical relation.

Teaching MaterialsCharts of Venn-diagram, geometrical instruments , flash cards

Note to the teacher1. Give the concept of cardinality of a set.

2. Describe the formula of n(A∪B) and n(A∪B∪C).

3. Give clear concept to solve the set problems with the help of Venn – diagram.

4. Set some extra assignment to the topics taught in the class.

Curriculum• Word problems of two and three sets using Venn diagram.

Specification Grid → Cognitive

Domain

↓

Topic

Knowledge (K) Comprehensive (C) Application (A) Higher ability(HA) Total no.

of

questions

Total

marks

Sets Each 1 mark Each 2 marks Each of 4 marks Each of 5 marks

- - 1 - 1 4

In previous grades, we have learnt about types of set, set relation, Venn- diagrams, operation of sets and simple problems of two sets by using Venn- diagrams. Now we will review some general problems related to basic concepts of set and further more we will study here word problems related to two or three sets by using Venn- diagrams.

School Mathematics Book - 102

Cardinality of set: The number of distinct elements present in the set is called cardinality. If A is a finite set then the number of elements in set A is denoted by n(A) which is the cardinal number of the set A. Thus, if A = {a,e,i,o,u}, then n(A) = 5

Use of Venn- diagram to find the number of elements in setCardinality relation two intersecting sets A and B with following informationn(A) = Number of elements in set A.n(B) = Number of elements in set B.Let, x = Number of elements in set A only.

= no(A) i.e. only in A not in B= n( A – B)= n(A) – n (A ∩ B)

y = Number of common elements of sets A and B = n (A ∩ B)z = Number of elements in set B only. = no(B) i.e. only in B not in A = n( B – A) = n(B) – n (A ∩ B)Now, n(A ∪ B) = x + y + z

= n(A) – n (A ∩ B) + n (A ∩ B) + n(B) – n (A ∩ B){ From the above relation }= n(A) + n(B) – n (A ∩ B)

Hence, 1. Number of elements lying in at least one of A and B. n(A ∪ B) = n(A) + n(B) – n (A ∩ B) or, n(A ∪ B) = no (A) + no(B) + n (A ∩ B) or, n(A ∪ B) = n(U) – n ( A ∪B )2. Number of elements lying in both A and B. n (A ∩ B) = n(A) + n(B) - n(A ∪ B) or, (A ∩ B) = n(A ∪B) - no(A) - no(B)3. Number of elements of A only. n(A-B ) = n(A ∪B) - n(B) or, no (A) = n(A) - n (A ∩B)

A B U

x y z


4. Number of elements of B only.n(B - A ) = n(A ∪B) - n(A)or, no (B) = n(B) - n (A ∩ B)

5. Number of elements of exactly one set no (A) + no (B)= n(A ∪ B) - n (A ∩ B)

6. Number of elements not in A and B. n ( A ∪ B ) = n (∪) – n(A ∪ B)7. Total number of elements under consideration

n(U) = n(A ∪ B) + n ( A ∪ B ) or, n (U) = n (A) + n (B)- n (A ∩ B) + n ( A ∪ B )

8. If the universal set contains only the elements of sets A and B , thenn ( A ∪ B ) = 0 , So, n(U) = n(A ∪ B)

Cardinality relations of union of three setsLet A , B and C be three overlapping subsets of universal set U. Then, the cardinal number of sets A, B and C are denoted by n(A), n(B) and n(C) respectively. Besides using the notations as in case of two sets, the number of elements in the union of three sets is denoted by n (A∪B∪C).n(A ∪ B ∪ C) = n {A ∪ (B ∪ C)} = n(A) + n(B ∪ C) - n{A ∩ (B ∪ C)}

= n (A) + n(B) + n(C) - n(B ∩ C) - n{(A ∩ B) ∪ (A ∩ C)}= n (A) + n (B) + n(C) - n(B ∩ C) - {n(A ∩ B) + n (A ∩ C) - n (A ∩ B) ∩ (A ∩ C)}= n (A) +n(B) + n(C) - n(B ∩ C) - n(A ∩ B) - n (A ∩ C) + n(A ∩ B ∩ C)∴ n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) -n(C ∩ A) + n(A ∩ B ∩ C)

Also,n(A ∪ B ∪ C) = no (A) + no(B) + n0 (C) + n0 (A ∩ B) +n0 (B ∩ C) +n0 (C ∩ A) + n(A ∩ B ∩ C) The relation given below are also equally applicable on solving problems. So keep these relation on your mind too.n(A) = n0 (A) + n0(A ∩B) + n0 (A ∩C) + n(A ∩B ∩C)n(B) = n0 (B) + n0(A ∩B) + n0 (B ∩C) + n(A ∩B ∩C)

no(A∩C)no(B∩C)

n(B∩C)n(A∩C)

no(A∩B)no(A)

no(C)

no(B)

A B U

C

n(A)n(B)

n(C)

n(A∩B)


n(C) = n0 (C) + n0(A ∩C) + n0 (B ∩C) + n(A ∩B ∩C)n(A) = n0 (A) - n (A ∩B) - n(A ∩C) + n(A ∩B ∩C)n0(B) = n(B) - n (A ∩B) - n(B ∩C) + n(A ∩B ∩C)n0(C) = n(C) - n (B ∩C) - n(A ∩C) + n(A ∩B ∩C)n(A ∩B) = n0 (A ∩B) + n (A ∩B ∩C) n(B ∩C) = n0 (B ∩C) + n (A ∩B ∩C)n(A ∩C) = n0 (A ∩C) + n (A ∩B ∩C) n0(A ∩B) = n(A ∩B) - n (A ∩B ∩C)n0(B ∩C) = n (B ∩C) - n (A ∩B ∩C)n0(A ∩C) = n(A ∩C) - n (A ∩B ∩C)

n( A ∪B ∪C ) =n(U) - n(A ∪B ∪C)

Example 1 : If n(U) = 100, n(A) = 60, n(B) = 65, n( A ∩B ) = 30find : a. n(A ∪B)

b. n(A ∩B) c. n0(B)

Solution: We have, n(U) = 100, n(A) = 60, n(B) = 65 and n( A ∪B ) = 30We know, a. n(A∪B) = n(U) - n( A ∪B ) = 100 - 30 = 70

Also, b. n(A∪B) = n(A) +n(B) - n(A∩B)or, n(A∩B) = n(A) + n(B) -n(A∪B)

= 60 + 65 - 70 = 125 - 70 = 55

Again, c. n0(B) = n(B) - n(A∩B) = 65 - 55 = 10

Worked Out Examples


Example 2 : If n(U) = 60 , n(A) = 25 , n(B) = 35 and n(A ∩B) = 5 , find n(A-B), n(B-A), n(A ∪B) and n( A ∪B ) . Find the maximum value of n(A∪B) and maximum value of n(A ∩B) if A ⊂B.Solution : Here, n (U) = 60, n (A) = 25, n (B) = 35 and n (A∩B) = 5 n( A – B ) = n(A) – n(A ∩B) = 25 – 5 = 20n( B – A ) = n(B) – n(A∩B) = 35 – 5 = 30n(A∪B) = n(A) + n(B) - n(A∩B) = 25 + 35 – 5 = 55n( A ∪B ) = n(U) - n(A∪B) = 60 – 55 = 5

If A ⊂B, then A∪B = B and A∩B = A

So, maximum value of n(A∪B) = 35 and maximum value of n(A ∩B) = n(A) = 25

Example 3: In a survey report of 1500 students of a school, 600 liked tea, 750 liked coffee and 430 liked both drinks. Find the number of students who like neither of the two drinks. Show the above information in Venn – diagram.

Solution: Let U = Set of all surveyed students, T = Set of tea likers, C = Set of coffee likers

We have, Total number of students n(U) = 1500 Number of students who liked tea, n(T) = 600Number of students who liked coffee, n(C) = 750Number of students who liked both tea and coffee, n(T∩C) = 430Number of students who did not like tea and coffee, n( T ∪C ) = ?We know,

n(T ∪C) = n(T) +n(C) - n(T ∩C) = 600 + 750 – 430 = 1350 – 430 = 920

∴n( T ∪C ) = n(U) - n(T ∪ C) = 1500 – 920 = 580

∴The number of students who liked neither of the two drinks is 580.

A B U

20 5 30

n(B) = 35n(A) = 25

5

T C U

170 430 320

580


Example 4 : In a survey of workers in a factory, 70% liked football and 50% liked volleyball. If there were no workers who did not like both games then.

i. Present the above information in Venn – diagram. i. What percentage were there who liked both games? ii. What percentage liked football only?

Solution: Let, U = Set of surveyed workers F = Set of football likers V = set of volleyball likers

We have, Percentage of workers who liked football, n(F) = 70%Percentage of workers who liked Volleyball, n(V) = 50%Percentage of workers who liked both games, n(F∩V) = ?Percentage of workers who liked football only, n(F0)=?We know,

n(U) = n(F∪V) + n( F ∪V ) 100% = n(F) + n(V) - n(F∩V)or, 100% = 70% + 50% -n(F∩V)or, n(F∩V) = 120% - 100%

n(F∩V) = 20% ∴Percentage of workers who liked both games is 20%. Also, percentage of workers who liked football only = n(F) - n(F∩V) = 70% - 20% = 50% ∴ Percentage of workers who liked football only is 50% .Example 5 : In a survey conducted among some people of group, it was found that 55% of them liked coffee, 80% of them liked milk and 15% of them liked none; then

i. Represent the above information in a Venn – diagram. ii. If there are 110 people who liked both of them. Find the number of people

participated in the survey. Solution:

Let, 'U','C' and 'M' be the set of surveyed people, set of coffee likers and set of milk likers respectively.Number of people who liked coffee n(C) = 55%Number of people who liked milk n(M) = 80%Number of people who do not like both n( C ∪ M ) = 15%

F V U

50% 20% 30%


Let, C = Set of coffee likers M = Set of milk likers

We know, n(U) = n(C∪M) + n( C ∪M )

or, 100% = n(C ∪M) + 15%or, n(C ∪M) = 100% - 15%∴ n(C ∪ M) = 85%Again, n(C ∪M) = n(C) + n(M) - n(C∩M)or, n(C ∩ M) = n(C) + n(M) -n(C ∪ M)

= 135% - 85%= 50%

Let the total number of people = x By question, 50% of x = 110

50x100 = 110

x = 110 × 10050

= 220

∴Total number of people participated in the survey is 220. Example 6 : In a group of 100 students, the ratio of students who like Geography and Sociology is 3:5. If 30 of them like both subjects and 10 of them like none of the subjects then by drawing Venn- diagram find how many students of them: i. like Geography only.ii. like Sociology only. Solution: Here, Let, U = Set of total group of students G = Set of students who like Geography S = Set of students who like Sociology total number of students n(U) = 100

Since the ratio of the student who like Geography and Sociology is 3:5, let us supposeNumber of students who like Geography, n(G) = 3x Number of students who like Sociology, n(S) = 5xNumber of students who like both Subjects, n(G ∩ S ) = 30Number of students who don’t like both Subjects, n( G ∪ S ) = 10Number of students who like Geography only, no(G) = ?

C M U

5% 50% 30%

15%


Number of students who like Sociology only, no(S) = ?We know, n(U) = n(G∪S) + n( G ∪ S )n(G∪S) = n(U) - n( G ∪ S )

= 100 – 10= 90

Now,n(G∪S) = n(G) + n(S)- n(G ∩S)Or, 90 = 3x + 5x – 30Or, 8x = 120

Or, x = 1208 = 15

Now, n(G)= 3x = 3 × 15 = 45 n(S) = 5x = 5 × 15 = 75 Number of students who like Geography only, no(G) = n(G) - n(G ∩ S)

= 45 – 30 = 15

Number of students who like Sociology only, no(S) = n(S) - n(G ∩S) = 75 – 30 = 45

Example 7 : In a survey one third children like only apple and 22 do not like apple at all. Also 2

5 of children like banana but 12 like none of them.i. Show the above data in Venn-diagram.ii. How many children like both types of fruits? Find it.

Solution : Here,Let A and B be the set of children who like apple and banana respectively. Let total number of children be x.Let, n(U) = x , no(A) = x

3 , n(A ) = 22, n(B) = 2x5 , n( A ∪ B ) = 12

Now we know thatno(B) = n(A ) - n( A ∪ B ) = 22 – 12 = 10Then,n (U) = no(A) + n(B) + n( A ∪ B )

x = x3 + 2x

5 + 12

Or, x - x3 - 2x

5 = 12

Or, 15x-5x-6x15 = 12

A B

12

U

15 8 10

G S

10

U

15 30 45


Or, 4x15 = 12

or, x = 12×154

∴x = 45Again, Now, no(A) = x

3 = 453 = 15 n(U) = no(A) + no(B) + n(A ∩ B) + n( A ∪ B )

n(B) = 2x5 = 2× 45

5 = 18 Or, 45 = 15+10 + n(A ∩ B) + 12∴ no (B) = n(B) - n (A∩B) Or, 45 = 37 + n(A ∩ B)= 18 - 8 = 10 Or, n(A ∩ B) = 45 – 37 = 8Thus, 8 children like both the fruits.Example 8 : Out of 120 students appeared in an examination, the number of students who passed in Mathematics only is twice the number of the students who passed in Science only. If 50 students passed in both subjects and 40 students failed in both subjects then

i. find the number of the students who passed in Mathematics.

ii. find the number of the students who passed in Science.

iii. Represent the above information along with result in a Venn- diagram.

Solution : Here, U be the set of all exam appeared students.

Let the set of students who passed in Mathematics and Science be M and S respectively. Then according to question,

n(U) = 120 , n(M ∩ S) = 50 , n( M∪S ) = 40

Let, no(M) = 2x, no(S) = x

Now , we know that ,

n(U) = no(M) + no(S) + n(M ∩ S) + n( M∪S )

Or, 120 = 2x + x + 50 + 40

Or, 120 = 3x + 90

Or, 3x = 120 - 90Or, 3x = 30 x = 10So, Number of students who passed Mathematics only = 2x = 2 × 10 = 20

M S

40

U

20 50 10


Number of students who passed Science only = x = 10 i. Number of students who passed Mathematics, n(M) = no(M) + n(M ∩ S)

= 20 + 50 = 70

ii. Number of students who passed Science, n(S) = no(S) + n(M ∩ S)

= 10 + 50 = 60

Example 9 : In a class of 75 students, 30 students like to drink milk but not tea and 25 students like to drink tea but not milk, 10 like none of these. How many students like both subjects ? Represent the information in Venn- diagram.

Solution : Here,

Let M and T represent the sets of the students of a class who like to drink milk and tea respectively and U be the set of all students in the class. Then n(U) = 75 , no (M) = 30 , no(T) = 25, n( M ∪T ) = 10 and n(M ∩ T) = ?

Now , we have

n(U) = no(M) + no(T) + n(M ∩ T) + n( M ∪T )

or, 75 = 30 + 25 + n(M ∩ T) + 10

or, 75 = 65 + n(M ∩ T)

or, n(M ∩ T) = 75 - 65

n(M ∩T) = 10

Thus, the number of the students who like to drink both milk and tea is 10.Example 10: In a survey of 100 people of Kritipur, it was found all liked at least one of the three local newspaper; Kritipur times, Chovar times and Pharping times. It was reported that 60 liked to read Kritipur times, 32 liked to read both Kritipur times and Chovar times 22 liked to read both Kritipur times and Pharping times, 20 liked to read both Chovar times and Pharping times, 15 liked Chovar times only and 10 liked Pharping times only. Draw a Venn- diagram to illustrate above information and then find

i. How many people like to read all three news papers ?

ii. How many people like to read Kritipur times only ?

Solution : Here, Let, U be the set of surveyed people of Kritipur.

Let K,C and P be the sets of people who read Kritipur times, Chovar times and Pharping times respectively. Let x be the number of people who like to read all newspaper.

M T

10

U

30 10 25


Here, n( K ∪C ∪P ) = 100, n(K) = 60 , n(K ∩ C) = 32 , n( K ∩ P) = 22 , n( C ∩ P) = 20 ,

no(C) = 15 , no(P) = 10, let n (K ∩C ∩P) be x

Let, n (K ∩C ∩P) be x.

Now, from the Venn- diagram

n(K) = no(K) + no(K ∩C) +no(C ∩P) + n(K ∩C ∩P)

or, 60 = no(K) + 32 – x + 22 – x + x

or, 60 = no(K) + 54 - x

Or, no(K) = 60 – 54 + x

no(K) = 6 + x

i. n (K∪C∪P ) = no(K) + no(C) + no(P) + no(K∩C) + no(C∩P) + no(P∩K) + n(K∩C∩P)

Or, 100 = 6 + x + 15 +10 + 32 - x + 20 – x + 22 – x + x

Or, 100 = 105 – x

Or, x = 105 – 100

x = 5

Thus, 5 people like to read all three newspapers.

i. Now from the Venn- diagram

no(K) = 6 + x = 6 + 5 = 11

So, 11 people like to read only Kritipur times.

Example 11 : In an examination, 30% students failed Nepali, 40% failed Science, 45% failed Mathematics, 15% failed Nepali and Science, 10% failed Nepali and Mathematics, 15% failed in Science and Mathematics and 5% failed in all three subjects.

i. What percent of the students passed in all subjects ?ii. What percent failed in only one subject ?iii. What percent failed exactly two subjects ?

Solution: Here, Let, U = Set of all students N = Set of Nepali failures S = Set of Science failures

K C U

6+x32-x

x

15

10

20-x22-x

P


M = Set of Mathematics failure.

Percentage of total failed students, n(U) = 100%

Percentage of students failed in Nepali, n(N) = 30%

Percentage of students failed in Science, n(S) = 40%

Percentage of students failed in Mathematics, n(M) = 45%

Percentage of students failed in Nepali and Science, n(N ∩S) = 15%

Percentage of students failed in Nepali and Mathematics, n(N ∩M) = 10%

Percentage of students failed in Mathematics and Science n(M ∩S) = 15%

Percentage of students failed in all three subjects n(N ∩M ∩S) = 5%

We know,

n(N ∪M ∪S) = n(N) + n(S) + n(M) - n(N ∩S) - n(N ∩M) - n(M ∩S) + n(N ∩M ∩S)

= 30% + 40% + 45% - 15% - 10% - 15% + 5%

= 120% - 40%

= 80%

i. Percentage of the students passed in all subjects

n(N ∪M ∪S ) = n(U) - n(N ∪M ∪S)

= 100% - 80%

= 20%

Students who failed only one subject

= no(N) + no(S) + no(M)

= 10% + 15% + 25%

= 50%

i. Students who failed exactly two subjects

= no(N ∩S) + no(S ∩M)+no(M ∩N)

= 10% + 10% + 5%

= 25%

N S U

10%10%

5%

15%

25%

10%5%

M 20%


Exercise (1)

1. a. If n(A) = 40, n(B) = 60 and n(A ∪B) = 80 then find

i. n(A ∩B) ii. no (A) iii. Show the above information in Venn diagram.

b. If A and B are two subsets of a universal set U in which n(U) = 70, n(A) = 40, n(B) = 20 and n(A ∪B ) = 15 then

i. Show the above information in a Venn diagram ii. Find n (A ∪B)

2 a. In a certain locality, 55 persons can understand Nepali and 40 persons can understand Newari. If 20 persons understand both the language, how many can understand at least one language ?

b. In a survey of 200 students, 30 liked neither to sing a song nor to dance, 60 liked to sing a song only and 50 liked to dance only then

i. Illustrate these information in a Venn- diagram.

ii. Find the total number of students who can sing a song.

c. In a class of 60 students, 15 students liked math only, 20 liked English only and 5 did not like any subject then

i. Represent the above information in a Venn- diagram.

ii. Find the number of students who liked both subjects.

iii. Find the number of students who like at least one subject.

d. A survey conducted among 560 youth of Kaski, it was found that 320 liked to play football,300 liked to play volleyball and 500 liked to play at least one of the games.

i. Find the number of youth who liked to play both games.

ii. Find the number of youth who do not like to play both games.

iii. Show the above information in a Venn- diagram.

3 a. In a survey of 350 students of a school , 200 liked Pokhara, 220 liked Janakpur, and 120 both places then

i. Show the above information in a Venn- diagram.

ii. Find the number of students who liked neither of two places.

b. Out of 150 people , it was found that 80 read daily news paper , 120 read weekly newspaper, 60 read all two news papers and remaining do not read any newspapers.

i. Draw a Venn- diagram to illustrate the above facts.

ii. How many are remaining who did not read any newspaper ?


c. In a survey of a community,55% of people like to listen to the radio, 65% like to watch the television and 35% like to listen to the radio as well as watch television.

i. Show the above information in a Venn diagram.

ii. Find the percentage of people who don’t like to listen the radio as well as to watch television.

d. In a survey of community 45% of the people like Dashain festival, 65% like Tihar festival and 20% like both festivals.

i. Show it in a Venn diagram

ii. What percentage of them don’t like both?

4. a. A survey conducted among 450 students of a school, the following information are found, 250 like orange, 280 like apple and 40 dislike both the fruits.

i. Find the number of students who like both the fruits.

ii. Find the number of students who like orange only.

iii. Show the above information in a Venn- diagram.

b. In a survey of 2400 tourists visited in Nepal, it was found that 1650 like to visit Kritipur, 850 liked to visit Kapan and 150 did not like to visit both places.


ii. How many were there who liked to visit both ?

iii. How many were there who liked to visit Kritipur only ?

c. 500 students of a hostel were asked what they would like tea or milk, 180 said they would like tea only,150 said they would like milk only and 100 said they would like neither tea nor milk.

i. How many liked to take tea ?

ii. How many liked to take milk ?

iii. How many liked tea or milk only ?

iv. Show the above information in Venn- diagram.

d. In a marked demand of 1500 consumers of pulses, it was found that 1200 purchased mungi daal, 1300 purchased mash daal and 1000 purchased both of them. Draw a Venn- diagram and find

i. How many consumers purchased neither of them ?

ii. How many consumers purchased mash daal only ?

iii. How many consumers purchased exactly one of them ?

5. a. In a survey conducted in a community, it was found that 80% people liked oranges,85% liked mangoes and 75% liked both. But 45 people liked none of them. By drawing a


Venn- diagram, find the number of people who were in the survey.

b. In a survey of a group of people, it was found that 70% of the people liked folk songs, 60% liked modern songs, 4000 liked both of them and 10% liked none of them.

i. Draw a Venn- diagram to illustrate the above information.

ii. Find the total number of people in the survey .

c. In a survey of a community, it was found that 55% like summer season, 20% like winter season, 40% do not like both seasons and 750 like both seasons. By using Venn- diagram find the number of people who like winter season.

6. a. In a group of 51 students , each likes red coloured dress or blue coloured dress. If the ratio of the number of the students who like red coloured dress only and blue coloured dress only is 4:3 and who like both is 16, find the number of students who like blue coloured dress.

b. In a group of 240 students, the ratio of number of students who like Mathematics and Science is 6:7. If 40 students like both of them and 20 students like neither of them, find the number of students who like Mathematics only and Science only using a Venn diagram.

c. In a survey of 300 students of class 10, the ratio of number of students who like English to the number of students who like Nepali is 2:5. If 70 students do not like both subjects and 1

6 times of the total students like both subjects. Then find,

i. The number of students who like only English.

ii. The number of students who like only Nepali.

d. In a survey, it was found that the ratio of the people who like milk and curd is 5:4. Out of which 40 people liked both, 30 liked milk only and 70 liked none.

i. Represent the above data in a Venn- diagram.

ii. Find the number of the people who participated in the survey.

e. In a group of 200 persons, 60 persons like poem, 20% like poem but not story. If the ratio of people who like story only and who dislike both is 4:3, find the percentage of persons who like only one.

7. a. In a survey conducted among 150 people of a group the number of people who liked tea only is twice the number of people who liked coffee only. If 70 people liked both and 50 people like none, then

i. Find the number of people who liked tea.

ii. Find the number of people who liked coffee.

iii. Show the result in a Venn- diagram.

b. In a group of 100 students, 80 like football game and 50 like volleyball game. If the number of students who like the both games is twice the number of students who do


not like both games then

i. Find the number of students who like both games.

ii. Find the number of students who do not like both games.

iii. Show the result in a Venn- diagram.

8. a. In a group of 85 students, 25 students like to read Nepali Novel only and 15 students like to read English Novel only. If 35 students do not like to read any of the two Novels, then


ii. Find the ratio of the students who read Nepali Novel to the students who read English Novel.

b. 250 students in a group were asked whether they like mango or apple. 80 students like mango but not apple and 60 students like apple but not mango. If 50 students do not like both the fruits, then


ii. Find the ratio of the students who like mango to the students who like apple.

9. a. If n(A)= 12,n(B)= 15, n(C)= 20, n(A∩ B)= 6, n(B ∩ C)= 8, n(A ∩ C) = 7, n(A ∩ B ∩ C) = 4 and n( A ∪B ∪C )= 4, illustrate the following information in a Venn diagram and find n(U).

b. If no (A) = 10, no (B)= 15, no C= 20, n(A ∩ B) = 25, n(B ∩ C) = 30, n(C ∩ A)= 35, n(U)= 130 and n( A ∪B ∪C ) = 5, find n(A ∩B ∩ C) by using Venn diagram.

10. a. In a survey of a group of people, it was found that 40 liked tea, 45 liked coffee. 30 liked milk, 25 liked coffee as well as tea, 20 liked tea as well as milk, 15 liked coffee as well as milk and 10 liked all the three. How many people were asked these questions ? Solve by drawing the Venn- diagram.

b. In a group of students. 20 study Physics, 21 study Chemistry, 18 study Biology, 7 study Physics only , 10 study Chemistry only, 6 study Physics and Chemistry only and 3 study chemistry and Biology only.


ii. How many students study all the subjects ?

iii. How many students are there all together ?

c. In a group of students, 20 study Biology,21 study Mathematics, 18 study Geography, 7 study Biology only, 10 study Mathematics only, 6 study Biology and Mathematics only and 3 study Mathematics and Geography only.

i. How many students study all subjects ?

ii. How many students are there altogether ?


Ans

wer

s (1) 1.a. 20,20 b. 55 2.a.75 b. 120 c. 20,55 d. 120,60

3a. 50 b. 10 c. 15% d. 10% 4.a.120,130 b. 250,1400 c. 250,220,330 d. 0,300,500 5a. 450 b. 10000 c. 1000 6a.31 b. 80,100 c. 30,150 d. 156 e. 60% 7.a. 90,80 b. 60,30 8a. 7:5b. 7:6 9a.30 b. 10 10.a.65 b. 2, 41 c. 2,41 11.a.30 b.3,4,7,36 12a.5% b.8% c. 11%,25%,61%

11. a. In a survey of 70 students, 24 liked only two of volleyball, football and basketball, 6 students liked all three. If 10 didn’t like any of these games, find the number of students who likes only one.

b. In a group of students,18 read Biology, 19 read Chemistry and 16 read Physics, 6 read Biology only, 9 read Chemistry only, 5 read Biology and Chemistry only and 2 read Chemistry and physics only.

i. How many read all three subject?

ii. How many read biology and physics only?

iii. How many read physics only?

iv. How many students are there all together?

12. a. Of the total candidates in an examination, 40% students passed in Maths, 45% passed in Science and 55% in Health. If 10% passed in Maths and Science, 20% in Science and Health and 15% in Health and Maths.

i. Draw a Venn diagram to show the above information.

ii. Calculate the percentage of students who passed in all three?

b. In a school result, 20% are passed in all three subjects,48% failed in Mathematics, 57% failed in English and 42% failed in Dance, 35% failed in Mathematics and English,25% failed in English and Dance, 15 % failed in Dance and Mathematics.

i. Draw Venn- diagram to show the above information.

ii. What percentage failed in all three subjects ?

c. In an examination, 48% students failed in Physics, 39% in Chemistry,33% in Biology,12% in Physics and Chemistry, 9% in Physics and Biology,13% in Chemistry and Biology and 3% in all three subjects.

i. What percent passed in all three subjects ?

ii. What percent failed in exactly two subjects ?

iii. What percent failed in exactly one subject ?

iv. Represent the above information in the Venn- diagram.


Group ‘B’ (1 x 4 = 4)1. In an examination, 58% students failed in English, 39% in Account and 25% in Statistics,

32% in English and Account, 19% in English and Statistics, 17% in Account and statistics and 13% in all three subjects.

i. What percent passed in all three subjects ?ii. What percent failed exactly in two subjects ?iii. What percent failed exactly in one subject ?iv. Show above information in a Venn- diagram.

Model - Question [F. M. 4]

Test Yourself F.M. 201. If n(A) = 8x + 20 , n(B)= 5x+30 , n (C) = 4x+40 , n(A∩B)= 5x, n(B ∩C)= 6x, n(C

∩A)=7x, n(A ∩B ∩C)= 2x +10, n(U) = 150, find the value of x. (50)2. Out of 250 tourists arriving to visit Nepal, 40% of them have already visited

Pokhara and 30% have visited Lumbini. Also 10% of them have visited both the places. Then, i. How many tourist hadn’t visited Pokhara and Lumbini. (100)ii. How many tourist have visited Lumbini only? (50)iii. Show the above information in Venn diagram.

3. In a class, it is found that 30 students speak Newari, 15 speak Newari and Maithali, 12 speak Newari and Bhojpuri, 25 speak Bhojpuri, 9 speak Bhojpuri only, 11 speak Maithali only, 5 speak Newari and Bhojpuri only and 10 speak neither of these languages. Illustrate the given information in a Venn diagram and find: i. total number of students in a class. (64)ii. the number of students who speak exactly two languages.(17)

4. Of the total candidates in an examination, 60% students passed in Mathematics, 55% in Science and 65% in English. If 25% passed in Mathematics and Science, 30% in Science and English and 35% in English and Mathematics. i. Draw a Venn diagram to show the above information. ii. Calculate the percentage of students who passed in all three subjects?

(10%)5. Out of the total candidates in an examination, 600 passed in Health,150 passed

in Health only, the number of students who passed Mathematics and Health only, Science and Health only and all the three subjects are represented by three consecutive numbers. Then find i. the number of students who passed all three subjects.(151)ii. the number of students who passed only two subjects.(299)


ArithmeticEstimated period : 25

2 Unit

Specific objectives :After the completion of this unit, the students will be able to

• develop the concept of marked price, discount and VAT and solve the problems related to them.

• solve the problems of money exchange related to VAT, commission, profit and loss.

• convert the currency of one country to currency of other country by using foreign exchange table fixed by Nepal Rastra Bank.

• find the compound interest on the basis of yearly system and half yearly system.

• solve the problems of population growth and compound depreciation.• explore, analyze and resolve real life problems on related areas.

Teaching Materials : Chart papers with formulae of discount, VAT, SP, MP, etc. Compound Interest, Depreciation and population growth etc, flash cards, VAT bills, vouchers of Banks.

Note to the teacher : 1. Discuss and explain the daily life problems with profit , loss , discount and

VAT.2. Have collection of chart papers for showing formula of compound interest,

population growth, flash cards and money cards3. Associate the daily life problems with discount and VAT.4. Deduce all the formulae with active participation of students so that they will

find it easy to remember and apply different condition.5. Ask the students to find difference between simple interest and compound

interest.6. Chain the idea of compound interest with population growth and compound

depreciation.

Curriculum• VAT and Discount


• Money exchange• Compound interest• Population growth• Depreciation

Specification GridCognitive Domain Knowledge

(K)Comprehensive (C)

Application (A)

Higher ability(HA)

Total no. of questions

Totalmarks

Topic Each 1 mark Each 2 marks Each of 4 marks Each of 5 marks

VAT and Money exchangeCompound interestPopulationgrowth and depreciation

1

2 1

1

4

14

Note : At least 2 marks questions are asked from each topic.

2.1. Discount and VAT

In the market, we see that the price of many articles is marked. This is called marked price (M.P.). It is also sometimes called list price or usual price. However you must have noticed bumper sale advertisements. In such sales to clear the stocks or for other reasons the seller sells the goods at prices lower than the normal marked price. This rebate is called discount. Thus, the deduction made on marked price is discount. It is generally given as percent of the marked price or list price.

To generate revenues for spending on public benefits, the central and local state government impose various kinds of taxes such as sales tax, entry tax, service tax etc. Value added tax (VAT) previously called sales tax is one such tax. The rate of value added tax in Nepal is 13% at present. Originally it was 10% . VAT is determined by adding taxes for transportation, insurance, commission for the agent etc. and other extra taxes for the imported goods. VAT is added on the price after deducting discount from the original price. In every fiscal year, the government fixes the rate of VAT and also fixes which goods are VAT free and which are not VAT free. It depends on the policy of the government. The rate of VAT is different in different countries.

In some countries, VAT is made free to foreign tourists for the product of their countries to encourage them and to buy the goods of the host nation. It is returned to them at the time of leaving the country.

VAT (Value Added Tax): The tax which is added while supplying goods or services is called VAT.


Thus, SP = MP – discountSP = MP – discount % of MP

Discount amount = MP – SP

Discount (%) = discount amountMP

×100 %

Price including VAT = Price after discount + VAT amount

VAT amount = SP with VAT – SP

VAT (%) = Amunt of VATSP

×100 %

Now, let’s see some worked out examples to illustrate the idea of previous mentioned terms and formula.

Marked Price

- Discount

Selling Price Cost Price

+ VAT

Price with VAT

+Cost Price -Loss Profit

Example 1: Find the amount of VAT at the rate of 13 % on a laptop whose selling price is Rs 52000.Solution : Here, Selling price of laptop = Rs 52000Rate of VAT = 13 %VAT amount = 13 % of Selling price

= 13100

× Rs 52000

= Rs 6760∴ Hence, the required VAT amount is Rs 6760.

CP MP

Selling Price (SP)

Customer price with VAT

+ Profit - Discount

+ VAT

- Loss + Commission

Worked Out Examples


Example 2: What is the net payment of the bill amounting Rs 1800 allowing a discount of 5 % .Solution : Here, Bill amount = Rs 1800Rate of discount = 5%∴Discount amount = 5 % of bill amount

= 5100

× Rs 1800

= Rs 90 Net payment = Bill amount – discount = Rs 1800 – Rs 90 = Rs 1710Hence, the net bill amount is Rs 1710.Example 3 : The marked price of a radio is Rs 5000. What will be the price of the radio if 10 % VAT is levied after allowing 15 % discount on it ?Solution : Here, Marked price of the radio (MP) = Rs 5000Rate of discount (R) = 15 % Rate of VAT = 10%Real price of the radio = ? We know, Discount amount = 15 % of Marked price

= 15100

× Rs 5000

= Rs 750Now, the selling price of the radio = SP + VAT = Rs 4250 + Rs 425 = Rs 4675Hence, the actual selling price of the radio is Rs 4675.Example 4 : After allowing 20 % discount on the marked price of an article, 13 % VAT is levied, then the price becomes Rs 2151.52. Find the marked price of the article.Solution : Here,Rate of discount (D) = 20 %Rate of VAT = 13 %Selling price with VAT = Rs 2151.52Marked price (M.P.) = ?

Selling price (SP) = MP – Discount = Rs 5000 – Rs 750 = Rs 4250VAT amount = 10 % of Selling price

= 10100

× 4250

= Rs 425


Let marked price (M.P.) be Rs x, then

Discount (D) = 20 % of x = 20100

x = Rs 20x100

∴ Selling price (S.P.) = M.P. - D

= x - 20x100

= 100x - 20x100

= 80x100

= Rs 4x5

Hence, the marked price of the article is Rs 2380.

Example 5 : After allowing 5 % discount on the marked price of a watch , 10 % VAT is charged on it , then its price becomes Rs 1672. Find the amount of discount.Solution : Here,Rate of discount (D) = 5 %Rate of VAT = 10 %Selling price with VAT = Rs 1672Discount amount (D) = ?Let marked price (M.P.) be Rs. x, then

Discount (D) = 5 % of x = 5100

× x = 5x100

Selling price (S.P.) = M.P. - D

= x - 5x100

= 100x - 5x100

= 95x100

= Rs 19x20

VAT amount = 10 % of S.P.

= 10100

× 19x20

= Rs 19x200


= 13100

× 4x5

= Rs 52x500

Now, we know, S.P. with VAT = S.P. + VAT

Or, 2151.52 = 4x5

+ 52x500

Or, 2151.52 = 400x+52x500

Or, 452x = 1075760

Or, x = 1075760452

Or, x = Rs 2380


Or, 1672 = 19x20

+ 19x200

Or, 1672 = 190x + 19x 200

Or, 1672 × 200 = 209x

Or, x = 1672 × 200209

Or, x = Rs 1600

Now discount amount (D) = 5x100

= 5×1600100

= Rs 80Hence, the discount amount is Rs 80.


Example 6 : If a tourist paid Rs 5610 for a carved window of a wood with a discount of 15% including 10 % value added tax (VAT), how much does he get back while leaving Nepal? Find it.Solution : Here,S.P. with VAT = Rs 5610Rate of discount = 15%Rate of VAT = 10%Let marked price (M.P.) of the window= Rs x

Discount (D) = 15 % of x = 15100

× x = 15x100

Selling price (S.P.) = M.P. - D

= x - 15x100

= 100x -5x100

= 85x100

= Rs. 17x20


= 10100

× 17x20

= Rs 17x200

Example 7 : The marked price of an article is Rs 2000 and the shopkeeper allows 10% discount. After levying VAT , if a customer pays Rs 2034 for it, what percent of VAT was charged?Solution : Here,Marked price of article (M.P.) = Rs 2000Rate of discount = 15%S.P. of article with VAT = Rs 2034Rate of VAT = ?

Discount (D) = 10 % of 2000 = 10100 × 2000 = 200

Now, selling price of article (S.P.) = M.P. – D = 2000 – 200 = Rs 1800We know, Amount of VAT = S.P. with VAT – S.P. without VAT = 2034 – 1800 = Rs 234


Or, 5610 = 17x20

+ 17x200

Or, 5610 = 170x + 17x 200

Or, 5610 = 187x200

Or, x = 5610 × 200187

Or, x = Rs 6000

Now, VAT amount (VAT) = 17x200

= 17 × 6000200

= Rs 510

Hence, the tourist gets the VAT amount Rs 510 back when he leaves Nepal.


Now, Percent of VAT = Amount of VATS.P. without VAT

× 100%

= 2341800

× 100%

= 13 %Thus, the VAT percent is 13%.Example 8 : A shopkeeper fixed the marked price of his camera 30% above his cost price. Allowing 15% discount on the marked price, the camera was sold. What percent of profit did he make?Solution : Here,Let cost price of a camera (C.P.) = Rs xBy question, marked price (M.P) = C.P. + 30% of C.P.

= x + 30100

× x

= 100x + 30x100

= 130x100

= Rs 13x10

Discount (D) = 15% of M.P.

= 15100

× 13x10

= Rs 39x200

Now, we know, S.P. = M.P. – D

= 13x10

- 39x200

= 260x - 39x200

= Rs 221x200

Hence, the profit percent is 10.5%.

We know, P = S.P. – C.P.

= 221x200

– x

= 221x - 200x200

= Rs 21x100

Now, Profit % = Profit amountC.P.

× 100%

=

21x100x × 100%

= 21x100

×

1x

× 100% = 10.5 %


Example 9 : The marked price of an article is 25% above its selling price and cost price is 30% less than its marked price. Find the discount percent and gain percent.Solution : Here,Let selling price(S.P.) of an article be Rs ‘x’Then, by question , marked price (M.P.) = x + 25% of x

= x + 25100

× x

= 100 x + 25 x100

= 125x100

= Rs 5x4

And cost price (C.P.) = 5x4

- 30% of 5x4

= 5x4

-

30100

× 5x4

= 5x4

- 3x8

= 10x - 3x8

= Rs 7x8

Now, we know, Discount = M.P. – S.P.

= 5x4

- x = 5x-4x4

= Rs x4

Also, we know, profit (P) = S.P – C.P.

= x - 7x8

= 8x - 7x4

= Rs x8

Now, discount % = Discount amountMarked price

× 100%

=

x4

5x4

100%


= x4

× 45x

× 100% = 20 %∴ Discount % is 20%

And, Profit % = Profit amountCost price

100%

=

x8

7x8

100%

= x8

× 87x

× 100% = 1007

% = 14 27

%

Hence, discount percent and profit percent are 20% and 14 27

% respectively.

Example 10 : A shopkeeper sold an article at a discount of 25 % and loses Rs 125. If he allows 10 % discount he gains Rs 250. Find the marked price and the cost price of the article.Solution : Here,Let marked price be Rs ‘x’1st case Discount = 25%Loss (L) = Rs 125 S.P. = M.P. – D = x - 25% of x

= x - 25100

× x

= 100x - 25x100

= 75x100

= Rs 3x4

We know, C.P. = S.P. + L

= 3x4

+ 125

= Rs 3x+5004

We know, C.P. = S.P. - G

2nd caseDiscount = 10%Gain (G) = Rs 250S.P. = M.P. – D = x - 10 % of x

= x - 10100

× x

= 100x - 10x100

= 90x100

= Rs 9x10


3x+5004

= 9x10

- 250

Or, 3x +5004

= Rs 9x+250010

Or, 36x – 10000 = 30x +5000 Or, 36 x – 30x = 5000 + 10000 Or, 6 x = 15000

Or, x = 150006

Or, x = Rs 2500

And, C.P. = 3 × 2500 + 5004

= 80004

= Rs 2000

Hence, marked price of an article is Rs 2500 and cost price is Rs 2000.Example 11 : Golchha supplier sold a tractor to Chaudhary supplier at Rs 700000 excluding 13% VAT. Chaudhary supplier sold the same tractor to a farm with Rs 10000 transportation charge, Rs 3500 local tax and profit of Rs 50000. What is the VAT amount that the farm has to pay at the rate of 13 % ?Solution : Here,For, Golchha supplier,Selling price (S.P.) = Rs 700000Rate of VAT = 13%S.P. with VAT = 700000 + 13% of 700000 = 700000 + 91000 = Rs 791000For Chaudhary supplier,Cost price (C.P.) = Rs 791000Transportation charge = Rs 10000Local tax = Rs 3500Profit = Rs 50000 Selling price of Chaudhary supplier = Rs ( 791000 +10000 + 3500 + 50000 ) = Rs 854500Now, VAT amount = 13 % of Rs 854500 = 854500 = Rs 111085Hence, the VAT amount paid by the farm is Rs 111085


Exercise (2.1)

1. a. Explain the term VAT. b. What is the present rate of VAT in Nepal ? c. Demonstrate the flow chart with the relation of M.P. ,D , S.P. , VAT and net S.P.2. a. Write the formula for finding the selling price when actual discount and marked price

are given. b. Write the formula for finding the selling price when marked price and discount(%) are

given. c. Establish the formula for finding the rate of discount when discounted amount and

marked price are given. d. Write the formula for finding the discount amount when marked price and selling

price are given.3. a. Find the discount, if the marked price of your pen is Rs ‘x’ and its SP is Rs ‘y’. b. If the marked price of a laptop is Rs. ‘p’ and selling price is Rs. ‘q’, then what is its

discount rate? c. Establish a relation to show the M.P of a mobile is Rs. ‘x’ and 10% discount is given and

sold at Rs. ‘y’. d. During festival Rs. ‘b’ is given as discount on M.P of a ball Rs. ‘a’ and c% VAT is levied

on it. How much should a customer pay for it?4. Find the missing one a. M.P. = Rs 180 , S.P. = Rs 160 , discount = ? b. M.P. = Rs 1600 , S.P. = Rs 1360 , discount = ? c. M.P. = Rs 2200 , S.P. = Rs 1980 , discount (%) = ? d. S.P. = Rs 1955 , M.P. = Rs 2300 , discount (%) = ? e. S.P. = Rs 4100 , discount = Rs 600 , M.P. = ? f. S.P. = Rs 1750 , discount = Rs 68 , M.P. = ? g. S.P. = Rs 5000 , VAT = 13 % , S.P. with VAT = ? h. S.P. = Rs 1450 , VAT = 13 % , S.P. with VAT = ? i. S.P. without VAT = Rs 18900 , S.P. with VAT = Rs 19500 , VAT amount = ? j. S.P. without VAT = Rs 4800 , S.P. with VAT = Rs 5200 , VAT (%)= ? k. S.P. without VAT = Rs 3200 , VAT = Rs 400 , VAT(%) = ? l. M.P. = Rs 4000 , VAT = 13 % , S.P. = ? ( In case of no discount ) m. M.P. = Rs 8900, VAT = 13 % , S.P. = ? ( In case of no discount )5. a. The marked price of an article is Rs 550. What will be the actual price of the article after

allowing 10 % discount ?


b. The cost of electric bulb is Rs 300.If the shopkeeper sells the bulb by allowing 10 % discount , at what price does he sell the bulb ?

c. The selling price of a watch is Rs 270. If the watch was sold at 10 % discount from the marked price , find the marked price.

d. After allowing 15 % discount on an article its price becomes Rs 2000 , what is the marked price ?

e. What percent of discount should be given to a bag costing Rs 1500 so that a customer can buy it for Rs 1200 ?

6. a. Find the actual selling price of the T- shirt whose marked price is Rs 2700 when 13 % VAT is levied.

b. The marked price of an article was fixed to Rs 1380 by increasing 15 % on its actual price. Find the actual price.

c. The price of an article with 13 % VAT is Rs 1356, find its price before adding VAT ?7. a. The price of a calculator is Rs 800 and 10 % discount is allowed while selling. i. Find the discount amount. ii. Find selling price of the calculator. iii. Find the price with 13 % VAT. b. The marked price of an article is Rs 2000 and 10 % discount is allowed. i. Find its selling price . ii. How much should a customer should pay for it including 15 % VAT ?8. a. If the marked price of a camera is Rs 3200 and a shopkeeper announces a discount of

8 %. How much will a customer have to pay for buying the camera if 10 % VAT was imposed on it.

b. The labeled price of a radio is Rs 5210 . If the shopkeeper allows 10 % discount and adds 10 % VAT on it , then

i. How much will a customer pay for the radio ? ii. Find the difference between marked price and selling price with VAT.9. a. After allowing 20 % discount on the marked price of a watch, the price of the watch will

be Rs 2376 when a VAT of 10 % is added, find its marked price. b. A foreigner bought an article at a discount of 15 % and paid Rs 17289 adding

13 % VAT on it. What was the marked price of the article? c. The price of an article is Rs 1485 including 10% VAT. Calculate the marked price, if a

discount of 10% is given.10. a. After allowing 10 % discount on the marked price of television and 15 % VAT was

levied on it. Then the price became Rs 16720, what amount was given in the discount? b. If a watch was sold for Rs 1656 after allowing 10 % discount on the marked price

adding 15 % VAT , find the discount amount.11. a. After allowing 15 % discount on the marked price of an article 13 % VAT is levied and

the price becomes Rs 19210.Find marked price and the VAT amount. b. A radio is sold at Rs 3616 after allowing 20 % discount on the marked price and 13 %

VAT is levied. Find the discount and VAT amount.


12. a. A man buys an article at 15% discount on the marked price and he pays 13% VAT. If the discount amount to Rs 1200, find the marked price and selling price including VAT.

b. A tourist buys a Nepalese Flag at a discount of 15% and pays 10% VAT. If he pays Rs. 170 for VAT, Find

i. Marked price of the Flag ii. Amount paid for the flag while buying13. a. A person buys an article at a discount of 15% and pays 13% VAT. If he pays

Rs 750 for VAT, find the marked price of the article and also the amount paid by him to buy the article.

b. A person buys an article at a discount of 13% and pays 16% VAT. If he pays Rs 261 for VAT, find the marked price of the article and also the amount paid by him to buy the article.

14. a. A tourist buys a statue at 20% discount on the marked price with 13% VAT. When he returned to his own country he got Rs. 1872 back for VAT at the airport, Find the marked price and selling price including VAT of the statue.

b. The marked price of an article is Rs 2400 and the shopkeeper levied the VAT after allowing 20% discount. If a customer pays Rs 2208, find the VAT rate.

15. a. A shopkeeper marks the price of an article 25% above the cost price. After allowing a discount of 10% on its marked price, it was sold at a gain of Rs 750. Find the marked price of the article.

b. The marked price of a watch was fixed 30% above the cost price. When it is sold allowing 20% discount on it, there was a gain of Rs 150. Find the marked price and the cost price of the watch.

16. a. A watch was sold on its marked price at a gain of 20%. But allowing 5% discount, there would have been gain Rs.140 only. At what price did the seller buy the watch?

b. When a discount of 15% is allowed on the marked price of an article, it is sold for Rs. 4250. Calculate the marked price, if the marked price is 25% above the cost price of the article. Calculate gain or loss percent too.

17. a. An article, after allowing a discount of 20% on its marked price was sold at a gain of 20%.Had it been sold after allowing 25% discount, there would have been gained Rs 125. Find the marked price of the article.

b. A shopkeeper marks the price of an article Rs 550 and gives a customer a discount of 10%. In this way he gains Rs 75 on the article. How much percent will the marked price more than the cost price?

18. A dealer sold a computer to a retailer for Rs 50000 excluding 13% VAT. Retailer sold it at a profit of Rs 5000. If retailer paid Rs 1500 as local tax before selling it to the customer, Find

a. How much VAT amount the customer had to pay? b. Find the selling price of the retailer.19. a. Taudaha suppliers sold a machine at Rs 180000 excluding 13% VAT to Pharping

suppliers. Pharping suppliers sold the machine including Rs 5000 transportation, Rs


1600 local tax and Rs 10000 profit to consumer. Now, if the consumer pays 13% VAT, what additional VAT amount should be paid by the consumer?

b. A dealer sold a water pump for Rs 480000 levying 13% VAT to the retailer. If the retailer spend Rs 2000 on transportation, Rs 3500 on local tax and Rs 18000 profit on original price and then supplied to a water pump centre, how much did the water pump centre pay for the VAT amount and the total amount for the pump at the present rate of VAT?

20. a. A wholesaler brought construction materials for Rs 320000 sold them to a retailer at 10% profit and levying 13% VAT. The retailer spend Rs 3000 for transportation and Rs 1500 for local tax. The retailer sold the materials to constructor with 8% profit for own cost price. How much did the constructor pay for the materials with 13% VAT?

b. A dealer bought a motor bike at Rs 250000 and sold it to a retailer with the profit of 5% and levying 13% VAT. The retailer spend Rs 4000 for transportation and Rs 1500 for local tax. If the retailer sold the motor bike to a customer with the profit of 10% and levying of 13% VAT, how much did the customer pay for the motor bike.

Ans

wer

s (2.

1)

1. b. 13% c. Show to your teacher 2.a.SP= MP- discount

b. SP=MP- discount % of MP c. Rate of discount = discount amountMP

×100%

d. Discount Amount = MP-SP 3.a. Discount=Rs.(x-y)

b. Discount rate = p - q

p × 100% c. y = 9x10

d. (a-b) (100+c)

100 4a.Rs.20

b. Rs.240 b. 10% d. 15% e. Rs.4700 f. Rs.1818 g. Rs.5650 h. Rs.1638.50 i. Rs.600j. 8.33% k. 12.5% l. Rs.4520 m. Rs.100575. a. Rs 495 b. Rs 270 c. Rs 300 d. Rs 2352.94e. 20% 6.a. Rs 3051 b. Rs 1200 c. Rs 12007. a. i. Rs 80 ii. Rs 720 iii. Rs 813.60 b. i. Rs 1800ii. Rs 2070 8.a. Rs 3238.40 b. i. Rs.5157.90 ii. Rs52.109.a. Rs 2700 b. Rs 18000 c. Rs 1500 10.a. Rs1615.45b. Rs 160 11.a. Rs 20000, Rs 2210 b. Rs 800, Rs 41612a.Rs 8000 ,Rs 7684 b. Rs 2000, Rs 1870 13.a. Rs 6787.33, Rs 6519.23 b. Rs 1875, Rs 1892.25 14.a. Rs 18000, Rs 16272 b. 15%15.a. Rs 7500 b. Rs 4875, Rs 3750 16.a. Rs 1200b. Rs 5000, 6.25% 17.a. Rs 1500 b. 30.95%18a. Rs 8190 b. Rs 71190 19.a. Rs 28600b. Rs 65455, Rs 568955 20.a. Rs 490918.104 b. Rs 375541.375


2.2 Money and ExchangeEvery country in the world has its own currency with its own value. The currency of USA is dollar, the currency of Japan is yen, the currency of England is Sterling pound, the currency of India is rupees (IC), the currency of Nepal is rupees (NC) and so on.The value of currency of any country is not fixed. It changes according to the economic development of the country and international trade policy. Now a days trade , transportation and travelling is not limited within a country.For these all exchange of money of different countries is most essential. Exchange of money means to convert currency of a country with the currencies of other countries. In our country Nepal, the central bank/ Nepal Rastra bank has the authenticity to fix the buying and selling rates of currencies of different countries every day. The buying rate is the rate at which Nepal Rastra bank or money dealers buy foreign currency, and the selling rate is the rate at which they sell that currency. Let’s study the following table which you can see in daily Gorkhapatra News paperExchange Rates fixed by Nepal Rastra bankCurrency Unit Buying /Rs Selling/RsIndian 100 160.00 160.15Open Market Exchange Rates

(For the purpose of Nepal Rastra Bank) Currency Unit Buying/Rs Selling / Rs

U.S. Dollar 1 102.85 103.45European Euro 1 121.76 122.47Sterling Pound 1 138.21 139.01Swiss Franc 1 104.31 104.92Australian Dollar 1 77.78 78.23Canadian Dollar 1 79.78 80.24Singapore Dollar 1 76.23 76.67Japanese Yen 10 9.16 9.21Chinese Yuan 1 15.55 15.64Saudi Arabian Riyal 1 27.42 27.58Qatar Riyal 1 28.25 28.41Thai Bhatt 1 3.15 3.17UAE Dirham 1 28 28.16Malaysian Ringgit 1 25.15 25.40South Korean Won 100 9.46 9.51Swedish Kroner 1 12.28 12.35Honking Dollar 1 13.17 13.25Kuwait Dinar 1 340.89 342.88Bahrain Dinar 1 272.55 274.14


Example 1 : If 1 US dollar = Rs 102.85, convert Rs 51425 into dollar.Solution : Here,Rs 102.85 = $ 1

∴ Re 1 = $ 1102.85

∴ Rs 51425 = $ 1102.85

× 51425

= $ 500

∴ Rs 51425 = $ 500Example 2 : If dollars($) 30 = Rs 3000 and Pound (£)20 = Rs 2800, how many Pounds can be exchanged from $ 75 ?Solution: Let x Pounds can be exchanged from $ 75 Using chain rule $ 30 = Rs 3000 Rs 2800 = £ 20 Let, x = $ 75Now, 30 × 2800 × x = 3000 × 20 × 75

Or, x = 3000 × 20 × 7530 × 2800

Or, x = £ 53.57∴ £ 53.57 can be exchanged from $ 75Example 3 : Prasuna needs $ 3000 to visit Thailand. If a bank charges 2% commission while exchanging it , how much Nepali currency did she need ? ($ 1 = Rs 103.45)Solution : Here $ 1 = Rs 103.45 ∴ $ 3000 = Rs 103.45 × 3000 = Rs 310350Amount of Bank commission = 2% of Rs 310350 = Rs 6207Require Nepali currency = Rs 310350 + Rs 6207 = Rs 316557Hence , she needs Rs 316557.

Worked Out Examples

Note: Chain rule If A = BB = CC = AThen A × B × C = B × C × A


Example 4 : Prabesh exchanged Rs 1000000 into dollars ($) for his foreign tour. But, due to some problem he had to cancel his tour . At the same time, Nepali currency devaluated by 10%, then how much Nepali rupees did he get as a profit . ( Initial rate : $1 = Rs 102.85 )

Solution : Here,

Rs 102.85 = $ 1

∴ Re 1 = $ 1102.85

∴Rs 1000000 = $ 1102.85

× 1000000

= $ 9722.90

So, He had $ 9722.90.

After devaluation of Nepali currency

1 $ = Rs 102.85 + 10% of 102.85

= Rs 102.85 + 10.285

= Rs 113.14

Again exchanging $ 9722.90 into Nepali currency using new rate

$ 9722.90 = Rs 9722.90 × 113.14

= Rs 1100048.91

∴Profit = Rs 1100048.91 – Rs 1000000

= Rs 100048.91

Hence , the required profit is Rs 100048.91.

Example 5 : 10 Tolas gold bought from American market at $ 400 per tola is sold in Nepal with 20% custom duty and 13% VAT. At what rate should he sold in Nepali market to make 15% profit? ($1= Rs 102.15)

Solution : Here,

Cost of 1 tola gold in American market = $ 400

Cost of 10 tolas gold in American market = $ 400 × 10 = $ 4000

Custom duty = 20% of $ 4000


= 20100

× 4000

= $ 800

Cost with custom duty = $ 4000 + $ 800 = $ 4800

VAT amount = 13% of $ 4800

= 13100

× 4800

= $ 624

Cost with custom duty and VAT = $ 4800 + $ 624

= $ 5424

Profit = 15%

Profit amount = 15% of $5424

= 15100

× $5424

= $813.6

Now SP of 10 tolas gold in Nepal = $5424 + $813.6

= $6237.6

= Rs 6237.6 × 102.15

= Rs 637170.84

∴ SP of 1 tola gold in Nepal = Rs 637170.8410

∴ The gold should be sold at the rate of Rs 63717.084 per tola.

Exercise (2.2)

1. a. If £ 1 = Rs 138 and $ 1 = Rs 102.15 then convert £ 525 into dollar.

b. If £ 1 = Rs 138 and $ 1 = Rs 102.15 then convert $ 6500 into Pound.

2. a. Sumit Yadav working in Saudi Arab earns 1500 Riyal in a month. If 1 Saudi Arabian Riyal = Rs 27.42, find his monthly earning in Nepali Rupees.


Ans

wer

s (2.

2) 1.a.$709.25 b. £4811.41 2a.Rs.41130 b. HKdollars 28046.70 3a.Rs.1500 b.Rs250000 4a.Bangkok to Kathmandu by Rs. 5980b. Through Nima by Rs.13500(NC) c. Through Sandesh by Rs.19125 5a.Rs.975600 b.$8626.425

b. How much Hongkong dollars can be obtained while exchanging Rs 375000 paying 1.5% commission to the bank. (If, Hong kong dollar= Rs 13.17)

3. a. If the buying and selling rate of U.S.is $ 1 = Rs 102.85 and 103.45 respectively, how much does Nepal Rastra Bank gain while selling $ 2500 ?.

b. Bikash exchanged Rs 2500000 into Pounds to visit UK but his tour was cancelled after two days. At the same time Nepali currency devaluated by 10%, how much profit did he make ? (Initial exchange rate (£ 1 = Rs 138.21).

4. a. The aeroplane fare from Kathmandu to Bangkok is Rs 25000 and from Bangkok to Kathmandu is 6000 Thai Bhatt. Which one is cheaper?

{1 Thai Bhatt = Rs 3.17}

b. Sonam who lives in Canada wants to send 1500 Canadian dollars to his wife in Nepal. One option for remit is direct from Canada to Nepal and other option is through his brother Nima who lives in Delhi. Which one is better remit ?

[1 Canadian dollar= Rs 79 (NC) , 1 Canadian dollar = Rs 55 (IC) and Rs 100(IC) Rs 160(NC)]

c. Prakash wants to send $ 2500 from USA to his mother in Nepal. There are two options for remit, direct from USA to Nepal or through Sandesh who lives in Darjeeling India, which one is better remit ? How much his mother gain ?

[$1= Rs 104.35 (NC), $1=Rs 70(IC) and Rs (IC)= Rs 160(NC)]

5. a. A merchant bought 15 pieces of gold in USA at the rate of $ 500 per piece and sold in Kathmandu with 10% tax and 13% VAT. Find the selling price of 15 pieces of gold. ( $1 = Rs 104.65 ).

b. A Shopkeeper bought 250 pieces pasmina sals in Kathmandu at the rate of Rs 2200 per piece and paid 10% tax. If he sold these in New York and gained 50% profit. Find the selling price of all the pasmina sals in US dollars

if $ 1 = Rs 105.20.


2.3 Compound interestInterest : When we take a loan from a bank or from a person we have to pay some charge along with the borrowed sum while returning it under the certain condition. Such extra sum of money paid along with the principal to clear the debt is called interest.There are two types of interest. They are

i. Simple interest: The interest which is calculated from the original principal at the end of the given time period having the given rate is called simple interest. It is calculated by using the formula,

Simple interest(I) = P × T × R100

where ‘P’ is principal’ T’ is time in years and ‘R’ is rate of interest.

ii. Compound interest: When the interest of a principal for one time period (Year or half year or quarter year) is added to the principal and the interest for the next time period is calculated from new principal (amount of the previous time period), It is called compound interest.

The process of calculating interest of a principal for one time period and adding to the principal and forming new principal for the next time period is known as compounding.

From above, it has become clear that compound interest is interest of interest. It is the result of reinventing interest in the next time period on the amount of the previous time period. In this way the principal is increased in every time period whereas it remains the same throughout the given time in simple interest.

Formula derivation to calculate the compounded interest.Let the original principal = Rs PAnnual rate of interest = R%Time interval = T years.Then,

Interest in the first year = P×T×R100

= P×l×R100

= PR100


And amount in the first year = P + I

= P + PR100

= P (1+ R100

) Interest in second year = P×T×R

100

= R

100( )1 +P × 1 × R

100

= PR100

R100

( )1 +

Amount in second year = P + I

= P R

100( )1 +

+ PR

100 R

100( )1 +

= P R100

( )1 + R100

( )1 +

= P R100

( )1 +2

Interest in third year = P×T×R100

= R

100( )1 +P2× 1 × R

100

= PR100

R100

( )1 +2

Amount in third year = P + I

= P R100

( )1 +2 + PR

100 R

100( )1 +

2

= P R

100( )1 +

2 R100

( )1 + = P R

100( )1 +

3


On proceeding we get,Compounded amount in’ T’ Years

CA = P R100

( )1 +T

and compound interest = Compound amount – Principal

= P R100

( )1 +T- P

= P { R100

( )1 +T-1}

For semi- annual (half yearly) compounding:If the interest is calculated twice a year at the interval of 6 months, it is called interest compounded semi annually or half yearly.

In this case, Rate = R2

%

Time = T Years= 2 T half year

Compound amount (CA) = P R200

( )1 + 2T

And compound interest (CI) = CA-P

= P R200

( )1 + 2T P

= P { R200

( )1 + 2T - 1}

Note: 1. When the interest is compounded yearly but the rate of interest are different in different years R1% in the first year,R2% in the second year,R3% in the third year then amount in 3 years is given by:

CA= PR1100( )1 +

R2100( )1 +

R3100( )1 +

and CI = CA-P. 2. When the interest is compounded yearly but the time is given T years and M

months, then the amount is calculated as,

CA= = P R100

( )1 +T MR

1200( )1 +

and CI = CA-P 3. When the interest is compounded half yearly but the time is given in 2T half

years and in months, then the amount is calculated as

CA= P R200

( )1 +2T MR

1200( )1 +

and CI = CA-P


Exercise 1 : Find the compound interest on Rs 5000 for 3 years at 8% per annum, compounded annually. Solution: Here, Principal (P) = Rs 5000. Interest for the first year = P×T×R

100

= 5000 × 1 × 8 100

= Rs 400. Amount at the end of the first year = P + I = Rs (5000 + 400) = Rs 5400.Principal for the second year = 5400

Interest for the second year = P×T×R100

= 5400 × 1 × 8 100

= Rs 432Amount at the end of the second year = Rs (5400 + 432) = Rs 5832.Principal for the third year = Rs 5832.

Interest for the third year = 5832 × 1 × 8100

= Rs 466.56 Amount at the end of the third year = Rs (5832 + 466.56) = Rs 6298.56 Therefore, compound interest is Rs (6298.56 - 5000) = Rs 1298.56 Example 2 : Find the compound amount and interest on Rs 2500 for 3 years at 6% per annum, compounded annually. Solution: Here, Principal (P) = Rs 25000. Time (T) = 3 yearsRate of interest (R) = 6%Compound amount (C.A.) = ?Compound interest (C.I.) = ?We know, Compound amount (C.A.) = P

R100

( )1 + T

= 2500 6

100( )1 +

3

Worked Out Examples


= 2500 × (1.06)3 = 2500 × 1.191016 = Rs 2977.54Again, we knowCompound interest (C.I.) = C.A. – P = 2977.54 – 2500 = Rs 477.54Thus, compound amount is Rs 2977.54 and compound interest is Rs 477.54.Example 3 : Find the half-yearly compound interest on Rs 5000 for 1 year at the rate of 8 % per annum.Solution: Here, Rate of interest(R) = 8% per annum Time (T) = 1 year Principal = Rs 5000. Compound interest (C.I.) = ?We know, for semi-annuallyCompound Interest (C.I.) = P

R200

{( )2T - 1}1 +

= 5000

4200

{( )2×1 - 1}1 +

= 5000 {(1.04)2 - 1} = 5000 {1.0816 - 1} = 5000 × 0.0816 = Rs 408Hence, the compound interest is Rs 408.Example 4 : Saroja borrowed Rs 150000 from Saroj at the rate of 21% per year. At the end of 9 months how much compound interest should she pay, if compounded half yearly ?Solution : Here,Principal (P) = Rs 150000Rate (R) = 21%Time (T) = 9 months = 6 months + 3 months

= 612

years + 3 months

∴ T = 12

years and M = 3


Compound interest (C.I.) = ?We know, for semi-annuallyCompound Interest (C.I.) = P

R200

MR1200

{( )- 1}1 + 1 +)2T(

= 150000 21200

3×211200{( )- 1}1 + 1 +)2×

12 (

= 150000 (1.105 × 1.0525 -1) = 150000 × 0.1630125 = Rs 24451.87Hence, the compound interest is Rs 24451.87

Example 5 : Vijay borrowed Rs 5000 and agreed to pay interest at the rate of 10 %, 12%, and 14% for the first, second and third year respectively. Find the total amount he had to pay after 3 years.

Solution : Here,Principal (P) = Rs 5000Rate for first year (R1) = 10%Rate for second year (R2) = 12 %Rate for third year (R3) = 14 %Time (T) = 3 YearsCompound amount (C.A.) = ?We know, Compound amount (C.A.) = P R1

100( )1 +

R2100

( )1 + R3100

( )1 +

= 5000 10100

( )1 + 12100

( )1 +

14100

( )1 +

= 5000 × 1.10 × 1.12 × 1.14 = Rs 7022.4Thus, the compound amount is Rs 7022.4.Example 6: At what rate percent of compound interest will Rs 625 amount to Rs 729 in 2 years?Solution: Here, Principal(P) = Rs 625 Compound amount(CA) = Rs 729 Time (T) = 2 years. Rate of interest(R)= 2 years


We have

CA= P R100

( )1 +T

Or, 729 = 625 R100

( )1 +2

Or, 729625

= R100

( )1 +2

Or, 729625

( )2 = R

100( )1 +

2

Or, 2725

= 1 + R100

Or, 2725

= 100+R100

Or, 2500+25R=2700 Or, 25R = 2700-2500 Or, 25R = 200

Or, R = 20025

∴ R= 8%.Thus, the rate of interest(R) is 8%.

Example 7 : At what time will the semi-annual compound amount of Rs 2400 at the rate of 10% per annum be Rs 2646? Find it.Solution: Here,Principal(P) = Rs 2400Compound amount(CA) = Rs2646Rate of interest(R) = 10%Time(T) = ?We have, CA= P R

200( )1 +

2T

Or, 2646= 2400 10200

( )1 +2T

Or, 26462400

= 10200

( )1 +2T

Or, 441400

= 2120

( )2T


Or, 2120

( )2= 21

20( )2T

Or, 2T = 2Or, T = 1Thus, the time is 1 year.Example 8: The simple interest on a sum of money in 2 years is Rs 36 less than the compound interest compounded annually. If the rate of interest is 12%, find the sum.Solution: Here, Time (T) = 2 yearsRate of interest (R) = 12%Let the principal (P) = Rs xWe know,Simple interest (I) = P × T × R

100

= x × 2 × 12100

= 24x100

= Rs 0.24xAccording to question

CI – I = 36Or, 0.2544x – 0.24x = 36Or, x(0.2544 – 0.24) = 36Or, x × 0.0144 = 36

Or, x = 360.0144

Or, x = 2500Thus, the principal is Rs 2500.Example 9: The compound amount of a sum of money in 3 years is Rs 13310 and in 4 years is Rs 14641. Find the sum and the rate of interest.Solution: Here,Let principal(P) = Rs PRate of interest = R%We know,

Compound interest (CI) = PR

100{( )T

- 1}1 +

= x12100

{( )2 - 1}1 +

= x × 0.2544 = Rs 0.2544x


Compound amount (CA) = P R100

( )1 +T

For 3 years,

13310 = P R100

( )1 +3 …………………(i)

For 4 years,

14641 = P R100

( )1 +4 ………………..(ii)

Dividing equation (ii) by (i) ,we get

1464113310

= R

100( )41 +

R100( )3

1 +

P

P

Putting the value of ‘R’ in equation (i)13310 = P 10

100( )1 + 3

Or, 13310 = P × 1.331

Or, P = 133101.331

Or, P = Rs 10,000Thus, the principal (P) is Rs 10,000 and rate of interest (R) is 10%

Example 10 : Sita borrowed Rs 170000 from Radha at the rate of 21% p.a. at the end of 1 year and 6 months,

i. How much simple interest will she have to pay?ii. How much compound interest compounded yearly will she have to pay?

Solution:Here,Principal (P) = Rs 170000Rate of interest (R) = 21%Time (T) = 1 year 6 monthsi. For simple interest

Time (T) = 1 year 6 months

= 1 + 612

= 32

years

Now, simple interest (I) = P × T × R100

= 32

100170000 × × 21

Or, 1110 = 1 + R

100

Or, 1110 =

100 + R100

Or, 1000 + 10R = 1100Or, 10R = 1100 – 1000Or, 10R = 100Or, R = 10%


= 1700 × 32

× 21

= Rs 53550∴ Simple interest is Rs 53550.

ii. For compound interestTime (T) = 1 year 6 months

= 1 year and 12 year

Simple interest in 1 year = P × T × R100

= 170000 × 1 × 21

100 = Rs 35700Amount in 1 year = Rs 170000 + Rs 35700 = Rs 205700

simple interest in 12

year = P × T × R100

= 12

100205700 × × 21

= Rs 21598.5∴ Compound interest at the end of 1 year and 6 months = Rs 35700 + Rs 21598.5

= Rs 57298.50Example 11: Mohan lent altogether Rs 6600 to Ram and Shyam for 2 years. Ram agrees to pay simple interest at 15% p.a. and Shyam agrees to pay compound interest at the same rate. If Ram paid Rs 112.50 more than Shyam as the interest. How much did he lent to each?Solution: Here,Let the sum lent to Ram = Rs xThen the sum lent to Shyam = Rs(6600 – x)For Ram,Principal (P) = Rs xTime (T) = 2 yearsRate of interest (R) = 15%Simple interest paid by Ram = P × T × R

100

= x × 2 × 15100


= 30x100

= Rs 0.3xFor Shyam,Principal (P) = Rs (6600 – x)Time (T) = 2 yearsRate of interest (R) = 15%Compound interest paid by Shyam= P

R200

{( )T -1}1 +

= (6600 –x) 15100

{( )2 -1}1 +

= (6600 – x) × 0.3225 = 2128.5 – 0.3225x

According to question,Interest paid by Ram – Interest paid by Shyam = Rs112.50Or, 0.3x – (2128.5 – 0.3225x) = 112.50Or, 0.3x – 2128.5 + 0.3225x = 112.50Or, 0.6225x = 112.50 + 2128.5Or, 0.6225x = 2241

Or, x = 22410.6225= Rs 3600

The sum lent to Ram is Rs 3600And the sum lent to Shyam = Rs (6600 – x) = Rs (6600 – Rs 3600) = Rs 3000Thus, the sum lent to Shyam is Rs 3000Example 12: The compound interest of a sum of money in 2 years and 4 years are Rs 1050 and Rs 2320.50 respectively. Find the rate of interest compounded annually and the principal.Solution:Here,Compound interest in 2 years = Rs 1050Compound interest in 4 years = Rs 2320.50We know, for yearly compounding

CI = P R

100{( )T

-1}1 +


For 2 years

1050 = P R

100{( )2

-1}1 +

Or, 1050 = P(a2 – 1) …………………(i)

Where a = 1 + R100

For 4 years

2320.50 = P R

200{( )4

-1}1 +

Or, 2320.50 = P(a4 – 1) …………………(ii)Where, a = 1 + R

100

Dividing equation(ii) by equation (i)

2320.51050 = P(a4-1)

P(a2 -1) Putting value of R in equation (i) 1050 = P{(1.1)2 – 1} Or, 1050 = P × 0.21

Or, P = 10500.21

Or, P = Rs 5000The rate of interest is 10% and principal is Rs 5000.Example 13 : Saroj took a loan of Rs 46875. If the rate of compound interest is 4 paisa per rupee per year, in how many year will the compound interest be Rs 5853Solution: Here, I = 4 paisa = Rs 4

100 P = Re 1 T = 1 yearRate of interest (R) = 1 × 100

P × T = 4 × 100

100 × 1 × 1 = 4%

Then, Principal (P) = Rs 46875Time duration (T) = ?Now, we have, CI = P

R100

{( )T -1}1 +

Or, 5853 = 46875 4

100{( )T

-1}1 +

Or, 2.21 = a2 + 1 Or, a2 = 1.21 Or, a = 1.1

Or, 1 + R100 = 1.1

Or, 100 + R100

= 1.1

Or, 100 + R = 110 Or, R = 110 – 100 ∴ R is 10%


Or, 585346875

= 104100

( )T -1

Or, 104100

( )T = 5853

46875 + 1

Or, (1.04)T = 5853 + 4687546875

Or, (1.04)T = 1.124864

Or, (1.04)T = (1.04)3

∴T = 3 So, the required time is 3 years.

1. a. The compound interest and the compound amount of a sum P in T years at R% p.a. are CI and CA respectively. Write down the relation among:

i. P,T,R and CI (Compounding annually) ii. P, T, R and CA (Compounding annually) iii. P,T,R and CI ( Compounding semi- annually) iv. P,T,R and CA ( Compounding semi annually) b. According to the compound interest ‘A’ is the amount, ‘R’ is the rate of interest per

annum and T is the time. What will be the principal?2. a. Find the simple interest and compound interest compounding annually for Rs 10000 in

1 year at the rate of 10% per annum. Are they same? Why? b. Find the compound amount and compound interest of the following for yearly

compounding. i. Principal(P)= 55000, Time(T) =2 years , Rate (R) =5% ii. Principal(P)= Rs 8000, Time(T)= 3 years , Rate (R)= 4.5% iii. Principal(P)= Rs 12000, Time(T)= 2 years , Rate (R)= 8% c. Find the compound amount and compound Interest of the following for half yearly

compounding. i. Principal(P)= Rs10000, Time(T)=2 years , Rate (R)= 6% ii. Principal(P)= Rs 4500, Time(T)= 1years , Rate (R)= 12% iii. Principal(P)= Rs 25000, Time(T)= 15months , Rate (R)= 10%3. a. Santosh borrowed Rs 13000 from Suresh at the rate of 21% per annum for 2 years. Find

the i. Simple interest he has to pay. ii. Compound interest he has to pay compounding annually.

Exercise (2.3)


b. Sita borrowed Rs 4800 from Ram at the rate of 10% per annum for 1 year. i. How much simple interest will sita have to pay? ii. How much compound interest compounded semi annually will she have to pay? c. A man deposited Rs 2000 in fixed deposit account of the bank for 2 years at the rate

of 12% per annum. The interest is compounded semi-annually. How much will be the amount and compound interest at the end of 2 years?

d. Hari borrowed Rs 150000 from Shyam at the rate of 21% per year. At the end of nine month how much compound interest should he pay compounded half yearly?

4 a. Find the difference between simple interest and yearly compound interest of Rs 6500 at the rate of 4.5% per annum for 3 years.

b. Find the difference between compound interest payable annually and semi annually for the sum of Rs 14000 at the rate of 12% per annum for 2 years.

c. A person borrowed Rs 16000 from a bank at 12.5% per annum simple interest and lent whole sum immediately to a shopkeeper at the same rate of compound interest. How much will he gain after 3 years?

5 a. The sum of annual compound interest and the semi -annual compound interest on a sum of money for 2 years at the rate of 20% per annum is Rs 18082. Find the principal.

b. The sum of the simple interest and compound interest on a sum of money for 2 years at the rate of 8% per annum is Rs 612. Find the principal.

c. The difference between the compound interest compounded semi-annually and compound interest compounded annually on a sum of money is Rs 220.25 at the rate of 10% per annum for 2 years. Find the sum.

6. a. Prakash lent altogether Rs 6000 to Anima and Elina for 2 years. Anima agrees to pay simple interest at 10% per annum and Elina agrees to pay compound interest at the rate of 8% per annum. If Elina paid Rs 50 more than Anima as the interest , find how much did he lend to each?

b. Divide Rs 21000 into two parts so that the amount of first part for 2 years is same as the amount of second part for 3 years at the rate of 10% p.a. compounded annually.

7. a. The compound interest on a certain sum for 2 years at 10% per annum is Rs 420. What would be the simple interest on the same sum at the same rate for the same time.

b. A sum of Rs 150000 amounts to Rs 262500 at a certain rate of simple interest in 5 years. Find the sum of money that amounts to Rs 198375 at the same rate of compound interest in 2 years.

8. a. The compound amount of a certain sum of money in 2 years and 3 years become Rs 8820 and Rs 9261 respectively. Find the sum and the rate of interest.

b. A sum of money amounts to Rs 19360 in 2 years and Rs 23425.60 in 4 years at the certain rate of compound interest compounded annually. Find the rate of interest and the sum.

9. a. The compound interest of a sum of money in 1 year and 2 years are Rs 400 and Rs 832 respectively. Find the rate of interest compounded yearly and the sum.

b. The compound interest of a sum of money in 2 years and 4 years are Rs 5460 and Rs 12066.59 respectively. Find the rate of interest compounded annually and the sum.

10. a. A person took a loan of Rs 40000. If the rate of compound interest is 5 paisa per rupee per annum, in how many years will the compound interest become Rs 6305.01?


b. According to the yearly compounded interest the compound interest on Rs 400000 at the rate of interest 6.5% per annum is Rs 53690.03? Find the time period.

11. a. At what percent rate of compound interest per annum will be compound interest on Rs 343000 be Rs 169000 in 3 years?

b. If the semi-annual compound interest of Rs 15625 for 1 year 6 months is Rs 1951, find the rate of interest.

12. a. The simple interest and compound interest of a sum of money in 2 years are Rs1000 and Rs 1050 respectively. Find the yearly rate of interest and the principal.

b. The compound interest on a sum of money for 2 years compounded yearly is Rs 1680 and simple interest on the same sum for the same period and at the same rate is Rs 1600. Find the sum and the rate of compound interest.

13. A invested Rs. 25000 for 3 years at the rate of 12%, simple interest per annum and B invested the same amount for the same time at the rate of 10% annual compound interest.

i. Calculate the interest received by A and B separately. ii. Without alterning time and rate, how much more or less money should 'A' have to

invest for equal interest?

Ans

wer

s (2.

3)

1.a.i. CI = P R

100{( )T

-1}1 + ii. CA = P R

100( )T

1 +

iii.CI = P R

200{( )2T

-1}1 + iv. CA = P R

200( )2T

1 +

b. P = R

100( )T

1 +

A

2a.Yes. For one time period compound interest and simple interest are same.b. i.CA=Rs.60637.5, CI = Rs.5637.5 ii. CA=Rs.9129.33, CI=Rs.1129.33 iii. CA=Rs.13,996.80,CI=Rs.1996.80 c. i. CA=Rs.11255.08, CI=Rs.1255.08 ii. CA=Rs.5056.2, CI=Rs.556.2 iii. CA=Rs.28251.56, CI=Rs.3251.56 3a. Rs.5460, Rs.6033.30 b. Rs.480, Rs.492c. Rs.2524.95, Rs.524.95 d. Rs.24451.884a. Rs.40.08 b. Rs.113.08c. Rs.781.25 5a. Rs.20,000b. Rs.1875 c. Rs.40,0006a. Rs.2588.43, Rs.3411.57 b. Rs.11000, Rs.10,0007a. Rs.400 b. Rs.1,50,0008a. Rs.8000, 5% b. 10% ,Rs.16,0009a. 8%, Rs.5,000 b. 10% ,Rs26,00010a. 3 Years b. 2 Years 11a.14.28% b. 8% 12a. 10%, Rs.5000 b. Rs.8000,10% 13.a. Rs 9000, Rs 8275 b. Rs 2013.89 less


2.4 Population growthPopulation means total number of people living in a place or a country. Generally the population has been found increasing in most of the places or countries.The process of increasing population is same as process of increasing compound amount. Just like in compound interest, every year population, increases from the previously increased population.So, if the population of a country or a place at a certain time = PoPopulation growth rate = R% per annumPopulation after T years = PT

Then PT = Po R

100( )1 +

T

There are many causes of population change, number of in-migrants, number of out- migrants, number of death etc. are the major causes.The actual population at the end of the given time period is given by

Then PT = Po R

100( )1 + T + Min – Mout – D where

Min = number of in-migrants Mout = number of out-migrants D = number of deaths

PT = P R1 - R2

100( )1 +

T Where R1 - R2 is not growth rate

PT = P R1100

( )1 + R2100

( )1 + R3100

( )1 +

Where R1, R2 and R3 are growth rates of first year, second year and third year respectively.

Example 1 : The population of a village is 20000. If the population growth rate of the village is 2.1 % per annum. What will be the population after 2 years ?Solution : HereOriginal population (P) = 20000Growth rate (R) = 2 %Time (T) = 2 years.Population after 2 years (PT) = ?We know,

Worked Out Examples


PT = Po

R100

( )T1 +

= 20000 2

100( )21 +

= 20000 ( 1. 02 )2

= 20000 1.0404 = 20808 ∴ The population after 2 years is 20808Example 2 : The population of a town before 2 years was 25000 and annual growth rate of population is 2 %. If the number of in-migrants and out- migrants at the end of 2 years were 1000 and 750 respectively and 650 died within this time period, find the present population of the town.Solution : Here,Initial population (Po) = 25000Population growth rate (R) = 2%Number of in- migrants ( Min ) = 1000Number of out- migrants ( Mout ) = 750Number of deaths (D) = 650Time (T) = 2 yearsPresent population (PT) = ?We know, Present population (PT) = P0

R100

( )T1 + + Min – Mout - D

= 25000 2

100( )T1 + + 1000 -750 – 650

= 25000 × (1.02)2 + 1000 -750 – 650 = 25000 × 1.0404 + 1000 -750 – 650 = 25610∴ The present population of the town is 25610.

Example 3 : The population of a village increases every year by 5%. At the end of 2 years, the total population of the village was 10000. If 1025 were migrated to other places, what was the population of the village in the beginning ?Solution : Here,Total population including migrated population after 2 years (PT) = 10000 +1025 = 11025Population growth rate (R) = 5%


Time (T) = 2 YearsInitial population of the village (P0 ) = ?We know,

(PT) = P0 R

100( )T1 +

Or, 11025 = P0 5

100( )21 +

Or, 11025 = P0 × (1.05)2

Or, 11025 = P0 × 1.1025

Or, Po = 110251.1025

Or, Po = 10000 ∴ The beginning population of the village was 10000.Example 4 : The population of the city before 3 years was 300000. If the annual growth rate of the population in the last 3 years were 2%, 4% and 5% respectively every year. Find the population of the city at the end of 3 years.Solution : Here,Population of the city before 3 years (Po) = 300000Annual growth rate of three years were R1= 2% , R2= 4% and R3= 5%Population of the city after 3 years (PT) = ?

We know , The population after 3 years (PT) = Po R1100

( )1 + R2100

( )1 + R3100

( )1 +

= 300000 2

100( )1 +

4100

( )1 + 5100

( )1 +

= 300000 × 1.02 × 1.04 × 1.05 = 334152∴ The population of the city at the end of the 3 years is 334152.Example 5 : In the beginning of 2065 B.S., the population of a town was 100000 and rate of population growth is 2% every year. If in the beginning of 2066 B.S., 8000 people migrated there from different places. What will be the population of the town in the beginning of 2068 ?Solution:Here,Population of the town in the beginning of 2065 (Po) = 1,00,000


Growth rate (R) = 2%Population of the town in the beginning of 2066(PT) = Po

R100

( )T1 +

= 1,00,000 × 2

100( )11 +

= 1,02,000Again considering the population of the town in the beginning of 2066 as P and the population in the beginning of 2068 as PT.

P = 102000 + 8000 = 110000

PT = P R

100( )T1 +

= 110000 2

100( )21 +

= 110000 × 1.0404 = 114444

∴ Population of the town in the beginning of 2068 is 1,14,444.Example 6 : 2 years ago the population of a town was 170000. The rate of annual growth is 3%. If the total number of deaths in last 2 years was 3500 and 4500 were in –migrants to the town, find the present population of the town.Solution: Here, The population of the town before 2 years (Po) = 170000The population growth rate (R) = 3%Total deaths in 2 years (D) = 3500Total number of in –migrants in 2 years (Min) = 4500Now,Present population (PT) = Po

R100

( )T1 + + Min – D

= 170000 3

100( )21 + + 4500 – 3500

= 170000 × ( 1.03)2 + 4500 – 3500 = 170000 × 1.0609 + 1000 = 180353 + 1000 = 181353Hence, the present population is 1,81,353.


Exercise (2.4)

1. a. If the population of a town P increases to PT in T years at the growth rate of R%, write down the relation between P, PT, T and R.

b. If the initial population P increases for the fist three years at the rates of R1%, R2 % and R3% respectively and becomes PT, then write the relation among PT, P, R1, R2 and R3.

2. a. The present population of a village is 30000. If it is increased at the rate of 10% per annum, what will be the population after 2 years?

b. The population of a city in 2011 AD was 50,000. If the rate of population growth was 3% per annum, what was the population in 2013? Find it.

c. Three years ago the population of a village was 50000. If the birth rate and death rate are 4.5% p.a. and 2% p.a. respectively, find the present population of the village.

3. a. The present population of a town is 80,000. If the population increases 2% by birth and 3% by migration. What will be the population of the town after 2 years?

b. The present number of students of a school is 2000. If every 10 students bring 1 student each year, find the number of students in the school after 2 years.

c. The present population of a village is 17640. If the annual population growth rate of population is 5%, what was the population of the village before 2 years.

4. a. The population of a town was increased from 45000 to 46818 in two years. Find the rate of population growth rate.

b. The population of a village was 10000 one year ago. The population at present is 10210. Find the population growth rate.

c. In how many years will the population of a village be 46305 from 40,000 at the growth rate of 5% per annum ?

5. a. The population of a village increases every year by 5%. At the end of three years, the total population of a village was 10000. If 1025 were migrated to other places, what was the population of the village in the beginning ?

b. The population of a village increases every year by 10%. At the end of two years, the total population of a town was 30000. If 5800 people were added by migration, what was the population of the town in the beginning ?

6. a. The population of a town before three years was 360000 and annual growth rate is 4% . If the number of in- migrants and out- migrants at the end of three years were 1850 and 2175 respectively. If 1525 people died within the time, find the present population of the town.


b. At the end of 2070 B.S., the population of a city was 20000 and the population growth rate is 2% every year. If at the end of 2072 B.S. 9192 people migrated there from different places. What will be the population of the town at the end of 2074 B.S. ?

c. The population of a town before three years was 600000. If the annual growth rate of the population in the last three years were 2% , 4% and 6% respectively, find the population of the town at the end of three years.

d. The number of bacteria in a sample of urine of a patient is 105. What will be the number of bacteria after it is increased with multiple growth rate of 3 %, 4% and 5.5% in Ist hour, 2nd hour and 3rd hour respectively.

e. The number of bacteria in a sample of urine of a patient is 106. What will be the number of bacteria after it is increased with multiple growth rate of 2 %, 3% and 4.5% in Ist hour, 2nd hour and 3rd hour respectively.

7. a. Three years ago, there were 1500 students in a secondary school. Since the past two years there is the rule a group of five students should bring a new student for enrollment was imposed to increase the number of students. What is the number of students at present ?

b. Three years ago, there were 2000 students in a secondary school. Since the past three years the rule a group of ten students should bring a knew student for enrollment was imposed to increase the number of students. What is the number of students at present ?

8. a. The population of Kathmandu was 1000000 at the end of year 2055 B.S. and the population growth rate was 4.5%. If 25000 men migrated here from other places at the end of the year 2056 B.S., What will be total population at the end of year 2058 B.S.?

b. The population of a district 2 years ago was 32000 and the growth rate of population was 3 % per year. If at the end of second year, the number of inhabitants leaving and entering the district are 1035 and 848 respectively and the number of death is 604, find it’s present population.

c. Three years ago the population of a town was 64000 and the growth rate of the population was 2.5% per year. If at the end of the second year, the number of inhabitants leaving and entering the district are 1025 and 750 respectively and the number of death is 1260, find the present population of the town.

Ans

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s (2.

4)

1.a. Show to your teacher 2.a. Rs 36300 b. Rs 53045c. Rs 53845 3.a. Rs 88200 b. Rs 2420 c. Rs 16000 4.a. 2% b. 2.1% c. 3 years 5.a. Rs 9524 b. Rs 20000 6.a. Rs 403101 b. Rs 31212 c. Rs 674669d. 1.130116 × 105 e. 1.097877× 106 7. a. Rs 2592 b. Rs 2420 8.a. Rs 1168467 b. Rs 33158 c. Rs 67,348


2.5 DepreciationWhen the assets like furniture, electronic items, vehicle etc. are used for some time, their values decrease continuously. Such decrement of values of liquid assets is called depreciation.Depreciation may be simple or compound but here, we will discuss about the compound depreciation. In compound depreciation, the amount of reduction is calculated every year from every depreciated value. So, the compound depreciation is calculated in the similar way of calculation of compound interest. It means in compound depreciation reduction in the value of the assets is not same. It is calculated from the depreciated value of the assets at the end of the previous year.Let the original value of goods= DoRate of depreciation = R%Value of goods after T years = DTTime period = T years

Then , DT = DoR

100( )T1 -

Similary, if the rate of depreciation is different in different years, then

DT = D0 R1100

( )1 - R2100

( )1 - R3100

( )1 -

Where, R1= rate of depreciation in first year. R2= rate of depreciation in second year. R3= rate of depreciation in third year.

Example 1 : A television costing Rs 11600 is depreciated at the rate of 15% per annum. What will be the cost of television after two years?Solution : Here,Initial cost of the television (Do) = Rs 11600Rate of depreciation ( R ) = 15 %Time period ( T ) = 2 yearsCost of the television after two years (DT) = ?We know, Cost of the television after two years (DT) = Do

R100

( )T1 -

= 11600 × 15100

( )T1 -

Worked Out Examples


= 11600 (0.85 )2

= 11600 0.7225 = 8381 ∴ Cost of the television after two years is Rs 8381.Example 2 : The value of a machine bought three years ago depreciates at the rate of 10% per annum and becomes Rs 92583 at present. Calculate the original value of the machine.Solution : Here,Present cost of the machine (DT) = Rs 92583Rate of depreciation ( R ) = 10 %Time period ( T ) = 3 yearsInitial cost of the machine (Do) = ?Cost of the machine after three years (DT) = Do

R100

( )T1 -

Or, 92583 = Do 10100

( )31 -

Or, 92583 = Do (0.90 )3

Or, 92583 = Do × 0. 729

Or, Do= 925830.729

∴ Do = 127000So, Initial cost of the machine is Rs 127000Example 3 : A man bought a mobile set for Rs 44100 and sold it for Rs 40000 after using for 2 years. Find the rate of the depreciation .Solution : Here,Initial cost of the mobile (Do) = Rs 44100Time period (T) = 2 yearsCost of the mobile after two years (DT) = 40000Rate of depreciation (R) = ?Cost of the mobile after two years (DT) = Do

R100

( )T1 -

Or, 40000 = 44100 × R

100( )21 -

Or, 4000044100

= R

100( )21 -

Or, 400441

= R

100( )21 -


Or, 2021

( )2=

R100

( )21 -

Or, 2021 =

R100

1 -

Or, 2021

= 100-R100

Thus, required rate of depreciation is 4.76 % . Example 4 : A factory bought for 400000 some years ago and now its value is Rs 196000. If the value of the factory depreciated at 30% per annum. When was the factory bought ? Solution : Here, Initial cost of the factory(Do) = Rs 400000Cost of the factory after T years(DT) = 196000Rate of depreciation (R) = 30 %Time period (T) = ?We know, Cost of the factory after T years(DT) = Do

R100

( )T1 -

196000 = 400000 30100

( )T1 -

Or, 196000400000 =

710

( )T

Or, 49100 = 7

10( )T

Or, 710

( )2=

710

( )T

Or, 2 = T∴ The factory was bought 2 years ago.Example 5 : A man purchased a microbus for Rs 3200000. He earned a profit of Rs 410000 in three years. If he sold it after three years at 10% per annum compound depreciation, find his profit or loss.Solution : Here,Initial cost of the Microbus (Do)= Rs 3200000Profit (P) = Rs 210000Time (T)= 3 yearsRate of depreciation (R) = 10 %

Or , 100 × 20 = 2100 – 21 ROr, 2000 -2100 = - 21 ROr, -100 = -21 R

Or, R = -100-21

R = 4.76


Profit or loss = ?We know, Cost of the Microbus after three years (DT) = Do

R100

( )T1 -

= 3200000 × 10100

( )31 -

= 3200000 × (0.9 )3

= 3200000 × 0.729 = 2332800Profit earned during three years = Rs 410000Total sum of money received = Rs 2332800 + Rs 410000= 2742800But amount paid = Rs. 3200000 ∴Loss = Rs 3200000- Rs 2742800 = Rs 457200

1. a. Write the relation among Do, T, R and DT , if Do represents initial price of a machine,T represents time, R represents rate of depreciation and DT represents price of the machine after T years.

b. If DT , D0 , R and T represent the price of a machine after T years, initial price,rate of depreciation and time in years respectively then write the relation of DT in terms of other variables.

c. Write the formula to calculate the price of an article after three years with rates of depreciation r1%, r2% and r3% successively.

2. a. The value of machine costing Rs 24000 is depreciated every year by 10%, what will be the value of the machine after one year ?

b. A watch costing Rs 2500 is depreciated in a year by 15%, find the price of the watch after one year.

c. The present price of a i-phone is Rs 95000.It is depreciated at 6% per year. Find the price after two years.

3. a. If the cost is depreciated at the rate of 10% per year, the cost of an article becomes Rs 92583 after three years. Find the original price of the article.

b. A laptop which is supposed to depreciate at the rate of 10% every year was sold at Rs 57000 after 2 years. At what price was it bought ?

Exercise (2.5)


4. a. The value of an article is depreciated from Rs 18000 to Rs 14580 in two years, find the rate of depreciation.

b. The price of a motorcycle was Rs 250000 in 2070 B.S. and in 2072 B.S., it is Rs160000. Find the annual rate of the compound depreciation.

5. a. A factory bought for Rs 400000 some years ago and now its value is Rs 196000. If the value of the factory is depreciated at 30% per annum, when was the factory bought ?

b. The value of a mobile set bought for Rs 50000 is depreciated at the rate of 10% per annum. In how many years will its value be Rs 40500 ?

6. a. A watch costing Rs 6000 is depreciated in two years at the rate of 30% per annum. Find the amount of depreciation.

b. The price of a scooty is Rs 140000. If it is depreciated at 10% yearly, find the amount of depreciation after two years.

7. a. A man purchased a secondhand car for Rs 640000 and spent Rs 18950 for repairing it .If the car was sold after three years at the rate of 5% compound depreciation. Find his loss.

b. A microbus was purchased on Rs 4000000. It earned Rs 1800000 profit during three years and its value is depreciated at the rate of 10% per annum. Find the amount of profit or loss at the end of three years

8. a. A press was established with an initial investment of Rs 600000. With in three years it earned Rs 200000.Assuming the compound depreciation in the assets at the rate of 10% per annum, the press was sold after three years. Find the net amount of profit.

b. A van costs Rs 1400000. Its value depreciates at the rate of 10% per year during first 2 years and at the rate of 5% in next two years. What will be the price of the van after 4 years.

c. The value of a piece of land increased by10 % in the first year and depreciated by 10% in the second year. Again increased by 15 % in the third year. If the initial value of the land was Rs 2500000, find the value after three years.

d. A man invested Rs 600000 on the gold. Its price was increased by 20% in the first year but its value is depreciated by 5% in second year and then 6% in the third year. What is the value of those gold on fourth year?

Ans

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1. Show to your teacher 2. a. Rs 21600 b. Rs2125 c. Rs 83942 3.a. Rs 127000 b. Rs 70370.37 4.a. 10% b. 20% 5.a. 2 years ago b. 2 years ago 6.a. Rs 3060 b. Rs 26600 7.a. Rs 93,983 b. profit= Rs 716000 8. a. Rs 37400 b. Rs 1023435 c. Rs 28,46,250 d. Rs 6,42,960


Model - Question [F. M. 12]

Group (A) (1 × 1 = 1)1. If M.P. = Rs 1600 , S.P.= Rs 1480 , find discount rate . (75%)

Group (B) (1 × 2 = 2)2. A laptop is brought from India at Rs 64000 I.C. Find its cost in the Nepali currency if 50%

custom and 13% VAT is added. (IC Rs 1 = NCRs 1.6) (Rs 173568)Group (C) (1 × 4 = 4)

3. There are 1000 students in Dhurbatara Boarding School in 2071 B.S. If each group of 10 students brought a new students for admission to increase the number of students in each session. Find the number of students in 2073 B.S. (1210)

Group (D) (1 × 5 = 5) 4. The compound interest on sum of money in 2 and 4 years are Rs 3816 and Rs 8602.79

respectively. Find the rate of interest compounded yearly and the sum. (12%, Rs15000)

Test Yourself F.M. 20

1. a. What is present VAT rate of NEPAL?

b. Write the formula to calculate CI for r1%, r2% and r3% for successive 3 years.

c. If DT, Do, R and T represent the price after T years, initial price, depreciation rate and time in year respectively. Then write the relation of DT in terms of other variables.

2. a. Amrit exchanged Rs6,00,000 into Pound Sterling. After one week, Nepalese rupees devaluated by 5%, find whether he has profit or loss initially, £1 = Rs 139.14. (Rs 30000 gain)

b. The population of a town was increased from 45,000 to 46,818 in two years, find the rate of population growth. (2%)

3. a. Ram suppliers sold a machine worth Rs1,50,000 excluding 13% VAT to Hari suppliers. Hari suppliers sold the machine including Rs4000 transportation fair, Rs1500 local tax and Rs7000 profit to the customer. What amount should the customer had paid for the machine including 13% VAT? (Rs 2,05,660)

b. The compound amount of a sum of money in 2 years is Rs8820 and in 3 years Rs9261. Find the sum and the rate of interest. (Rs 8000, 5%)

4. Two years ago the population of a town was 64000 and the growth rate of population was 2.5% per year. At the end of the second year, the number of inhabitants leaving and entering the town are 1025 and 848 respectively and the number of deaths is 604, find the present population. (66459)


MensurationEstimated period : 24

3 Unit

Specific objectives After the completion of this unit, the students will be able to

• find the area of triangles.• find the lateral surface area, total surface area and volume of the triangular

prism.• find the surface area and volume of cylinder, sphere, hemisphere, cone and

pyramid.• find the surface area and volume of combined solids made by cylinder, cuboid

pyramid, sphere hemisphere, cone etc.

Teaching materialsSolid models of prism, pyramid, cone, sphere, hemisphere etc. formula chart.

Note to the teacher1. Review the concept of different types of triangles, basic concept of area, pe-

rimeter and volume.2. Introduce the formula to find area of triangles. Derive the formula to find the

area of scalene triangle.3. Show the net of triangular prism to find lateral surface area, area of cross-sec-

tion and total surface area.4. Give the concept of volume as the product of area of base × height.5. Introduce cylinder, sphere, pyramid and cone with the help of solid model.

Obtain the formula to calculate curved surface area, area of base, lateral sur-face area and volume.

Curriculum• Area of different triangles• Practical problems of triangular region• Problems on lateral surface area, total surface area and volume of triangular

prism.• Problems on curved surface area, total surface area and volume of cylinder,

sphere and hemisphere.


• Problems on curved surface area, total surface area and volume of combined solid made by cylinder, sphere ,hemisphere and cone (any 2).

• Problems on curved surface area, total surface area and volume of different pyramids.

Specification GridCognitive domain

Knowledge(K)

Comprehensive(C)

Application(A)

Higher ability(HA)

Total no.of questions

Totalmarks

Topic Each 1 mark

Each2 marks

Each of 4 marks

Each of 5 marks

Plane SurfaceCylinder and SpherePrism and Pyramid

1 3

1

1

6

16

Note : At least 3 marks questions are asked from each topic.

3.1 Area of triangleThere are different types of triangles such as right angled triangle, equilateral triangle, isosceles triangle and scalene triangle. In this chapter we calculate the area of these triangles.1. Area of triangle when base and height are given:Let us take triangle ABC, in which AD is perpendicular to BC. So, AD is height or altitude of triangle and BC is base.

Here, Area of triangle ABC = 12 (base × height)

= 12

(BC × AD)

= 12

(b × h)

2. Area of right angled triangle :ABC is a right angled triangle having right angle at B. Two sides AB and BC contain-ing right angle are consider as height and base to find the area.

Area of triangle ABC = 12 (base × height)

= 12 (BC × AB)

= 12

(b × h)


3. Area of an equilateral triangle

Let us consider an equilateral triangle ABC in which AB = BC = AC

Let AB = BC = AC = a

Let AD ⊥ BC

Here, BD = CD = a2

In right angled triangle ABD,

AD = AB2 - BD2

AD = a2 - ( a2 )2

= a2 - a2

4

= 3a2

2 = 32

a

Now area of triangle ABC = 12 × base × height

= 12 × a × a

2 3

= 34 a2

Therefore, Area of an equilateral triangle = 34 a2

Where ‘a’ is the length of each side of an equilateral triangle. 4. Area of an isosceles triangle

Let us take an isosceles triangle ABC having AB=ACLet AB = AC = a and base BC = bLet AD ⊥ BCHere, BD = CD = b

2 (∵AD⊥ BC)In right angled triangle ABD,

AD = AB2 - BD2

= a2 - ( b2 )2

= a2 - b2

4


= 4a2 - b2

4

= 12

4a2 - b2

Now, Area of triangle ABC = 12 (base × height)

= 12 (BC × AD)

= 12 × b × 1

2 4a2 - b2

Therefore, Area of an isosceles triangle = b4 4a2 - b2

Where ‘a’ represents the length of equal sides ‘b’ represents the length of base of the triangle.

5. Area of scalene triangle when three sides are given Let us take triangle ABC having sides AB, BC and AC. Let AD ⊥ BC.

Let BC = a, AC = b and AB = c, AD = h Let DC = x , then BD = a-xAlso, let perimeter of triangle be P i.e. 2s.Then,P = AB + BC + AC

2s = a + b + c

s = a+b+c2 ………….. (1)

Now, in right angled triangle ABDc2 = h2 + (a-x) 2

or, h2 = c2 – (a – x) 2 ……………. (2)In right angled triangle ACD b2 = h2 + x2

or, h2 = b2 – x2 …………………. (3)From equation (2) and (3)b2 – x2 = c2 – (a – x) 2

b2 – x2 = c2 – a2 + 2ax – x2

2ax = a2 + b2 – c2


x = a2 + b2 - c2

2a …………. (4)

Now from equation (3)

h2 = b2 – a2 + b2 - c2

2a( )2

Or , h2 = a2 + b2 - c2

2a(b + ) a2 + b2 - c2

2a(b - )

Or ,h2 = 2ab + a2 + b2 - c2

2a( ) 2ab - a2 - b2 + c2

2a( )

Or ,h2 = (a+b)2 - c2

2a{ } c2 - (a-b)2

2a{ }

Or ,h2 = (a+b+c) (a+b-c)2a{ } (c+a-b) (c-a+b)

2a{ }

Or , h2 = 2s(2s-2c) (2s-2b) (2s- 2a)4a2{ }

Or, h2 = 2 × 2 × 2 ×2× s(s-a) (s-b) (s-c)4a2{ }

Or, h2 = 16s (s-a) (s-b) (s-c)4a2{ }

Or, h2 = 16s(s-a) (s-b) (s-c)4a2

Or, h

= s (s-a) (s-b) (s-c)2

a

Now, Area of ∆ABC = 12 base × height

= 12 × ( BC × AD)

= 12

× a × s(s-a) (s-b) (s-c)a

2

= s(s-a) (s-b) (s-c)

Area of scalene triangle = s(s-a) (s-b) (s-c)Thus, if three sides of a triangle are known, its area can be obtained by using above mentioned formula.


Area of quadrilateralsThe diagonal of a quadrilateral divides the quadrilateral into two triangles. So, sum of these triangles gives the area of quadrilateral.

Now, Area of ∆ ABD = 12 × base × height

= 12 × BD × AQ

Then, Area of ∆ BCD = 12 × base × height

= 12 × BD × PC

Area of quadrilateral ABCD = Area of ∆ ABD + Area of ∆ BCD

= 12

× BD × AQ + 1

2 × BD × PC

= 12 × BD ( AQ + PC )

= 12 × one diagonal (sum of length of perpendiculars drawn on it from the remaining

vertices).Similarly, we can find the area of the following quadrilaterals .

a. Area of trapezium = 12

× height × ( sum of parallel sides )

b. Area of kite = 12 × product of length of diagonals

c. Area of rhombus = 12 × product of length of diagonals

Worked Out Examples

Example 1 : Find the area of the following figures.


Solution: a. Base of ∆ ABC (BC) = 10 cm

Height of ∆ ABC (AD) = 6 cm

Area of ∆ ABC = 12 × base × height

= 12 × 10cm × 6cm

= 30 cm2

b. In right angled ∆ ABC, AB2 + BC2 = AC2 (by Pythagoras theorem) Or, AB2 + 82 = 102

Or, AB2 = 100 – 64 Or, AB = 36 AB = 6 cm

Now, area of ∆ ABC = 12 × base × height

= 12

× 8cm × 6cm

= 24 cm2

c. Three sides of a scalene triangle are, a = 6 cm, b = 5cm and c = 4 cm

∴ s = a + b + c2

= 6 cm + 5 cm + 4cm2

= 7.5 cm

Now, area of ∆ PQR = s(s-a) (s-b) (s-c)

= 7.5 (7.5 - 6) (6.5 - 5) (7.5 - 4)

= 7.5 × 1.5 × 2.5 × 3.5

= 98.4375 = 9.92 cm2


d. Two equal sides of an isosceles triangle ABC (a) = 5 cm and base (b) = 6 cmArea of an isosceles triangle = b

4 4a2 - b2

= 64 4×52 - 62

= = 12cm2

Example 2 : Find the area of the following figures.

Solution:a. Three sides of ∆ BCD are a = 8 cm, b = 6 cm and c = 4 cm

s = a + b + c2 = 8 + 6 + 4

2 = 182 = 9 cm

Area of ∆ BCD = s(s-a) (s-b) (s-c)

= 9 (9-8) (9 - 6) (9 - 4)

= 9 × 1 × 3 × 5

= 11.61 cm2

Now, area of parallelogram ABCD = 2× ∆ BCD ( Since diagonal bisects the parallelogram) = 2 × 11.61 cm2

= 23.22 cm2

Thus, the area of parallelogram ABCD is 23.22 cm2.b. Three sides of ∆ PQR are a = 5 cm, b = 4 cm and c = 2 cm

s = a + b + c2 = 5 + 4 + 2

2 = 5.5 cm

Now area of ∆ PQR = s(s-a) (s-b) (s-c)

= 5.5 (5.5 - 5) (5.5 -4) (5.5 -2)

= 5.5 × 0.5 × 1.5 × 3.5

= 3.8 cm2


Again, three sides of ∆ PSR are, a = 4 cm, b = 3 cm and c = 2 cm

s = a + b + c2 = 4 + 3 + 2

2 = 4.5 cm

Area of ∆ PSR = s(s-a) (s-b) (s-c)

= 4.5 (4.5 - 4) (4.5 - 3) (4.5 - 2)

= 4.5 × 0.5 × 1.5 × 2.5

= 2.9 cm2

Now, area of quadrilateral PQRS = Area of ∆ PQR + Area of ∆ PSR = 3.8 cm2 + 2.9 cm2

= 6.7 cm2

c. Area of right angled triangle ABC = 12 × base × height

= 12

(BC × AB)

= 12

(16 × 12)

= 96 cm2

In right angled triangle ABCAB2 + BC2 = AC2

AC2 = 144 + 256

AC = 400 ∴ AC = 20 cmNow three sides of triangle ADC are a = 13 cm, b = 20 cm and c = 21 cm

s = a + b + c2 = 13 + 20 + 21

2 = 27 cm

Area of triangle ADC = s(s-a) (s-b) (s-c)

= 27 (27 - 13) (27 - 20) (27 - 21)

= 27 × 14 × 7 × 6

= 126 cm2

Now, area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ADC = 96 cm2 + 126 cm2

= 222 cm2

Hence, the area of quadrilateral ABCD is 222 cm2.


Example 3.a. If area of an equilateral triangle is 4 3 cm2, find its perimeter.Solution : Here,

Area of equilateral triangle = 4 3 cm2

Perimeter of triangle = ?We know, Area of an equilateral triangle = 3

4 a2 where a is the length of each side of equilateral triangle. Or, 4 3 = 3

4 a2

Or, a2 = 16 ∴ a = 4 cmNow, perimeter of equilateral triangle = 3 a = 3 × 4 = 12 cmThus, the perimeter of an equilateral triangle is 12 cm.b. The area of an issoscles triangle is 240 cm2 and its base is 20 cm. Find the length

of its equal sides.Solution : Here,

Area of an issoscles triangle = 240 cm2 Length of the base of an issoscles triangle (b) = 20 cm.Length of equal sides of an isosceles triangle(a) = ?We know,Area of an issoscles triangle (A) = b

4 4a2 - b2

Or, 240 = 204 4a2 - 202

Or, 240 × 420

= 4a2 - 202

Or, 48 = 4a2 - 400Squaring both sides, we get 2304 = 4a2 - 400

Or, 2304 + 400 = 4a2

Or, a2 = 27044

Or, a = 26 cmThus, the length of equal sides of an issoscles triangle is 26 cm.


Example 4: If the diagonal of a rhombus are 10 cm and 6 cm respectively. Find the area of the rhombus.Solution: Here,

ABCD is a rhombus having diagonals BD and ACDiagonal BD (d1) = 10 cmDiagonal AC (d2) = 6 cm

Area of rhombus = 12

product of diagonals

= 12

d1 × d2

= 12

10cm × 6cm

= 30 cm2

Thus, the area of rhombus is 30 cm2.

Example 5 : If the ratio of the sides of a triangle is 5:7:8 and perimeter is 80 cm, find the area of triangle.Solution: Here,

Let three sides of triangle be 5x cm, 7x cm, and 8x cm respectively.Perimeter of triangle = 80 cmor, 5x +7x + 8x = 80or, 20x = 80

or, x = 8020

= 4 cmNow, 5x = 5 × 4 = 20 cm 7x= 7 × 4 = 28 cm8x = 8 × 4 = 32 cm

Here, s = Perimeter2 = 40 cm

Now, a = 20 cm, b = 28 cm and c = 32 cm

Area of triangle = s(s-a) (s-b) (s-c)

= 40 (40 - 20) (40 - 28) (40 - 32)

= 40 × 20 × 12 × 8

= 76800


= 277.12 cm2

Thus, the area of triangle is 277.12 cm2.

Example 6 : The perimeter of a right angled triangle is 24 cm and its area is 24 cm2. Find the sides of triangle.Solution: Here,Let us take a right angled triangle ABC having a right angle at B.Let AC = b (hypotenuse)

BC = a (base)AB = c (perpendicular)

Perimeter of triangle ABC = 24 cmor, a + b + c = 24 cmor, a + c = 24 – b cm ……… (1)

Area of triangle ABC = 12

× base × height

or, 24 cm2 = 12

ac

or, ac = 48 …….. (2)Also, b2 = a2 + c2 (by Pythagoras theorem)

b2 = (a + c) 2 - 2ac

or, b2 = (24 - b) 2 - 2 × 48

or, b2 = 576 - 48b + b2 - 96or, 48b = 480

or, b = 10 cm∴ Hypotenuse (b) = 10 cm

∴ The required three sides of triangle are 10 cm, 8 cm and 6 cm.Example 7 : The area of triangle is 24 cm2. If its perimeter is 24 cm and its one side is 10 cm, find the length of remaining sides.Solution :

Let the length of remaining sides of triangle are ‘a’ and ‘b’Then, perimeter of the triangle (s) = 24 cm The length of one side (c) = 10 cm Now, a + b + c = 24

A

BC

Now from equation (1) a + c = 14

or, a + 48a

= 14

or, a2 + 48 = 14a or, a2 - 14a + 48 = 0 or, a2 - 8a - 6a + 48 = 0 or, a(a - 8) (a - 8) = 0 or, (a - 8) (a - 6) = 0

Either a = 8 cmOr, a = 6 cm


Or, a + b = 24 – c Or, a + b = 24 - 10 Or, a + b = 14 Or, b = 14 - a

Also, s = 242 = 12 cm

According to formula,

A = s(s-a) (s-b) (s-c)

Or, 24 = 12(12 -a) (12 - 14 + a) (12 - 10)

Or, 24 = 12 (12 - a) (a - 2) × 2

Or, 24 = 24 (12 - a) (a - 2)

When a = 6 cm then b = 8 cm and when a = 8 cm then b = 6 cm.

Squaring both sides. We get Or, 576 = 24 (12 - a) (a - 2)

Or, 57624

= 12a - 24 - a2 + 2a

Or, 24 = 14a – 24 – a2

Or, a2 – 14a + 48 = 0Or, a2 – 8a – 6a + 48 = 0Or, a(a – 8) – 6(a – 8) = 0Or, (a - 8) (a – 6) = 0Either, a – 8 = 0 or, a – 6 = 0 a = 8 a = 6 a = 8 or 6

Exercise (3.1)

1. a. What is the area of an equilateral triangle whose one side is ‘a’ unit long.

b. Write the formula of the area of an isosceles triangle whose equal side is ‘x’ unit and base side is ‘y’ unit long.

c. If the base and height of a right angled triangle are ‘b’ cm and ‘h’ cm respectively, write down the formula to find the area.

d. If length of three sides of a triangle are ‘a’ cm , ‘b’ cm and ‘c’ cm respectively, write down the formula to find the area.

e. What do ‘a’ and ‘b’ represent in the formula A = b4 4a2 - b2

f. In s(s-a) (s-b) (s - c) , what does ‘s’ mean ?

2. a. The length of each side of an equilateral triangle is ‘a’ unit, then show that the

area of triangle is 34

a2.

b. The length of two equal sides of an issoscles triangle is ‘a’ cm and base is ‘b’

then prove that the area of the triangle is b4 4a2 - b2

.3. Find the area of the following figures.



4. a. Find the area of the triangles whose sides are i. 13 cm , 14 cm , 15 cm ii. 13 cm , 20 cm , 21 cm iii. 12 cm , 16 cm , 20 cm iv. 7 cm , 24 cm , 25 cm 5. a. Find the area of an equilateral triangle having each side 8 cm. b. If the area of an equilateral triangle is 9 3cm2, find each side of triangle. c. Find the perimeter of equilateral triangle whose area is 25 3cm2. d. If the perimeter of an equilateral triangle is 27 cm, find its area. e. The perimeter of a triangle is 14 cm and its two sides are 4 cm and 5 cm. Find the area

of triangle. f. The perimeter and the length of the base of an isosceles triangle are 25 cm and 9 cm

respectively. Calculate the area of the triangle.6. a. Find the area of a parallelogram whose adjacent sides are 10 cm and 8 cm and one of

the diagonal is 12 cm. b. Find the area of a parallelogram whose adjacent sides are 12 cm and 10 cm and one of

the diagonal is 15 cm.7. a. A rhombus shaped piece of cloth has its diagonals 12 cm and 8 cm. Find the area of the

piece. b. A rhombus has each side 10 cm and length of one diagonal 14 cm. Find the area of the

rhombus. 8. a. The area of an isosceles triangle is 110 cm2 and its base is 20 cm, find the measure of its

equal sides. b. Find the equal sides of an issoscles triangle whose area is 12 cm2 and its base is 8 cm.9. a. The perimeter of a triangle is 360 cm. If the ratio of sides is 4:5:3, find the area of

triangle. b. The sides of a triangle are in the ratio of 3:6:8. If its perimeter is 340 cm, find the sum of

the longest and the shortest side.10. a. The perimeter of a triangle is 54 cm. if its area is 126 cm2 and its one of the three side is

21cm, find the length of the remaining sides. b. A triangle has its perimeter 84 cm and area 336 cm2. If one side of the triangle is 30 cm,

find the length of other two sides.11. a. The perimeter of a triangle is 29 cm. The ratio of the two sides is 3:4 and third side is 8

cm. Find the area of triangle. b. The perimeter of a triangle is 24 cm. The ratio of the two sides is 3:5 and third side is 10

cm. Find the area of triangle.12. a. If two sides of a triangle are 7 cm and 11 cm. If the perimeter of triangle is 26 cm, find

the area of triangle. b. If two sides of a triangle are 5 cm and 7 cm. If the perimeter of triangle is 17 cm, find the

area of triangle.


13. a. If each side of an equilateral triangle is increased by 4 cm then its area increase by 6 3 cm2. Calculate the length of each of its sides and its area.

b. If each side of an equilateral triangle is increased by 5 cm then its area increase by 103 cm2. Calculate the length of each of its sides and its area.

14. a. Find the area of figure PQRS. Also find the length of QR.

b. Find the area of given polygon.

15. a. The perimeter of a right angled triangle is 36 cm and hypotenuse is 15 cm. Find its area.

b. The perimeter of a right angled triangle is 60 cm and hypotenuse is 26 cm. Find its area and other two sides.

16. a. The perimeter of a right angled triangle is 12 cm and area is 6 cm2 . Find the sides of the triangle.

b. The perimeter of a right angled triangle is 24 cm and area is 24 cm2 . Find the sides of the triangle.

17. A hand fan is made by stitching 10 equal size triangular strips of two different types of paper. The dimensions of equal strips are 25 cm , 25 cm and 14 cm. Find the area of the hand fan.

18. An umbrella is made by stitching 8 triangular pieces of cloth each measuring 1.5m, 1.5m and 90 cm. Find the cost of cloth used to make an umbrella at the rate of Rs. 675 per sq. metre


Ans

wer

s (3.

1)1.a. 3

4 a2 sq. units b. y4

4x2 - y2 sq.units c. 12

bh sq uints

d. s(s-a) (s-b) (s - c) cm2 e. ‘a’ equal sides and ‘b’ base of an isosceles triangle

f. Semi - perimeter 3.a. 48cm2 b. 24cm2

c. 30cm2 d. 29.76cm2 e. 17.32cm2

f. 9 3cm2 g. 79.96cm2 h.117.2cm2

i. 72.57cm2 j. 32cm2 k. 45cm2

4. a. 84cm2 b.126cm2 c. 96cm2

d. 84cm2 5.a. 16 3 cm2 b.6cm

c. 30cm d. 35.07cm2 e. 9.16cm2

f. 29.76 cm2 6.a.79.37cm2 b. 119.63cm2

7. a. 48cm2 b.99.97cm2 8.a. 14.8 cm

b.5cm 9. a.5400 cm2 b. 220cm

10.a.13 cm and 20 cm b.26 cm and 28 cm 11.a.36 cm2

b. 22.94 cm2 12a. 27.93cm2 b.12.49cm2

13.a. 1cm, 34 cm2 b. 1.5cm, 9 3

16 cm 14.a.60cm2, 7.21cm

b. 6.81 cm2 15.a. 54cm2 b.24cm,10cm, 120cm2

16.a. 3cm ,4cm, 5cm b.a. 10cm ,8 cm, 6 cm 17. 1680 cm2

18. Rs 3477.10

3.2 PrismA solid having opposite faces congruent and parallel and rectangular lateral faces is called prism. The perpendicular distance between the opposite faces is the height of prism. Total surface area of the prism is the sum of area of bases and lateral surface area. Volume of the prism is product of area of base and height.

Triangular prismA triangular prism consists of two triangular bases and three rectangular surfaces. Thus in a triangular prism, there are two congruent and parallel


triangles and three rectangles obtained by joining the corresponding vertices of the triangles at the bases.In the given figure, triangle ABC and triangle DEF are the triangular bases. ADEB, ACFD and BCFE are the three rectangular surfaces. BE = l is the length (height) of the prism. Area of triangular base is also known as area of cross section and area of three rectangular surfaces is known as lateral surface area.Let area of two triangular bases = 2∆

1. lateral surface area = Area of three rectangular surfaces= al + bl + cl= (a + b + c ) l= p × l (where p is perimeter of triangular base)

2. Total surface area of the prism = Area of two triangular bases + lateral surface area = 2∆ + p × l

3. Volume of the prism = Area of triangular base × length (height)ActivityDesign a net of a prism and explore to verify the formulae of the + lateral surface area and the total surface area.

Example 1 : Find lateral surface area, total surface area and volume of the following figures.

Solution:a. In right angled triangular base ABC,

AB2 + BC2 = AC2 (by Pythagoras theorem)Or, 42 + BC2 = 52

Or, BC2 = 25 – 16Or, BC = 9 ∴ BC = 3cmLateral surface area = Perimeter of triangular base × length = (3 cm + 4 cm + 5 cm) × 8 cm

Worked Out Examples


= 12cm × 8 cm = 96 cm2

Area of triangular base ABC = 12 × base × height

= 12 × 3 × 4

= 6 cm2

Total surface area of prism = area of two triangular bases + lateral surface area = 2 × 6cm2 + 96 cm2

= 108 cm2

Volume of prism = area of triangular base × length = 6 cm2 × 8 cm = 48 cm3

Hence, LSA,TSA and volume of prism are 96 cm2, 108 cm2 and 48 cm3 respectively.

b. Base is an equilateral triangle having side (a) = 6 cm

Area of triangular base = 34 × a2

= 34 × 62

= 9 3 cm2

∴ Length of prism (l) = 10 cmLateral surface area = Perimeter of triangular base × length = 3 × 6 cm × 10 cm = 180 cm2

Total surface area = Area of 2 triangular bases + lateral surface area

= 2 × 9 3 cm2 + 180 cm2

= 211.176 cm2

Volume of prism = Area of triangular base × length

= 9 3 cm2 × 10 cm = 155.88 cm3

Hence, LSA,TSA and volume of prism are 180 cm2, 2110176 cm2 and 155.88cm3 respectively.


Example 2 : If the volume of the given triangular prism is 240 cm3, find the value of x.Solution: Here,

Volume of given triangular prism = 240 cm3

Length of prism (l) = 12 cm

Area of triangular base = VolumeLength

= 240 cm3

12 cm

= 20 cm2

Area of triangular base = 12

× base × height

Or, 20 cm2 = 12

× x × 8

Or, x = 204

∴ x = 5 cmThus, the value of x is 5 cm.

Example 3 : The length of the sides of the base of a triangular prism is 12 cm , 16 cm and 20 cm. If its LSA is 768 cm2 , what is its volume ?Solution : Here,The base of the prism is a triangle with sides 12 cm , 16 cm 20 cm. LSA is 768 cm2.

Let ‘h’ be the height of the prism.Lateral surface area = Perimeter of base × height of the prism. 768 = ( 12 + 16 + 20 ) × h

Or, h = 76848

∴h = 16 cmSo, the height of the prism(h) = 16 cmNow, for the volume of the prism.

Semi perimeter (s) = 12 + 16 + 202 = 24 cm

Let the sides of the base be a = 12 cm , b = 16 cm and c = 20 cmNow,Area of base triangle = s(s-a) (s-b) (s - c)

= 24 (24 - 12) (24 - 16) (24 - 20)

= 24 × 12 × 8 × 4


= 96 cm2

Then, volume of the prism = base area × height of the prism = 96 × 16 = 1536 cm3

Thus, the volume of the prism is 1536 cm3.Example 4 : The base of the right- prism is an equilateral triangle having perimeter 24 cm. If the length of the prism is 14 cm. Find its total surface are and volume.Solution : Here,

Perimeter of an equilateral triangle = 24 cmLength of the prism = 14 cmLet each side of an equilateral is ‘a’ cmNow, we knowPerimeter of an equilateral triangle(P) = a + a + a 24 = 3a ∴ a = 8 cmNow,Lateral surface area of prism = Perimeter of base × height of prism = 24 cm × 14 cm = 336 cm2

Area of an equilateral triangle (A) = 34 × a2

= 34 × 82

= 16 3 cm2

Now, Total surface are of the prism (TSA) = 2×Area of base+Lateral surface area

= 2 × 16 3 + 336 = 391.4 cm2

And, Volume of the prism (V) = Area of base × height of prism

= 16 3 × 14 = 387.98 cm3

Hence, total surface area of prism is 391.4 and volume of the prism is 387.98 cm3.


Example 5 : The base of a triangular prism is an equilateral triangle. Its lateral surface area is 120 cm2 and volume is 20 3 cm3. Find the height and side of the base of the prism.Solution : Here,

Lateral surface area(LSA) = 120 cm2

Volume of prism(V) = is 20 3 cm3

Let each side of an equilateral triangle is ’a’ cm and height be ‘h’ cm.Now we knowLateral surface are (LSA) = Perimeter × height 120 = 3a × h

Or, h = 1203a …………….(I)

Again, We knowVolume of the prism = Area of triangular base × height

20 3 = 34 × a2 × h

Or, 20 3 = 34 × a2 × 120

3a (Putting the value of h from (I) ) Or, a = 2 ∴ Each side of an equilateral triangle (a) = 2 cmPutting the value of a in equation (i), we get

h = 1203×2 = 20 cm

Thus, each side of an equilateral triangle is 2 cm and height of prism is 20 cm.Example 6 : Find the lateral surface area and total surface area of a right prism of volume 240 cm3 and base is a right angled triangle of sides 8 cm and 6 cm contain-ing right angle.Solution : Here,

Volume of the prism(V) = 240 cm3

Two sides of a right angled triangle containing right angle are p= 8 cm and b= 6 cm. Let the third side of the right angled triangle is ‘h’ We know, h2 = p2 + b2

= 82 + 62

= 64 + 36 = 100 ∴ h = 10 cm


Now, Area of triangular base (A) = 12 × 8 × 6 = 24 cm2

Volume of the prism (V) = Area of triangular base × height of the prism 240 = 24 × h

Or, h = 24024 = 10 cm

Height of the prism (h)= 10 cmThen, Lateral surface area (LSA) = Perimeter of base × height = ( 10 + 8 + 6 ) × 10 = 24 × 10 = 240 cm2

And total surface area of the prism (TSA) = 2 × Area of base + LSA = 2 × 24 +240 = 48 + 240 = 288 cm2

Hence, Lateral surface area (LSA) is 240 cm2 and total surface area of the prism (TSA) is 288 cm2

1. a. Define the term triangular prism. b. What is cross section part of the prism ? c. Mention the relation of cross section and base of the prism. d. Mention any one charecteristics of a prism. e. Write the formula of : i. Lateral surface area of triangular prism. ii. Total surface area of triangular prism. iii. Volume of the prism.2. a. Find the volume of a prism whose area of base is ‘A’ and height ‘h’. b. Find the volume of a prism whose area of base is y cm2 and height is x . c. Find the volume of the equilateral triangle base prism whose length of base

side is ‘a’ and height of prism ‘h’. d. Find the volume of the prism whose height of the prism is ‘h’ and sides of the

base triangle is p cm, b cm q cm.

Exercise (3.2)


e. What is the lateral surface area of a triangular prism having sides of base are a cm, b cm c cm and length of the prism is d cm.

3. a. The length of the sides of the base of a triangular prism is 6 cm , 8 cm and 10 cm .

If its LSA is 384 cm2 , what is its volume ?

b. The length of the sides of the base of a triangular prism is 3 cm , 4 cm and 5 cm

If its LSA is 192 cm2 , what is its volume ?

4. a. The base of the right- prism is an equilateral triangle having perimeter 12 cm. If the length of the prism is 7 cm. Find its total surface are and volume.

b. The base of the right- prism is an equilateral triangle having perimeter 18 cm. If the length of the prism is 8 cm. Find its total surface are and volume.

5. a. The base of a triangular prism is an equilateral triangle. Its lateral surface area is 240 cm2 and volume is 40 3 cm3 . Find the height and side of the base of the prism.

b. The base of a triangular prism is an equilateral triangle. Its lateral surface area is 360 cm2 and volume is 60 3 cm3 . Find the height and side of the base of the prism.

6. a. Find the lateral surface are and total surface area of a right prism of volume 120 cm3 and base is a right angled triangle of sides 4 cm and 3 cm containing right angle.

b. Find the lateral surface are and total surface area of a right prism of volume 360 cm3 and base is a right angled triangle of sides 12 cm and 9 cm containing right angle.

7. Find the area of cross section, lateral surface area, total surface area and volume of the following prisms


8. a. If the height and perimeter of the given triangular prism are equal, find the volume of prism.

b. If the volume of the given prism is 1800 cm3 find the length of prism.

c. The volume of the given triangular prism is 255 cm3. Find the value of x.

9. a. A prism 20 cm high, has the base an equilateral triangle of side 10 cm. Find its volume


b. A prism 30 cm high, has the base an equilateral triangle of side 15 cm. Find its volume

10. a. If the base of a prism is an equilateral triangle of side 8 cm and its volume is 144 3cm3, find the height of the prism.

b. If the base of a prism is an equilateral triangle of side 12 cm and its volume is 216 3cm3, find the height of the prism.

11. a. The ratio of sides of a triangular prism is 3:4:5. If lateral surface area and height of the prism are 432 cm2 and 18 cm respectively, find the area of the base.

b. The ratio of sides of a triangular prism is 3:3:4. If lateral surface area and height of the prism are 120 cm2 and 4 cm respectively, find the area of the base.

12. a. The area of a rectangular faces of a triangular prism is 540 cm2. If the ratio of perimeter of base height is 5:3. Find the height of the prism.

b. The area of rectangular faces of a triangular prism is 960cm2. If the ratio of perimeter of base and height is 5:3, find the perimeter of base and height of the prism.

13. a. The area of cross section of a triangular prism is 63 cm2 and its volume is 1890 cm3. If the perimeter of triangular base is 24 cm, find

i. Lateral surface area

ii. Total surface area

b. The area of cross section of a triangular prism is 252 cm2 and its volume is 7560 cm3. If the perimeter of triangular base is 96 cm, find

i. Lateral surface area

ii. Total surface area

14. a. The base of a prism is an equilateral triangle. If its lateral surface area is 240cm2 and the volume is 80 3 cm3, find its height and length of each side.

b. The length of equal sides of an isosceles right angled triangular prism having volume 2000cm3 is 20cm, find the height of prism.

c. The base of a prism is an equilateral triangle. If its lateral surface area is 120 cm2 and the volume is 40 3cm3, find its height and the length of each side.

15. The base of the given triangular prism is a right angled issoscles triangle. If the volume and length of the prism are 750 cm3 and 15 cm respectively.


Find

i. Lateral surface area of the prism.

ii. Total surface area of the prism.

Ans

wer

s (3.

2)

2a. Ah cu.units b.xycm3 c. 34 a2hcm3

d. s(s-p) (s-b) (s - q) h cm3 where s = p + b+ q2

e. (a+b+c)dcm2

e.i. ph where p = perimeter of base triangle

ii. 2∆+ph where ∆ is area of base triangle iii.area of base triangle ×height

3a. 384cm3 b. 96cm3 4a. 97.86cm2 ,28 3cm3

b. 175.18cm2, 72 3cm3 5a. a = 2cm, h = 40cm

b.a = 2cm, h=60cm 6a.240cm2, 252cm2 b.240cm2, 348cm2

7a.6cm2, 180cm2, 192cm2, 90cm3

b. 5.33cm2, 130cm2, 140.66cm2,53.3cm3 c. 24cm2,288cm2,336cm2, 288cm3

d. 9 3cm2, 270cm2, 301.18 cm2, 233.83cm3

e. 14.98cm2, 493.63cm2, 523.59cm2, 389.19cm3

f. 96cm2 , 2160cm2 , 2352cm2 , 4320cm3 8.a.1299.03cm3

b. 30cm c.5.32cm 9a.866.02cm3

b. 2922.83cm3 10a. 9cm b. 6cm

11a. 24cm2 b. 40.24cm2 12a.18cm

b. 40cm, 24cm 13a.i. 720cm2 ii. 846cm2

b.i. 2880cm2 ii. 3384cm2

14a. a = 4cm, h=20cm b. 10cm c.10cm, 4cm

15.i. 512.13cm2 ii. 612.13cm2


3.3 CylinderCylinder is a prism consisting of two parallel congruent circular bases and curved surface area. It is also known as circular prism.

Let us take a rectangular piece of paper having length l and breadth b and bend it as shown in the figure. We will get a figure having two circular faces and a curved sur-face. It is a cylinder.From the above figure,Curved surface of cylinder = circumference of circular base × height = 2 pr × h = 2 prhWhere ‘r’ is radius of circular base and ‘h’ is height of the cylinderTotal surface area = Area of two circular bases + Curved surface area = 2× pr2+ 2prh = 2pr(r + h)And, volume of cylinder = Area of circular base × height = pr2 × h = pr2 hVolume of material contained by a hollow cylinderThe figure shows a hollow cylinder like a pipe, flute etc. Let the external radius = R Internal radius = r Height = h External volume of cylinder = pR2h Internal volume of cylinder = pr2hVolume of material contained by cylinder = External volume – Internal volume = pR2h - pr2h = ph (R2 – r2) = ph (R + r) (R - r)

R

r

h


Example 1 : Find curved surface area, total surface area and volume of the following cylinders.

a. Solution: Here, Diameter of circular base (d) = 21 cm

Radius of circular base (r) = 212 = 10.5 cm

Height of cylinder (h) = 15 cm Curved surface area = 2prh = 2 × 3.14 × 10.5 × 15cm2

= 989.1 cm2

∴ Total surface area = 2pr(r + h) = 2 × 3.14 × 10.5 × (10.5 + 15) cm2

= 1681.47 cm2

∴ Volume = pr2h = 3.14 × 10.5 × 10.5 × 15cm3

= 5192.17 cm3

Thus, curved surface area of cylinder is 989.1 cm2, total surface area is 1681.47cm2 and volume is 5192.17 cm3

b. Solution :Here,Circumference of circular base C= 154 cmOr, 2pr = 154

Or, 2 × 227 × r = 154

Or, r = 154 × 744

∴ r = 24.5 cm

Worked Out Examples


Now, curved surface area = 2prh

= 2 × 227 × 24.5 × 12 cm2

= 1848 cm2

Total surface area of cylinder = 2pr(r + h)

= 2 × 227 × 24.5 × (24.5 + 12)cm2

= 5621 cm2

Volume of cylinder = pr2h

= 227

× 24.5 × 24.5 × 12 cm3

= 22638 cm3

Thus, curved surface area of cylinder is 1848 cm2, total surface area is 5621 cm2 and volume is 22638 cm3.

Example 2 : The volume of cylindrical can is 1.54 liters. If the height is 20 cm, find its area of base and radius.Solution: Here,

Volume of cylindrical can (V) = 1.54 liters = 1.54 × 1000cm3

= 1540 cm3

Height of can (h) = 20 cmVolume = 1540 cm3

Or, pr2h = 1540Or, pr2 × 20 = 1540

Or, pr2 = 154020

= 77Therefore, area of circular base is 77 cm2.Again pr2 = 77

Or, 227

× r2 = 77

Or, r2 = 77 × 722

Or, r2 = 24.5Or, r = 24.5∴ r = 4.95 cmThus, radius of circular base (r) is 4.95 cm.


Example 3 : The height and radius of a cylinder are equal and its curved suface area is 4928 cm2. Find the circumference of the base of the cylinder.Solution : Here,Let r and h be the radius and height of the cylinder respectively such that r = hCurved surface area of the cylinder (CSA) = 4928 cm2

We have,CSA of cylinder = 2prh

Or, 4928 = 2 × 227 × r × r

Or, r2 = 784 ∴ r = 28 cmNow,Circumference of the base = 2pr

= 2 × 227 × 28

= 176 cmThus, the circumference of the base of the cylinder is 176 cm.Example 4 : The external and internal radii of a cylindrical vessel 56 cm high are 4.5 cm and 3 cm respectively. Find the volume of the material contained by the vessel.Solution: Here,

External radius of vessel (R) = 4.5 cmInternal radius of vessel (r) = 3 cmHeight of vessel (h) = 56 cm

Volume of material contained by cylinder = External volume – Internal volume = pR2h - pr2h = ph (R2 – r2)

= 227

× 56 {(4.5) 2 – (3)2}

= 22 × 8(20.25 – 9) = 1980 cm3

Thus, the volume of the material contained by the vessel is 1980 cm3.

Example 5 : The volume and curved surface area of a cylindrical wooden log are 1078 cm3 and 308 cm2 respectively. Find the radius of base and height of the log.

Solution: Here,


Volume of log = 1078 cm3

Or, pr2h = 1078 cm3 ………. (1)Where r is radius of circular base and h is height of log.

Curved surface area of log = 308 cm2

Or, 2prh = 308 ………… (2) Dividing equation 1 by 2

pr2h2prh

= 1078308

Or, r = 3.5 cmTherefore, radius of base = 7 cmPutting value of r in equation 1, we get

227 × 7 × 7 × h = 1078

Or, h = 107822 × 7 ∴ h = 7cmTherefore, height of the log is 7 cm.

Example 6: The adjoining figure is a half portion of a solid cylinder obtained by splitting vertically through its height. Find the curved surface area, total surface area and the volume of the solid.Solution: Radius of semicircular base (r) = 14 cmHeight of solid (h) = 56 cm

Curved surface area = 2prh2

= 227 × 14 × 56 cm2

= 2464 cm2

Total surface area = Area of two semicircular base + curved surface area + area of rectangular surface

= 2pr2

2 + prh + hd where d is diameter of circle

= 227 × 14 × 14 + 2464 + 56 × (2 × 14)

= 4648 cm2

h

hr =14cm

56cm

r

r


Volume of solid = pr2h2

= 72

22× 14 × 14 × 56

= 17248 cm3

Thus, curved surface area of solid is 2464 cm2, total surface area is 4648 cm2 and volume is 17248 cm3.

1. a. What is a cylinder ?

b. What is right circular cylinder?

c. Write the formula of the cylinder for

i. Cross section area ii. Curved surface area iii. Total surface area iv. Volume

d. If the base area of a cylinder is ‘a’ cm2 and height is ‘b’ cm, then find the volume of the cylinder.

e. What is the total surface area of a cylinder whose radius is ‘x’ cm and height is ‘y’ cm ?

f. What is the volume of the cylinder whose height is equal to diameter of the base.

2. Find curved surface area, total surface area and volume of the following cylinders.

Exercise (3.3)

12 cm

7cm

14cm

20cm

10cm

16cm

d. e. f.

3. a. The radius of the cylinder is 21 cm. If the height of the cylinder is 21 cm , find

i. Curved surface are ii. Total surface are iii. Volume

b. The radius of circular base of a cylinder 8 cm high is 7 cm, Calculate


i. Base area ii. Curved surface area iii. Total surface area iv. Volume

4. a. The volume of a cylinder is 1550 cm3 and area of circular base is 155 cm2, find the height of the cylinder.

b. The volume of the cylinder is 1760 cm3 and height of the cylinder is 20 cm, find the area of circular base.

c. The circumference of the base of the cylindrical drum is 88 cm. If the sum of its radius and the height is 25 cm, find total surface area.

d. The perimeter of the base of a cylinder is 44 cm and height is 15 cm. Find the total surface area.

e. A cylindrical drum is 70 cm high. If it can hold 49.5 l of water, find the curved surface area of the drum.

f. The volume of a cylindrical can is 1.54 l. If the area of the base is 77 cm2, find its height.

5. a. A solid cylindrical wooden log of length is 5 cm is cut into two halves along the length. Find the total surface area of each part if diameter of log is 1.4 cm.

b. A solid cylindrical wooden log of length 20 cm and diameter 14 cm is cut into two halves along the length. Find the surface area of each part.

6. a. The radius and the height of the cylinder are in the ratio of 5:7. If the volume of the cylinder is 550cm3, find its height.

b. The radius and the height of the cylinder are in the ratio of 7: 9. If the volume of the cylinder is 693 cm3, find radius of the cylinder.

7. a. The curved surface area of a cylinder is 23

of its total surface area. If the total

surface area is 1212 cm2, find its radius and height.

b. The area of curved surface of a solid cylinder is equal to 13

of the total surface area

of the same cylinder. If total surface area is 462 cm2, find the volume of the cylinder.

8. a. A hollow cylindrical plastic pipe has inner radius 2.5 cm and thickness 0.5 mm. Find the volume of plastic in pipe if the length of the pipe is 21 cm.

b. An iron pipe 105 cm long has an internal diameter of 5 cm. The thickness of iron is 1 cm . Find the volume of the iron in the pipe.

9. a. The volume and curved surface area of a cylinder are 770 cm3 and 440cm2 respectively. Find the radius and height of the cylinder.

b. The curved surface area and volume of a cylindrical pole are 264 m2 and 924 m3 respectively. Find the diameter and height of the pole.

10. a. The sum of height and radius of the base of a cylinder is 34 cm and total surface area is 2992 cm2. Find the radius of base.


b. The curved surface area of the cylinder whose height is equal to radius of the base is 1232 cm2. Find the total surface area of the cylinder.

11. a. The volume of the cylinder is half of its total surface area. If the radius of its base is 7 cm, find the volume.

b. The volume of the cylinder is half of its total surface are. If the diameter of its base is 28 cm, find the volume.

12. a. 100 circular plates, each of radius 7 cm and thickness 0.5 cm, are placed one above the other to form a right cylinder. Find total surface area and volume of the cylinder so formed.

b. 50 circular plates, each of radius 7 cm and thickness 0.5 cm, are placed one above other to form a solid right cylinder. Find i. Total surface area and ii. Volume of the cylinder so formed.

13. a. The external and internal diameter of hollow tube made up of iron is 20 cm and 18 cm. If its length is 16 cm, find the volume of iron.

b. The external and internal diameter of hollow tube made up of iron is 22 cm and 20 cm. If its length is 20 cm, find the volume of iron.

14. The ratio between the curved surface area and total surface area of right circular cylinder is 1 : 2. If total surface area is 616 cm2. Find the volume of the cylinder.

15. A water storage tank is made by using 50 circular rings having diameter 6 feet and height 1.5 feet each. How many cubic feet of water can be stored in the tank ? What is the cost of making the tank at the rate of Rs 2500 per ring ? If 1 cubic foot is equivalent to 28.317 litres, find the capacity of the tank in litres.

Ans

wer

s (3.3

)

1c.i. πr2 ii.2πrh iii.2πr(r+h) iv.πr2h d.abcm3 e.2πx(x+y)cm2

f. πd3

4 2a. 440cm2, 748cm2 , 1540cm3

b. 2310cm2, 3003cm2, 12127.5cm3 c. 616cm2, 693cm2, 1078cm3 d. 132cm2, 254.5cm2, 231cm3 e. 440cm2, 874cm2, 1540cm3 f. 251.43cm2, 612.57cm2, 1005.71cm3 3a.i.2772cm2, 5544cm2, 29106cm3

b.i.201.14cm2 ii.352cm2 iii. 754.28cm2 iv. 1408cm3

4a. 10cm b. 88cm2 c.2200cm2 d. 968cm2 e. 6600cm2 f. 20cm 5a. 19.54cm2 b. 874cm2

6a.7cm b. 5.5cm 7a. 8.01cm, 16.03cm b. 539cm3 8a.16.665cm3 b. 1980cm3 9a. 3.5cm, 20cm b. 14m , 6m10.a. 14 cm b 2464cm2 11a. 179.67cm3 b. 663.38 cm3 12a. 2508cm2 , 7700cm3 b.1408cm2, 3850cm3 13.a.955.43 cm3 b. 1320cm3 14. 1078cm3 15. 2121.43cu.ft., Rs.125000, 60072.53 litres


3.4 SphereSphere is a solid object having smooth curved surface so that each and ev-ery point on its surface is equidistant from a fixed point inside it. The fixed point is called center of the sphere and the distance between center and any point on the surface is called radius of sphere. Globe, volleyball, shot put are some examples of sphere.The figure shown alongside is a sphere. O is the center and P is the point on the surface. OP is the radius of sphere.

When a sphere is cut along its diameter, two half spheres will be formed. The half spheres are called hemisphere. The cross section of hemisphere is called the great circle. The radius of the sphere and that of the great circle is same.Area and volume of a sphereLet us take a hollow tennis or rubber ball and cut off it into two equal halves. Again cut off each into two equal halves and continue the process so that they can be flattened and spread on the square graph paper. With the help of graph paper we can find the approximate area of each half. By adding the area occupied by two halves on the graph, we can get the total surface area of the sphere.By an experiment it is known that if we divide the area obtained above by the square

of diameter of same sphere, the result obtained is approximately equal to 3.14 or 227 or p .

So, Surface area of sphere(Diameter)2

= p

Or, Ad2 = p where A is the surface area of sphere and is

its diameter. Or, A = pd2

Or, A = p(2r)2

Thus, surface area of sphere (A) = 4pr2

To find the volume of a sphere, we find diameter of sphere first.

PO


Let us place a sphere on a meter scale and press it from two sides by two wooden piece as shown in the figure.The diameter of the sphere can be read on the scale. Let it be 7 cm.Fill up the measuring cylinder with water to some level and mark it. Immerse the above sphere into the water. The water level rises up. Read the water level. The difference of two levels is the volume of sphere.We will get the volume of above sphere with 7 cm diameter equal to 179.5 cm3.Let us take a ratio

6νd3

= 6 × 179.573 = 3.14 = p

Or, 6νd3 = p

Or, 6v = pd3

Or, v = 16

pd3

Or, v = 16

p(2r)3

Or, v = 43 pr3

∴Volume of a sphere = 43 pr3 or 1

6 pd3

Area and volume of hemisphereHemisphere is half of sphere. It has a plane circular surfaceand a curved surface. The radius of the hemisphere is equal to the radius of the sphere.Curved surface area of hemisphere = 1

2 × 4pr2

= 2pr2

Volume of a hemisphere = 12 × volume of a sphere

= 12 × 4

3 pr3

= 23

pr3 or 112 pd3

T.S.A. of hemisphere = 3pr2

Volume of material contained by a hollow sphereVolleyball, football, basketball are hollow spheresLet external radius of a hollow sphere be Rand internal radius be r.

O Rr


Example 1 : Find surface area and volume of the given spheres.

a. Solution : Here,Diameter of sphere (d) = 14 cm

Radius of circle (r) = d2

= 7 cm

Surface area of sphere = 4pr2

= 4 × 227 × 7 × 7cm2

= 616 cm2

Volume of sphere = 43 pr3

= 43 ×

227 × 7 × 7 × 7cm3

= 1437.33 cm3

Thus, the surface area of sphere is 616 cm2 and volume is 1437.33 cm3

b. Solution : Here, Circumference of great circle of sphere = 44 cm

Worked Out Examples

d = 14 cmc = 44 cm

External volume = 43 pR3

Internal volume = 43 pr3

Volume of material contained by the hollow sphere = External volume – Internal

volume = 43 pR3- 4

3 pr3 = 43 p(R3 – r3)


C = 44

Or, 2pr = 44

Or, 2 × 227 × r = 44

Or, r = 44 × 7 2 × 22 = 7

∴ Radius of circle = 7 cm

Surface area of sphere = 4pr2

= 4 × 227 × 7 × 7 cm2

= 616 cm2

Volume of sphere = 43 pr3

= 43 ×

227 × 7 × 7 × 7cm3

= 1437.33 cm3

Thus, the surface area of sphere is 616 cm2 and volume is 1437.33 cm3.Example 2: a. Find the total surface area of the hemisphere having radius 7 cm.Solution : Here, Radius of hemisphere (r) = 7 cmTotal surface area of the hemisphere (TSA) = ?We know,

Total surface area of the hemisphere (TSA) = 3pr2

= 3 × 227 × 72

= 462 cm2

Thus, total surface area of hemisphere is 462 cm2.b. The total surface area of sphere is 22176 cm2. Find the diameter of the sphere.Solution : Here,The total surface area of sphere (TSA) = 22176 cm2

Diameter of the sphere (d) = ?We know,The total surface area of sphere (TSA) = 22176 cm2


Or, 4pr2 = 22176

Or, 4 × 227 × r2 = 22176

Or, r2 = 22176 × 788

Or, r2 = 1764 Or, r = 42 cm

Now, diameter of sphere = 2r = 2 × 42 cm = 84 cmThus, the diameter of the sphere is 84 cm.

c. The volume of the solid sphere is 9p2

cm3. Find the surface area of the solid.Solution : Here,

The volume of the solid sphere (V) = 9p2

cm3

Total surface area (TSA) = ?We know,

The volume of the solid sphere (V) = 9p2

Or, 4pr3

3 = 9p

2

Or, r3 = 278

Or, r = 32

∴ r = 32

cm

Now, surface area of sphere = 4pr2

= 4 × 277 × 3

2( )2

= 34.71 cm2

Thus, surface area of sphere is 34.71 cm2.Example 3 : Find the curved surface area, total surface area and volume of the following hemispheres.


a. Solution : Here, Diameter (d) = 21 cm

Radius (r) = d2 = 10.5 cm

Curved surface area = 2pr2

= 2 × 227 × (10.5)2

= 693 cm2

Total surface area of hemisphere (TSA) = 3pr2 = 3 × 227 × (10.5)2 = 1039.5 cm2

Volume = 23 pr3

= 23

× 227 × 10.5 × 10.5 × 10.5cm3

= 2425.5 cm3

Hence, curved surface area is 693 cm2, total surface area is 1039.5 cm2 and volume is 2425.5 cm3.

b. Solution : Here,Circumference of great circle = 88 cm

Or, C = 88

Or, 2pr = 88

Or, 2 × 227 × r = 88

Or, r = 88 × 72 × 22 = 14

∴ Radius is 14 cm.

Now, Curved surface area = 2pr2

= 2 × 227 × 14 × 14 cm2

= 1232 cm2

a. b.c = 88 cm

d = 21 cm


Total surface area of hemisphere (TSA) = 3pr2 = 3 × 227 × (14)2 = 1848 cm2

Volume of sphere = 23

pr3

= 23 ×

227 × 14 × 14 × 14cm3

= 5749.33 cm3

Hence, curved surface area is 1232 cm2, total surface area is 1848 cm2 and volume is 5749.33 cm3.

Example 4 : The total surface area of a solid sphere is 616 cm2. It is cut into two equal halves. Find the total surface area of each hemisphere so formed.

Solution: Here, Total surface area = 616 cm2

Or, 4pr2 = 616

Or, 4 × 227 × r2 = 616

Or, r = 616 × 7 88 = 7

Radius of sphere = 7 cm.

Now, total surface area of each hemisphere = 3pr2

= 3 × 227 ×7 × 7 cm2

= 462 cm2

Hence, total surface area of each hemisphere is 462 cm2.Example 5 : Three metallic spheres of radii 3 cm, 4 cm and 5 cm are melted and a single sphere is formed. Find the surface area of new sphere.

Solution: Here , Radius of 1st sphere (r1) = 3 cm

Volume of 1st sphere (v1) = 43 p(r1)

3

Radius of 2nd sphere (r2) = 4 cm

Volume of 2nd sphere (v2) = 43 p(r2)

3

Radius of 3rd sphere (r3) = 5 cm

Volume of 3rd sphere (v3) = 43 p(r3)

3


Let, R and V be radius and volume of the new sphereWe know that,Volume of new sphere = sum of volume of three spheres

Or, 43 p(R) 3 = 4

3 p(r1)3 + 4

3 p (r2)3 + 4

3 p(r3)3

Or, 43 p(R) 3 = 4

3 p( r13 + r2

3 + r33)

Or, R3 = 33 + 43 + 53

Or, R3 = 216∴ R = 6 cm

Surface area of new sphere = 4 pR2

= 4 × 227

× 6 × 6cm2

= 452.7 cm2

Hence, surface area of new sphere is 452.7 cm2.Example 6 : The diameter of a solid spherical metallic ball is 12 cm . If it is melted and drawn into a cylindrical wire of length 288 cm, find the thickness of wire.

Solution: Here,Diameter of metallic ball = 12 cm

Radius of ball = d2 = 12

2 = 6 cm

Volume of metal in the ball = 43

pr3

= 43

p× 6 × 6 × 6cm3

= 288 pcm3

Volume of cylindrical wire = volume of metal in the ball

Volume of wire = 288 pcm3

Or, pr2h = 288 p,where h is the length of wireOr, r2 × 288 = 288Or, r2 = 1Or, r = 1 cmThickness of wire = 2r =2 cm Hence, the thickness of the wire is 2 cm.


1. a. What is sphere ?

b. What is hemisphere ?

c. Find the formula of :

i. Curved surface area of hemisphere

ii. Total surface area of sphere.

iii. Volume of sphere.

2. a. What is the total surface area of a sphere of radius 1 cm?

b. What is the total surface area of a sphere of diameter ‘y’ cm?

c. What is the volume of sphere of radius ‘x’ cm?

d. What is the volume of sphere of diameter ‘y’ cm?

e. What is the diameter of hemisphere having total surface area ‘p2’ cm2?

3. a. Find surface area and volume of the sphere having the following radii

i. 7 cm ii. 14 cm iii. 21 cm iv. 21.5 cm

b. Find circumference and area of great circle of the sphere having the following radii.

i. 7 cm ii. 14 cm iii. 21 cm iv. 21.5 cm

4. Find i. curved surface area ii. total surface area iii. volume of following solid hemisphere.

5. a. Find the total surface area of hemisphere if its volume is 19,404 cm3.

b. Find the circumference of sphere whose volume is 137521 cm3.

6. a. If the radius of a sphere is 21 cm, find its surface area and volume.

b. If the diameter of a solid sphere is 42 cm, find its surface area and volume.

c. Find total surface area and volume of a sphere having circumference of great circle 88 cm.

Exercise (3.4)

i. ii. iii.

c = 44cm

14 cm42 cm


d. The surface area of sphere is 36p cm2, find its diameter.

e. If volume of a sphere is 38808 cm3,find its radius.

f. The volume of a marble is p6 cm3, find its diameter.

7. a. Find the circumference of the base of a hemisphere whose volume is 486 pcm3.

b. The total surface area of the hemisphere is 1848 cm2, find its radius.

c. Find the diameter of a hemisphere whose volume is 6174 p cm3.

d. Find the curved surface area of a hemisphere whose total surface area is 147 p cm2.

8. a. The surface area of sphere is p cm2. If its radius is doubled, by how much does its surface area increase?

b. The surface area of sphere is 4 pcm2. If its radius is doubled, by how much does its surface area increase?

c. The surface area of sphere is p sq. cm. If its radius is doubled, by how much percent does its surface area increase?

9. a. If the radius of the sphere is made thrice, by how many times does its volume increase?

b. If the radius of the sphere is reduced to one-third, by how many times does its volume decrease?

10. a. A sphere is formed by joining two hemispheres of total surface area is 231 cm2 each. Find the surface area of sphere.

b. A sphere is formed by joining two hemispheres of total surface area is 462 cm2 each. Find the surface area of sphere.

c. A solid metallic sphere of radius 14 cm is cut into two halves . Find the total surface area of the two solid hemispheres so formed.

d. A solid metallic sphere of radius 5.6 cm is cut into two halves . Find the total surface area of the two solid hemispheres so formed.

11. a. Three iron spheres having radii 3 cm, 4cm and 5 cm respectively are melted to form a new sphere. Find the diameter of sphere so formed.

b. Three iron spheres having radii 6 cm, 8cm and 10 cm respectively are melted to form a new sphere. Find the diameter of sphere so formed.

12. a. The surface area of a sphere is equal to curved surface area of a cylinder. If the height and radius of the cylinder are 24 cm and 12 cm respectively, find the volume of sphere.

b. A hemispherical bowl of internal diameter 18 cm is full of water. This water is poured into cylindrical bottles of radius is 1.5 cm and height 4 cm. Find the number of bottles to fill the whole water of the bowl.


13. a. A solid metallic sphere of radius 3.5 cm is melted and drawn into a cylindrical wire of thickness 0.2 cm. Find the length of wire.

b. How many solid spheres each 3 cm radius can be made from a solid metallic cylinder of a diameter 4 cm and height 36 cm.

c. Solid metal sphere of radius 21 cm is dropped into a cylindrical jar partly filled with water. If the radius of the drum is 1.4 m, how much will the surface of water be raised?

14. a. If the radius of a sphere is doubled, what will be the percentage change in volume?

b. The ratio of the surface area of two sphere is 16:9. Find the ratio of their volume.

c. Find the difference in a. volume b. surface area of a sphere, if its diameter is made one – third of its original size.

Ans

wer

s (3.

4)

1c.i.2πr2 ii. 4πr2 iii. 43

πr3 or 16

πd3 2a.4πcm2

b. πy2cm2 c. 43

πx3cm3 d. 16

πy3cm3 e. 2p3π

cm

3a.i. 616 cm2, 1437.33 cm3 ii. 2464cm2, 11498.67cm3,

iii. 5544cm2, 38808cm3, iv. 5811.14cm2, 41646.52cm3

b.i. 44cm, 154cm2 ii. 88cm, 616cm2

iii. 132cm, 1386cm2 iv. 135.14cm, 1452.78cm2

4.i. 1232 cm2, 1848 cm2, 5749.33 cm3 ii. 2772 cm2, 4158 cm2, 19404 cm3

iii. 308 cm2, 462 cm2, 718.67 cm3

5a. 4158cm2 b. 15.71cm 6a. 5544cm2, 33808cm3

b. 5544cm2, 33808cm3 c. 2464cm2, 11498.67cm3

d. 6cm e. 21cm f. 1cm 7a. 56.57cm

b.14cm c. 42cm d. 308 cm2 8a. 3πcm2

b. 12πcm2 c. 300% 9a. 27 times b. 27times

10a. 308cm2 b. 616cm2 c. 3696cm2 d. 591.36cm2

11a. 12cm b. 24cm 12a. 7241.14cm3 b.54

13.a. 57.17m b. 4 c. 6.3mm 14. a.700%

b. 64 : 27 c. 26162

πd3, 89

πd2


Square base pyramidA square based pyramid is shown in the figure.ABCD is a square base of each side ‘a’. V is the vertex of pyramid. Triangle ΔVAB, ∆VBC, ∆VDC and ∆ VAD are four congruent triangular faces.The perpendicular drawn from the vertex to the base is the height of pyramid.Here, VO = h is the height. The perpendicular drawn from the vertex to opposite side of each triangular face is the slant height. VP = l is the slant height.Lateral surface area = Sum of area of triangular facesSo, Lateral surface area of square based pyramid = Area of 4 triangular faces

= 4 × 12 (base × height)

= 4 × 12

× al

= 2al Area of square base = a2

∴ Total surface area = Area of square base + Lateral surface area

3.5 Pyramid A pyramid is a solid with polygonal base and triangular faces with common vertex. Line drawn from vertex to center of the base is called the height of the pyramid. When the height is perpendicular to the base, the pyramid is called right pyramid. The perpendicular drawn from the vertex to any side of its base is called the slant height for the face consisting that side.The polygonal base of pyramid may be a triangular, rectangular, square, pentagonal, hexagonal etc.Here, we discuss about the total surface area, area of triangular surfaces and volume of square based, rectangular based and equilateral triangular based pyramid.

slant height

height

vertext (apex)

face

base


= a2 + 2al∴ Total surface area = a2 + 2al

∴ Volume of pyramid = 13 × Area of square base × height

= 13 × a2 × h

∴ Volume of square base pyramid = 13 a2h

Rectangular base pyramidABCD is a rectangular base of the pyramid having Length ‘a’ and breadth b. VM and VN are slant heights along length and breadth respectively. Let VM = l1 and VN = l2. VO = h is the height of the pyramid. Here, ∆VAB and ∆VDC are the triangular faces along the length.∆VBC and ∆VDA are the triangular faces along the breadth.

Area of two triangular faces along the length = 2 × 12 (base × height )

= 2 × 12 × (AB × VM)

= al1Area of two triangular faces along the breadth = 2 × ( 1

2 × base × height )

= 2 × 12 × (BC × VN)

= bl2Area of rectangular base = a × b = abTotal surface area of pyramid = Area of rectangular base + Lateral surface area = al1 + bl2Total surface area of pyramid = ab + al1 + bl2Volume = 1

3 × Area of base × height

= 13 ab × h

Equilateral triangular base pyramid (Tetrahedron)ABC is an equilateral triangular base having the length of each side ‘a’. ∆VBC, ∆VAC and ∆VAB are triangular faces which are congruent to the base. VP is the slant height (l) and VO is the height (h) of pyramid.


Example 1 : Find the total surface area and volume of the following figures.

Solution:a. Side length of square base (a) = 8 cm Slant height (l) = 10 cm Height of pyramid (h) = 12 cm Total surface area = a2 + 2al = 82 + 2 × 8 × 10 = 64 + 160 = 224 cm2

Worked Out Examples

Area of equilateral base = 34 × a2

Lateral surface area = area of 3 triangular faces

= 3 × 12 (BC × VP) = 3

2 al∴ Total surface area = Area of equilateral triangular base + Area of triangular faces

= 34

a2 + 32 al

∴ Total surface area = 34

a2 + 32 al

∴ Volume = 13 × Area of equilateral triangular base × height

= 13 × 3

4 a2 × h = 312

a2h

∴ Volume = 312 a2h


Volume of pyramid = 13 × Area of square base × height

= 13 × 82 × 9

= 192 cm3

Hence, total surface area and volume of pyramid are 224 cm2 and 192 cm3.b. Length of rectangular base (a) = 12 cm

Breadth of the rectangular base (b) = 8 cmSlant height along length (l1) = 9cmSlant height along breadth (l2) = 8 cmHeight of pyramid (h) = 10 cmNow total surface area of pyramid = ab + al1 + bl2 = 12 × 8 + 12 × 9 + 8 × 8 = (96 + 108 + 64) cm2

= 268 cm2

Volume of pyramid = 13

× Area of rectangular base × height

= 13

× a × b × h

= 13

× 12 × 8 × 6

= 192 cm3

Hence, total surface area and volume of pyramid are 268 cm2 and 192 cm3.c. Length of equilateral triangular base (a) = 8 cm

Slant height (l) = 12 cmHeight of pyramid (h) = 10 cm

∴Total surface area of pyramid = 34 a2 + 3

2 al

= 34 × 82 + 3

2 × 8 × 12

= 16 3 + 144

= 171.71 cm2

∴Volume of the pyramid = 13 × 3

4 a2h


= 312

× 8 × 8 × 10

= 92.37 cm3

Hence, total surface area and volume of pyramid are 171.71 cm2 and 92.37cm3.

Example 2 : a. A square base pyramid has base area 100 m2 and total surface area 360m2. Find area of triangular faces. Solution: Here,Area of square base of pyramid = 100 m2

Total surface area = 360 m2

We know that,Total surface area of pyramid = Area of base + Area of triangular facesOr, 360 = 100 + Area of triangular facesOr, Area of triangular faces = 360 – 100 = 260 Thus, area of triangular faces is 260 cm2.b. The slant height and the length of base of a square base of a square base pyramid

are 8 cm and 6 cm respectively. Find the total surface area of pyramid.Solution : Here, Slant height (l) = 8 cm Length of square base (a) = 6 cm Total surface area of pyramid (TSA)= ?We know,Total surface area of pyramid (TSA) = Area of square base + Area of triangular faces = a2 +2al = 62 + 2 × 6 × 8 = 36 + 96 = 132 cm2

Thus, total surface area of pyramid (TSA) is 132 cm2.c. The volume of square based pyramid is 750 m3and height is 10 cm. Find the

total surface area of the pyramid.Solution : Here, Volume of the pyramid (V) = 750 m3

Height of the pyramid (h) = 10 cm


We know,

Volume of pyramid (V) = 13 Area of square base × height

Or, 750 = 13 × a2 × 10

Or, a2 = 750 × 310

Or, a2 = 225 Or, a = 15 cm Now, total surface area of pyramid = a2 + 2al = 152 + 2 × 15 × 12.5 = 600 cm2

Thus, total surface area of pyramid is 600 cm2.

Example 3 : The alongside figure is a square based pyramid whose length of the base and slant height are in the ratio 3:5. If its total surface area is 624 cm2, find its vertical height.

Solution: Here, Let length of square base (a) = 3x cmSlant height (l) = 5x cmTotal surface area of pyramid = 624 cm2

Or, a2 + 2al = 624Or, (3x) 2 + 2 × 3x × 5x = 624Or, 39x2 = 624

Or, x2 = 62439

Or, x = 16 = 4∴ a = 3x = 3 × 4 = 12 cm l = 5x = 5 × 4 = 20 cm

OQ = 12 BC = 1

2 × 12 = 6 cm

In right angled ∆ OPQ,

OP = PQ2 - OQ2

Now, slant height (l) = a2h2 + ( )2

= 152102 +( )2

= 156.25 = 12.5 cm


= 202 - 62

= 364 = 19.07 cm = 19 cm Height of pyramid (h) is 19 cm.

Example 4 : Given figure is a square based pyramid. If the volume of the pyramid is 384 cm3 and side of the base is 12 cm, find the total surface area.Solution: Here,

In the given figure OP is the height of pyramid.

Let OP = h, OM = 12 × (AB) = 1

2 × 12cm = 6cm

Volume of pyramid = 384 cm3

Or, 13 a2h = 384

Or, 13

× 12 × 12 × h = 384

Or, h = 384 × 312 × 12

= 1152144

= 8 cmHence, total surface area of pyramid is 384 cm2.

Example 5 : Find the volume of the given triangular base pyramid.

Solution: Here,Sides of triangular base are a = 6 cm, b = 5 cm, c = 5cm

s = a+b+c2 = 6+5+5

2 = 8 cm

Area of triangular base = s(s-a) (s-b) (s-c)

= 8(8-6) (8-5) (8-5) = 8 × 2 × 3 × 3 = 12 cm2

Height of pyramid (h) = 15 cm

∴ Volume of pyramid = 13

× Area of base × height

= 13 × 12 × 15

= 60 cm3 Hence, volume of pyramid is 60 cm3.

In right angled ∆ POM

OP2 + OM2 = PM2

Or, 82 + 62 = PM2

Or, PM2 = 100Or, PM = 10 cmSlant height (l) = 10 cmNow, total surface area of pyramid = a2 + 2al= 122 + 240 = 384 cm2


1. a. What is pyramid ?

b. Write the formula for finding volume of a square based pyramid in terms of area and vertical height.

c. Establish the relation among slanting height(l), vertical height(h) and side of square base(a).

d. What is the total surface area of square based pyramid having slanting height (l), height (h) and length of side (a).

e. Write the formula of volume of square base pyramid having length of square base ‘a’ cm and height ‘b’ cm.

f. Write the formula of volume of an equilateral triangular based pyramid having length of side of triangular base ‘m’ cm and height ‘n’ cm.

2. a. The height of square based pyramid having the area of its base 25 cm2 is 11.5 cm. Find the volume of the pyramid.

b. The height of square based pyramid having the area of its base 100 cm2 is 42 cm. Find the volume of the pyramid.

c. A pyramid of base area 30 cm2 has each triangular face of each area 12 cm2. Find total surface area.

d. A pyramid of base area 90 cm2 has each triangular face of each area 25 cm2. Find total surface area.

3. a. The volume of a squared base pyramid of height 12cm is 400 cm3. Find the length of side of its base.

b. A squared base pyramid has total surface area 624 cm2 and lateral surface area 480 cm2, find the area of its base.

c. If the volume of the square based pyramid is 384 cm3 and the side of its base is 12 cm, find the slant height.

d. Total surface area of square based pyramid is 144 cm2 and length of the side of square base is 8 cm, find the height of the pyramid.

4. Find total surface area and volume of the given pyramids.

5. In a square based pyramid, the ratio of vertical height and slant height is 4:5 and the total surface area is 96 cm2, find the volume of a pyramid.

Exercise (3.5)


6. The total surface area of the given square based pyramid is 240 cm2 and the side of the square base is 10 cm, find the slant height of the pyramid.

7. The slant height and total surface area of the given square based pyramid are 13 cm and 360 cm2 respectively. Find the perimeter of the base and volume of the pyramid.

8. ABCD is a square base of a pyramid. AO = 6 cm and PC = 10cm. Find the volume of the pyramid.

9. Find total surface area and volume of given equilateral triangular based pyramid.

10. Find the total surface area and volume of rectangular based pyramid.

6cm8cm

12cm

14cm

A

BC

D

P

Ans

wer

s (3.

5)

1b. 13 area of base ×height c. l2 = h2 + ( a

2 )2 d. a2 + 2al

e. 13 a2b f. 3

12 m2n 2a. 95.83cm3 b. 1400cm3

c.66cm2 d. 165cm2 3a. 10cm b. 144cm2

c. 10cm d. 3cm 4a. 360cm2 , 400cm3 b.736cm2,1082.02cm3 5. 48cm3 6. 7cm 7. 40cm 400cm3 8. 192cm3 9. 242.35 cm2 , 166.28cm3 10. 393.07 cm2, 336 cm2


3.6 Cone

A cone is a solid object having circular base and a smooth curved surface that symmetrically ends at a point in the space. The point is called the vertex of the cone. The line joining the vertex to the center of base is called height of the cone. The line joining the vertex to any point on the circumference of the circular base is called generator or slant height.

Surface area of cone

Curved surface area and total surface area Let us take a hollow cone made up of paper. Cut the cone along its slant height. We get a sector whose radius is r and arc length l. Let’s divide the sector into 10 equal parts as shown in the figure.These parts are cut and arranged as shown in the last figure.

The last figure is approximate rectangle having length pr and breadth l.Now area of rectangle = length × breadth

= pr × l

∴ Curved surface area of cone = prl

∴ Area of circular base = pr2

Total surface area = pr (r + l)

We know that cone is a circular pyramid. Area of circular base is pr2 and height is h.

The volume of pyramid = 13 × Area of circular base × height

Volume of a cone (circular pyramid) = 13 × Area of circular base × height

= 13

× pr2 × h = 13 pr2 × h

The following examples will illustrate the use of the above formulae for solving problems.

pr

l


Example 1 : If the height of the cone is 9 cm and its base area is 23 cm2, find the volume of the cone.Solution : Here,Height of the cone(h) = 9 cmBase area of the cone(A) = 23 cm2 Now, we know that

Volume of the cone (V) = 13

× pr2 h

= 13

× A × h

= 13

× 23 × 9

= 23 × 3 = 69 cm3

Hence, the volume of cone is 69 cm3.

Example 3 : How many meters of cloth 4 m wide will be required to make a conical tent when base radius is 12 m and height is 16 m?Solution : Here, Breadth of the cloth (b) = 4 mBase radius of the cone (r) = 12 mHeight of the cone (h) = 16m Let length of the cloth be’x’ mNow , we knowTotal surface area of the cloth = l × b = x × 4 = 4x cm2

∴ Total surface area of the cloth = 4 x cm2

For a cone, we know l2 = h2 + r2

= 162 + 122

= 256 + 144

Worked Out Examples

Example 2 : The total surface area of a cone is 4928 cm2. If the sum of the radius of the base and slant height of the cone is 49 cm, find the radius of the base.Solution : Here,Total surface area of cone (TSA) = 4928 cm2

sum of radius and slant height (r+l)= 49 cmNow, we knowTotal surface area of cone (TSA) = pr (r + l )

Or, 4928 = 227

× r × 49

Or, r = 4928 × 722 × 49

∴ r = 32 cmHence, the radius of the base is 32 cm.


= 400 ∴ l = 20 m.We know,

curved surface area of conical tent = pr l

= 227

× 12 × 20

= 52807

m2

∴ Total surface area of conical tent = 84487

m2

Now, we knowTotal surface area of cloth = curved surface area of conical tent

Or, 4 x = 52807

Or, x = 528028

∴ x = 188.57 m.Thus, the length of the cloth is 188.57 cm.Example 4: The radius and slant height of a cone are in the ration of 3:5. If its total

surface area is 21127

sq.cm, find the slant height.

Solution: Here,

Total surface area of cone = 21127

cm2

r : l = 3 : 5

or, r = 3l5

We know,

Total surface area of the cone (TSA) = 21127

Or, pr (r + l) = 21127

Or, p× 3l5

( 3l5

+1) = 21127

Or, 227

× 3l5

× ( 3l + 5l5

) = 21127

Or, l(8l) = 2112 × 7 × 5 × 57 × 22 × 3


Or, 8l2 = 800 Or, l2 = 100 ∴l = 10 cmHence, length of the slant height is 10 cm. Example 5 : Find total surface area and volume of the following cones.

Solution: Herea. Height of cone (h) = 8 cm

Slant height (l) = 10 cm In a right circular cone,h2 + r2 = l2

or, 82 + r2 = 102

or, r2 = 100 – 64or, r = 36 = 6 cm

Total surface area = pr (r + l)

= 227 × 6 × (10 + 6)

= 301.71 cm2

Volume of the cone = 13 pr 2h

= 13

× 227 × 6 × 6 × 8

= 301.71 cm3

Thus, total surface are of the cone is 301.71 cm2 and volume is 301.71cm3.

b. Diameter of circular base (d) = 12 cm

Radius of circular base (r) = 122 = 6 cm

Height of cone (h) = 10 cmLet slant height = l cmThen l2 = r2 + h2

Or, l2 = 62 + 102

Or, l = 136 = 11.66 cmNow total surface area = pr (r + l)

= 227

× 6 × (11.66 +6)

= 333.01 cm2

Volume of the cone = 13 pr 2h

= 13

× 227 × 6 × 6 × 10

= 377.14 cm3

Thus, total surface are of the cone is 330.01 cm2 and volume is 377. 14 cm3


Example 6 : Find the curved surface area, total surface area and volume of the given cone.

Solution:Diameter of circular base (d) = 12 cm

Radius of circular base (r) = 122 = 6 cm

Slant height (l) = 12 cmLet height of pyramid = h cmWe know,l2 = r2 + h2

or, 122 = 62 + h2

or, h2 = 144 – 36or, h = 108 = 10.39 cm

Curved surface area = pr l

= 227

× 6 × 12

= 226.28 cm2

∴ Total surface area = pr (r + l)

= 227

× 6 × (12 + 6)

= 339.42 cm2

∴Volume of cone = 13 pr 2h

= 13 × 22

7 × 6 × 6 × 10.39

= 391.85 cm3

Hence, the curved surface area of cone is 226.28 cm2, total surface area is 339.42 cm2 and volume is 391.85 cm3

Example 7 : The curved surface area and the total surface area of a cone are 550 cm2 and 704 cm2 respectively, find its slant height.

Solution: Here,Curved surface area of a cone = 550 cm2

Or, prl = 550 ………….. (1)Total surface area = 704 cm2

Or, prl + pr2 = 704


Or, pr 2 = 704 – 550

Or, r2 = 154 × 7 22

Or, r = 49 = 7 cmPutting value of r in equation 1

prl = 550

Or, l = 550 × 722 × 7

l = 25Thus, the slant height of cone is 25 cm.

Example 8 : Find the depth of a conical flask having diameter of circular base 14 cm and that can hold 600 ml of water.

Solution: Here,Diameter of circular base (d) = 14 cm

∴ Radius of circular base (r) = 142 = 7 cm

Capacity of flask = 600 ml∴ Volume of flask = 600 cm3 (∴ 1 l = 1000 cm3 and 1l = 1000 ml)Volume of cube (V) = 600cm3

Or, 13 pr2h = 600

Or, 13 × 22

7 × 7 × 7 × h = 600

Or, h = 600 × 322 × 7 = 11.68 cm

∴ Depth of the conical flask is 11.68 cm.

Exercise (3.6)

1. a. Write the formula to find volume of cone.

b. Establish the relation of radius ‘r’, slant height ‘l’ and vertical height ‘h’ of the cone.


c. If ‘x’ is the radius of the circular base, ‘y’ is vertical height and ‘z’ is slant height, then write the formula of total surface area of cone.

2. a. Find curved surface area and total surface area of cone.

i. r = 4 cm , l = 7 cm ii. r = 5.5 cm , l = 9.5 cm iii. r = 14 cm , l = 17 cm

b. Find the volume of the cone with

i. r = 3 cm , h = 8 cm ii. r = 6.5 cm , h = 8.5 cm iii. r = 15 cm , h = 17 cm

3. a. If the height of the cone is 6 cm and its base area is 18 cm2, find the volume of the cone.

b. If the height of the cone is 10 cm and its base area is 25 cm2, find the volume of the cone.

4. a. The total surface area of a cone is 2464 cm2. If the sum of the radius of the base and slant height of the cone is 36 cm, find the radius of the base.

b. The total surface area of a cone is 5025 cm2. If the sum of the radius of the base and slant height of the cone is 64 cm, find the radius of the base.

5. a. How many meters of cloth 3 m wide will be required to make a conical tent when base of radius is 13 m and height is 17 m?

b. How many meters of cloth 7 m wide will be required to make a conical tent when base of radius is 18 m and height is 11 m?

c. If the height of a conical tent is 12 m and its radius is 5 m. How much canvas is required for making the tent?

6. a. The radius and slant height of a cone are in the ration of 6:10. If its total surface

area is 42247 sq.cm, find the slant height.

b. The radius and slant height of a cone are in the ration of 4:6. If its total surface

area is 21127 sq.cm, find the slant height.

c. The curved surface area of the cone is 4070 cm2 and its diameter of the base is 70 cm. What is its slant height ?

d. The circumference of the base of a solid cone is 42π cm. If the sum of its slant height and its radius of its base is 49 cm. What will be the slant height of the cone ? Find it.

e. The total surface area of a cone with a base of radius 5 cm is 282 cm2. Find the slant height of the cone.


7. Find the curved surface area, total surface area and volume of the following cones.

8. a. The radius of the base and height of the cone are in the ratio of 2:3 and its volume is 242p cm3, find its height.

b. The radius of the base and height of the cone are in the ratio of 4:5 and its volume is 324 pcm3, find its height.

9. a. The curved surface area of a cone is 550 sq. cm and the radius of the base is 7 cm. Find the volume.

b. The curved surface area of a cone is 770 sq. cm and the radius of the base is 11 cm. Find the volume.

10. a. The circumference of base of a cone is 88 cm and slant height is 30 cm. Find the curved surface area of the cone.

b. The circumference of base of a cone is 176 cm and slant height is 60 cm. Find the curved surface area of the cone.

11. If the height of the given solid cone (OA) = 14 cm and volume is 528 cm3, find the area of base.

12. The diameter of cone is equal to height. If its slant height is 5 5 cm, find the volume of the cone.

13. A conical pot can hold 1.54 liters of water. If the radius of the circular base is 7 cm, find the height of the cone.

a. b. c.

d. e. f.


Ans

wer

s (3.

6)

1.a. 13 πr2h b. l2 = r2+h2 c. πx(x+z) 2a.i. 88cm2, 138.29cm2

ii. 164.21cm2, 259.29cm2 iii. 748cm2, 1364cm2 b.i.75.43cm3

ii. 376.23cm3 iii. 4000.14cm3 3a. 36cm3 b. 83.33cm3

4a. 21.78cm b. 24.98cm 5a. 291.44m b. 170.48 m

c. 204.28 cm2 6a. 10 2 cm b.9.30 cm c. 37cm

d. 28 cm e. 12.94 cm 7a. 204.29cm2, 282.86cm2,314.29cm3

b. 188.57cm2, 301.71cm2, 301.7cm3 c. 550 cm2, 704 cm2, 1232cm3

d. 902.8cm2, 1518.8cm2, 3080cm3 e.550cm2, 704cm2, 1232cm3

f. 2310cm2, 3696cm2, 12936 cm3 8a. 11.78 cm

b.11.5 cm 9a.1232cm3 b.2454.57cm3 10.a. 1320cm2

b. 5280 cm2 11. 113.14 cm2 12. 261.9cm3 13. 30cm

14.a. 14cm b. 7cm 15.a. 452.57cm2 b. 990cm2

16. 462cm2, 622.36cm3

14. a. The total surface area and the curved surface area of a cone are 968 cm2 and 352 cm2 respectively. Find the radius of the cone.

b. The total surface area and the curved surface area of a cone are 704 cm2 and 550 cm2 respectively. Find the radius of the cone.

15. a. The ratio of actual height and slant height of a cone is 3:5 and the volume is

28167 cm3. Find the total surface area of the cone.

b. If the circumference of the base of a right circular cone is 66 cm and sum of its slant height and radius is 30 cm, find its total surface area.

16. Find total surface area and volume of the adjoining solid cone.


3.7 Combined SolidsIn this topic, we are going to study the problems about the surface area and volume of combined solids. The combined solids may be of a cone and a hemisphere, a cylinder and a cone, a prism and a cone and so on.Example 1 : Find the curved surface area, total surface area and volume of the following objects.

a. Solution:Radius of circular base(r) = 5cmHeight of cylinder (h) = 24cmSlant height of cone(l) = 13cmHeight of cone(h1) = l2 - r2

= 132 - 52

= 144 = 12cm

Curved surface area of solid object = curved surface area of cylindrical portion + curved surface area of conical portion

= 2πrh + πrl = πr (2h + l)

= 227

× 5 (2 × 24 + 13)

= 1107 × 61

= 958.57cm2

Total surface area of the solid = Area of circular base of cylinder + CSA of cylinder + CSA of cone = πr2 + 2πrh + πrl

= 227 × 52 + 2 × 22

7 × 22

7 × 5 × 24 + 22

7 × 5 × 13

= 227 × 5 (5 + 2 × 24 + 13)

= 1107

× 66

= 1037.14 cm2

80 cm

94 cm

31 cm14 cm

a. b. c.


Volume of solid = Volume of cylinder + Volume of cone

= πr2h + 13

× πr2h1

= πr2 (h + 13

h1)

= 227

× 5 × 5(24 + 1

3× 12)

= 5507

× 28

= 2200 cm3

Thus, curved surface area of solid is 958.57 cm2 , total surface area is 1037.14 cm2 and volume is 2200 cm3.

b. Solution:Diameter of circular base (d)= 14 cmRadius of circular base (r) = 7 cmTotal height of solid = 31 cmHeight of the cone (h) = Total height of solid – Radius of base = 31 cm – 7 cm = 24 cm

Slant height of cone (l) = r2 + h2

= 52 +242

= 49 + 576

= 625 = 25 cmCurved surface area of solid = CSA of hemisphere + CSA of cone = 2πr2 + πrl = πr(2r + l)

= 227

× 7 (2 × 7 + 25)

= 22 × 39 = 858 cm2

Here, total surface area and curved surface area of the solid is same. Therefore total surface area is also 858 cm2.

Volume of solid = Volume of hemisphere + Volume of cone

= 23

πr3 + 23

πr2h


= 13

πr2 (2r+h)

= 13

× 227

× 72 (2 × 7 + 24)

= 1543

× 38 = 1950.66 cm3

Thus, curved surface area of solid is 858 cm2, total surface area is 858 cm2 and volume is 1950.66 cm3.

c. Solution:Total height of solid 94 cmHeight of the cylindrical portion (h) = 80 cm ∴ Radius of base = height of hemisphere (r)= 94 cm – 80 cm = 14 cm Curved surface area of solid = CSA of cylinder + CSA of hemisphere = 2πrh + 2πr2

= 2 πr(h + r)

= 2 × 227 × 14(80 + 14)

= 88 × 94 = 8272 cm2

Total surface are of solid = Area of circular base of cylindrical portion + CSA of cylindrical portion + CSA of hemispherical end = πr2 + 2πrh + 2πr2

= πr(r + 2h + 2r)

= 227

× 14 (14 + 2 × 80 + 2 × 14)

= 44 × 202 = 8888cm2

Volume of solid = volume of cylinder + volume of hemisphere

= πr2h + 23 πr3

= πr2( h + 23 r)

= 227 × 14 × 14 (80 + 2

3 × 14)

= 22 × 28 (80 + 9.33) = 55027.28 cm3

Thus, curved surface area of solid is 8272 cm2, total surface area is 8888 cm2 and volume is 55027.28 cm3.


Example 2 : A figure is a solid composed of cylinder with hemisphere at one end. If the total surface area and the height of the solid are 770 cm2 and 14 cm respectively, find the height of the cylinder.

Solution : Here, Total surface area of solid (TSA) = 770 cm2

Total height of the solid = 14 cm Let height of the cylinder be ‘h’ and ‘r’ be the radius of the hemisphere.

Then, h + r = 14 cmOr r= 14-h ……… (i)We know, Total surface area of given solid = Surface area of cylinder + Surface area of hemisphere

770 = 2 prh + pr2+ 2pr2

or, 770 = 2prh + 3pr2

or, 770 = pr(2h + 3r )

or, 770 = 227 × (14- h ) {2h + 3(14-h)}

or, 770 × 722 = (14 – h) ( 2h +42 – 3h )

or, 245 = (14 – h) ( 42 –h )or, 245 = 588 – 14h – 42h + h2

or, 245 = 588 -56h + h2

or, h2 – 56 h + 588 -245 = 0or, h2 – 56 h + 343=0or, h2 – 49 h- 7h + 343 = 0or, h(h- 49) – 7( h – 49) = 0or, (h- 49 ) (h- 7) = 0Either h- 49 = 0 or h – 7 = 0 or , h = 49 (which is not possible) Hence, the height of the cylinder is 7 cm.

14 cm


1. Find the volume of the following combined solids:

Exercise (3.7)

2. Find the total surface area of the following combined solids

3. Find the curved surface area, total surface area and volume of the following combined solids:

a. b.

c.

c. 7cm

31 cm

28cm 28cm

80cm

12cm

88cm

a.

24cm

14cm

c.

54cm

10.5cm

b.

31 cm

24 cm

d.

12cm13cm

10cm

24cm


4. The given object is made up of a hemisphere and a cylinder. The radius of hemisphere is equal to the radius of a cylinder. The height of the cylinder is 80cm and the volume of a hemisphere is 144π cubic cm. Find the total surface area of the solid object.

5. Given figure is a solid composed of a cylinder with hemisphere at one end. If the curved surface area and whole height of the solid are 528 sq. cm and 14 cm respectively, find the height of the cylinder.

6. Given solid is made up of a cone and a cylinder. The base area of the cylinder is 100 sq cm and the height of the cylinder is 3 cm. If the volume of the whole solid is 600 cubic cm, find the height of the whole solid.

7. Find the total surface area and volume of the solid formed by two cones standing on the same base as shown in the figure.

8. A solid cylinder of radius 6 cm and height 15 cm is melted and recast into a cone of diameter 12 cm, Find the height of the cone.

9. The figure alongside is a tent in shape of cylinder up to a height of 4cm and radius 14 m. Above cylindrical part, it is a cone with height 6m. Calculate the total surface area of the tent.

10. The cone having radius of circular base 7 cm and height 12 cm is completely filled with water. If the water is poured into a cylindrical tank having radius of circular base 6 cm, up to what height does the surface of water rise in the cylinder?

11. a. If the total surface area and the height of the solid made up of cylinder with hemisphere on the top are 770 cm2 and 14 cm respectively, find the height of the cylinder.

b. If the total surface area and the height of the solid made up of cylinder with hemisphere on the top are 2464 cm2 and 21cm respectively, find the height of the cylinder.

80cm

14cm

3cm

80cm7cm

17cm

6m

4m14m

Ans

wer

s (3.

7) 1.a. 388.08cm3 b. 1232cm3 c. 1576.96 cm3 d. 94.29 cm3

e. 8250 cm3 2.a. 3318.86 cm2 b.4603.5 cm2 c. 858cm2 d. 858cm2 3a. 581.42 cm2, 660 cm2, 1257.14 cm3 b. 858cm2, 858cm2, 1950.67cm3 c.2103.16cm2 , 2103.16cm2 , 8624cm3 4. 3356.57cm2 5. 8cm 6. 12cm 7.2140.6cm2 , 4901.82cm3 8. 45cm 9. 1022.19cm2 10. 5.44 cm 11.a. 7cm b. 7 cm


3.8 Application of MensurationIn earlier classes, we have learnt to find the area and volume of different objects like cube, cuboid, cylinder, sphere, cone and hemisphere. When we see around us we are using different objects which are the combination of these solids such as an ice cream cone full of ice cream and hemispherical on one end of cylinder, a capsule which is in the form of a cylinder and two hemispherical end, sharpened pencil etc. In this topic, we shall find the surface area and volume of combinations of solids, we shall also find the estimation for carpeting ,painting and costing etc.

Worked Out Examples

Example 1 : What is the capacity of a tank having length 15 m, breadth 7 m and height 8 m ?

Solution : Here,

Length of the tank (l) = 15 m

Breadth of the tank (b) = 7 m

Height of the tank (h) = 8 m

Volume of the tank (V) = ?

We, know

Volume (V) = l × b × h

= 15 × 7 × 8

= 840 m3

∴ The volume of the tank is 840 m3.

We have,

Capacity of tank = Volume of the tank = Volume of the water = 840 m3

Now, we know,

1 m3 = 1000 l of water

840 m3 = 1000 × 840 l of water = 840000 litres of water.

Hence, the capacity of the tank is 840000 litres .

Example 2 : The cost of painting the outside surface of closed cylindrical oil tank at Rs 2 per square centimeter is Rs 792.The height of the tank is 6 times the radius of the base of the tank.Calculate the volume of the tank.

Solution : Here, Let radius of base of the tank r cm , height of the tank (h) = 6 r cm


Outer surface area of cylindrical tank = Total cost of paintingRate of painting

2 pr ( r + h) = 7922

Or, 2 pr (r + 6r) = 396

Or, 2 × 227 × r × 7r = 396

Or, r2 = 9 cm

∴ r = 3 cm

∴ h = 6r = 6 × 3 cm = 18 cm

Now, volume of the tank = pr2h

= 227 × 32 × 18

= 509.14 cm3

Thus, volume of the tank is 509.14 cm3.

Example 3 : The adjoining figure shows the kitchen garden of a house. Estimate the cost of fencing with 5 rounds of fencing wire. Per meter cost of wire is Rs 80. Labour wage is Rs 850 per labour per day. If it takes 4 days to complete the fencing for 2 labours, find the cost of fencing the garden.

Solution: Here, Perimeter of kitchen garden = PQ + QR + SR + PS

= 20m + 15m + 18m + 10m = 63mThe length of wire required for 5 rounds of fencing = 5 × 63m

= 315mThe cost of wire = Rs 80 × 315 = Rs 25200The wages of 2 labours for 1 day = Rs 2 × 850 = Rs 1700The wages of 4 labours for 1 day = Rs 4 × 1700 = Rs 6800 Total cost of fencing the garden = Rs 25200 + Rs 6800 = Rs 32000

Hence, the cost of fencing the garden is Rs 32000.

R

Q

S

P

18m

15m10m

20m


Example 4: An umbrella shaped tent is made by stitching 10 triangular pieces of clothes which are equal in area. If the three sides of triangular pieces are 10 m, 16 m and 20 m. Find the area of cloth required to make an umbrella and total cost for making an umbrella if the rate of cloth is Rs 25 per square meter.

Solution : Here,

Let the length of sides of triangle are ‘a’ m , ‘b’ m and ‘c’ m

Then, a = 10 m , b = 16 m , c = 20 m.

We know, s = a + b+ c2

= 10 + 16 + 202

= 23 m

Now,Area of each triangular piece of cloth = s(s - a) (s - b) (s - c)

= 23 (23 -10) (23 - 16) (23 - 20)

= 23 × 13 × 7 × 3

= 6279

= 79.24 m2

∴ Total area of cloth to make an umbrella = 10 × 79.24 m2 = 792.40 m2.

Now, cost of making 1 m2 umbrella = Rs 25

∴ cost of making 792.40 m2 umbrella = Rs 19810.

Hence, total area of cloth for making an umbrella is 792.40 m2 and total cost is Rs 19810.

Example 5 : Prashant Mali made an arrangement for accommodation for 100 guests in his birthday party. A conical tent is built such that each person has 2 square meter space on the ground and 10 cubic meter of air to breathe in average. What should be the height of the tent ? Also, calculate the cost of carpet to cover the floor of the tent at the rate of Rs 250 per square meter.

Solution : Here,

Area of circular base of conical tent(A) = 100 × 2 = 200 m2

Volume of tent (V) = 100 × 10 = 1000 m3

∴ Height of tent = 3vA

= 3 × 1000200

= 15 mAgain, Area of carpet to cover the floor of the tent = Base area of tent


= 200 m2

Now,

Cost of carpeting 1 m2 area = Rs 250

Cost of carpeting 200 m2 area = Rs 250 × 200

= Rs 50000Hence, the height of the tent is 15m and cost of carpeting is Rs 50000.

Example 6 : A man wants to build a cylindrical water tank having diameter 3.5 m .If the height of the tank is 4 m, how much water can it hold? Also find the cost of plastering the tank at the rate of Rs 175 per square meter.

Solution : Here,

Diameter of cylindrical water tank (d) = 3.5 cm

Radius of cylindrical water tank (r) = 3.52

cm = 1.75 cm

Height of cylindrical water tank (h) = 4 mNow, we knowVolume of tank = pr2h

= 227 ×(1.75)2 × 4

= 38.5 m3

= 38.5 × 1000 l = 38500 l

Again,

Total area for plastering = Area of base + Area of curved surface = pr2 + 2 prh

= pr (r + 2h)

= 227

× 1.75 (1.75 + 2 × 4)

= 53.625 cm2

Now,

Cost of plastering of 1 m2 area = Rs 175

∴ Cost of plastering of 53.625 m2 area = Rs 175 × 53.625 = Rs 9384.38

Hence, the capacity of the water tank is 38500 l and total cost of plastering is Rs 9384.38


1. a. What is the capacity of a tank having length 6 m , breadth 4 m and height 1.5 m ?

b. What is the capacity of a tank having length 7.5 m , breadth 4.5 m and height 3.5 m ?

2. a. A well has 30 wheels each of radius 3.5 m and height 1.5 m. What is the volume of water that the well can hold if 2 of the wheels are to left unfilled.

b. A well has 40 wheels each of radius 8 m and height 5 m. What is the volume of water that the well can hold if 5 of the wheels are to left unfilled.

3. a. A cylindrical water tank of radius 7 m and height 10 m has a hemisphere on its top. Find the volume of water that can be hold by the tank.

b. A cylindrical water tank of radius 9 m and height 12 cm has a hemisphere on its top. Find the volume of water that can be hold by the tank.

4. An umbrella shaped tent is made by stitching 20 triangular pieces of clothes which are equal in area. If the three sides of triangular pieces are 12 m, 18 m and 22 m. Find the area of cloth required to make an umbrella and total cost for making an umbrella if the rate of cloth is Rs 30 per square meter.

5. Sumip Chaudhary made an arrangement for accommodation for 250 guests in his birthday party. A conical tent is built such that each person has 2 square meter space on the ground and 10 cubic meter of air to breathe in average. What should be the height of the tent ? Also , calculate the cost of carpet to cover the floor of the tent at the rate of Rs 350 per square meter.

6. A man wants to build a cylindrical water tank having diameter 7 m .If the height of the tank is 8m, how much water can it hold? Also find the cost of plastering the tank at the rate of Rs 195 per square meter.

7. The adjoining figure shows the compound of a house. Calculate the cost of fencing the compound with 6 rounds of fencing wire . If the cost of 1 m wire is Rs 55 and it takes 3 Labours for 2 days for fencing when the labour charge is Rs 950per labour per day

Exercise (3.8)

R

Q

S

P

18cm

15cm10cm

20cm


8. The figure alongside shows compound of a house. The length of diagonal BD = 10 m. AM = 10 m and CN = 15 m. Calculate the cost of plastering the compound at the rate of Rs 250 per square meter.

9. A tent in the shape of square based pyramid is 4 m tall and its slant height is 5 m. Calculate the cost of canvas required to make its triangular faces at the rate of 500 per square meter. Also calculate the cost of plastering its floor at the rate of Rs 650 per square meter.

10. A tent house as shown in the figure is 8 m long. It has two doors of size 2m × 1 m and a window of size 1m × 0.5 m . Find the cost of canvas required to build the tent house at the rate of Rs 750 per square meter. Also calculate the cost of carpeting the floor at the rate of Rs 795 per square meter.

11. A tent is in the shape of a right circular cylinder upto the height of 4m and then takes the shape of right circular cone with maximum height 28 m above the ground. Calculate the cost of painting the inner surface of the tent at the rate of Rs. 30 per square meter, if the radius of the circular base is 7m.

12. The adjoining parachute has 32 ropes of 32 feet each of length to be joined with a man and a semicircular shaped umbrella with circumference 44 feet. Find the volume of the parachute and area of the cloth used as the umbrella.

13. A cylindrical tank having 1.4 m inner diameter and 1.5 m height is constructed in the school to collect the 5kg water. If the upper part of the tank is a cone of vertical height 0.36m, how much water can be hold by the tank? Find it.

Ans

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8)

1a. 36m3 b. 118.125m3 2a.1617m3 b.56320m3

3a. 2258.67 m3 b. 4582.28 cm3 4. 2158.51cm2, Rs.64755.54

5. 15 m, Rs.175000 6. 308m3, Rs.41827.5 7. Rs 26490 8. Rs 31250

9. Rs 30000, Rs 23400 10. Rs 92007.68, Rs 38160 11. Rs 21780

12. 2321.29 ft3, 308 ft2 13. 2494.80 litres


Model - Question [F. M. 16]Group – A (1 × 1 = 1 )

1. Write the formula to find the area of scalene triangle with sides p cm, q cm and r cm and its perimeter 2s cm.

Group - B ( 3 × 2 = 6 )2. a. If the height of a triangular prism is 21.5cm and perimeter of triangular base is 16 cm,

find the lateral surface area of the prism (344 cm2) b. The sum of the height and radius of the cylinder is 34cm and total surface area is 2992

cm2 find its i. curved surface area ii. Volume (1760cm2, 12320 cm3) c. If the total surface area of a hemisphere is 462cm2, find its volume. (718.67 cm3)

3. Find the total surface area of the given solid. (1539.99cm3)

Group D (1×5 = 5)4. Find the total surface area of given combined solid (858 cm2)

40cm 13cm10cm

14cm31cm

Test Yourself F.M. 251. Establish the relation among slant height(l), vertical height(h) and side of square

base(a).2. If the total surface area and curved surface area of cone are 704cm2 and 550cm2

respectively, find the radius of the cone. (7 cm)3. a. Find i.CSA ii.TSA ii. Volume of the given solid. (858 cm2, 1950.66 cm3)

b. The iron pipe 105 cm long has an internal diameter of 5 cm. The thickness of the iron is 1 cm. Find the volume of iron in the pipe. ( 1980 cm3)

4. a. If each side of an equilateral triangle is increased by 2 cm then its area is

increased by 3 3 cm2. Calculate the length of each its sides and the area.

(2 cm , 3 cm2 ) b. The height of the cylinder and slant height of the cone are twice of the

base radius in the given figure. If the total surface area of the whole solid is 1078 cm2, find the total cost to paint its curved surface area at the rate of Rs 3.50 per cm2. (924 cm2, Rs 3234)

14cm31cm


AlgebraEstimated period : 36

4 Unit

Specific objectivesAt the end of this unit, the students will be able to

• find H.C.F. and L.CM. of algebraic expression by factorize method.• simplify the algebraic fraction.• simplify and solve algebraic expression • solve simple radical and surds involving based on four simple operations.• solve word problems of simultaneous linear equations of two variables or

quadratic equations.

Teaching materials:Chart papers, Models, Geometry box, Scissor

Note to the teacher:1. Give clear concept of HCF and LCM and state the relation between them.2. Give the clear concept for simplifying algebraic fractions.3. Discuss the rules of indices.4. Introduce surds and their simplification.5. Solve many problems to clear the concept of word problems of simultaneous

equations.

Contents• HCF and LCM• Simplification of algebraic fraction.• Indices• Surds and radical equations• Verbal problems• Simultaneous linear equations• Quadratic equations.


Specification GridCognitive Domain

Knowledge(K)

Comprehensive(C)

Application(A)

Higher ability(HA)

Total no. of questions

Totalmarks

Topic

H.C.F. and L.C.M. Each 1 mark Each 2 marks Each of 4 marks Each of 5 marks

SurdsIndicesRational FractionsEquations

1 5 2

1

9

24

Note : At least 3 marks questions are asked.4.1 H.C.F. of Algebraic expressionsThe highest common factor(HCF) of two or more algebraic expressions is the expression of highest dimension which divides each of them without remainder.

Let us consider two monomial expressions 4x3y2 and 6x2y3

Here, all the possible factors of 4x3y2 = 2×2×x×x×x×y×y

Also, the possible factors of 6x2y3 = 2×3×x×x×y×y×y

The factors common to both the expressions are 2,x,x,y and y. The product of these common factors is 2.x.x.y.y i.e. 2x2y2. It is the highest common factor (H.C.F.) of both the expressions. Therefore, the H.C.F. of 4x3y2 and 6x2y3 is 2x2y2.

The H.C.F. divides both the expressions exactly.

Similarly, let us consider algebraic expressions; 3a4 b3-9a2b3 and 6a3b- 3a2b2 - 3ab3.

On factorization of the given expressions,

First expression = 3a4 b3- 3a2b3

= 3a2b(a2 - b2)

= 3a2b(a+b)(a - b)

Second expression = 6a3b - 3a2b2 - 3ab3

= 3ab{2a2 - ab - b2}

= 3ab{2a2 - (2 - 1)ab - b2}

= 3ab{2a2 - 2ab + ab - b2}

= 3ab{2a(a - b)+b(a - b)}

= 3ab(a-b) (2a+b)

The common factors of both the expression are 3,a,b and (a-b). The product of these common factors is 3ab(a-b). Therefore, the Highest Common Factors (H.C.F.) of the given expressions is 3ab(a-b).


Steps for finding the H.C.F. of two or more than two algebraic expression:From the above discussion we can easily find the H.C.F. of two or more than two algebraic expression by following the given steps:

1. Factorize each of the expression.2. Take all the common factors of each of the expression.3. Express all the common factors in the product form.4. Thus, H.C.F. = common factors of all the expressions.5. If there is no any common factor to all the polynomials, then the H.C.F. is one.

Example 1 : Find the H.C.F. of (a + b) and (a + b) (a – b)

Solution : Here,First expression = ( a + b )Second expression = ( a + b) (a – b)Common factor = ( a + b )∴ H.C.F. = ( a + b )Hence, the required H.C.F. is ( a + b )

Example 2 : Find the H.C.F. of a3-4a and a4-8a.Solution : Here,

First expression = a3–4a = a(a2–4) = a(a2–22) [∵ a2–b2 = (a+b)(a–b)] = a(a+2)(a–2)

Second expression = a4 – 8a = a(a3– 8) = a(a3–23) [∵ a3-b3 = (a–b)(a2+ab+b2)] = a(a–2) (a2+2a+22) = a(a–2) (a2+2a+4)

Common factors = a × (a – 2)∴ H.C.F. = a(a–2)

Hence, the required H.C.F. of given expressions is a(a–2).

Worked Out Examples


Example 3 : Find the H.C.F. of x3y+y4 and x4+x2y2+y4.Solution : Here,

First expression = x3y+y4

= y(x3+y3) = y(x+y)(x2-xy+y2) [∵ a3+b3 = (a+b)(a2–ab+b2)]Second expression = x4+x2y2+y4

= x4+y4+x2y2 = (x2)2 + (y2) + x2y2 = (x2 + y2)2 - 2x2y2+x2y2 [∵ a2+b2 = (a + b)2 - 2ab] = (x2 + y2)2 - x2y2 = (x2 + y2) - (xy)2 [∵ a2–b2 = (a+b)(a–b)] = (x2 + xy + y2) (x2 - xy + y2)Common factor = (x2 - xy + y2)∴ H.C.F. = (x2 - xy + y2)Hence, the required H.C.F. of given expressions is (x2 - xy + y2)

Example 4 : Find the H.C.F. of a2+2ab+b2– c2, b2+c2+2bc– a2 and c2+2ca– b2+a2

Solution : Here,First expression = a2+2ab+b2-c2

= (a+b)2- c2

= (a+b+c) (a+b-c)Second expression = b2+c2+2bc - a2

= b2+2bc +c2 - a2

= (b+c)2 - a2

= (b+c+a) (b+c-a) = (a+b+c) (b+c-a)

∴ H.C.F. = (a+b+c)

Example 5: Find the H.C.F. of m6 – 1 , m5 + m3 + m , m3 +2m2 + 2m + 1Solution : Here,

First expression = m6 – 1 = ( m3 )2 – (1)2

= ( m3 – 1 ) ( m3 + 1 ) = ( m – 1 ) ( m2 + m + 1 ) ( m +1 ) ( m2 – m + 1 )

Third expression = c2+2ca– b2+a2

= c2+2ca+a2 - b2

= (c+a)2 - b2

= (c+a+b) (c+a-b) = (a+b+c) (c+a-b)


Second expression = m5 + m3 + m = m (m4 + m2 + 1 ) = m { (m2)2 + (1)2 + m2 } = m { ( m2 + 1 )2 – 2 m2 + m2 } = m { ( m2 + 1 )2 – m2 } = m ( m2 + m +1 ) ( m2 – m +1 )Third expression = m3 +2m2 + 2m + 1 = m3 + 13 + 2m ( m + 1 ) = ( m +1 ) ( m2 – m + 1 ) + 2m ( m + 1 ) = ( m +1 ) { m2 – m + 1 + 2m } = ( m +1 ) ( m2 + m + 1 } ∴ Common factor = ( m2 + m + 1 ) ∴ H.C.F. is ( m2 + m + 1 }

Exercise (4.1)

1. a. What do you mean by H.C.F. of algebraic expressions ?

b. In which case, the H.C.F. of algebraic expression is 1 ?

c. What is the H.C.F. of x2 and x3 ?

d. What is the H.C.F. of x2 and y2 ? e. What is the H.C.F. of 4x2 and 6x3 ?

2. Find the H.C.F. of the following expressions.

a. (a + b) (a – b) and (a + b) (a2 – ab + b2 )

b. 3x ( x + 5 ) ( x – 5 ) and 3x2 ( x + 5 ) ( x – 5 )

c. 2x2(x+2)(x–2) and 4x(x+2)(x+3)

d. 4xy2(x–1)(x+2) and 6x2y(x–1)(x–4)

3. Find H.C.F. of the following expressions.

a. x2–9 and 3x+9 b. x2–4 and 2x+4

c. x2–y2 and x2+2xy+y2 d. 4x2– 100 and 4x+20

e. a3–1 and a4+a2+1 f. 4x3+8x2 and 5x3–20x

g. a2–b2–2a+1 and a2–ab–a h. p4–16 and p2–p–6


i. 16 x4-4x2-4x-1 and 8x3-1 j. a3–a2+a–1,2a3–a2+a–2

k. 8a3+b3, 16a4+4a2b2+b4 l. m4+m2n2+n4,m2n3+m5

4. Find the H.C.F. of the following expressions .

a. x3–1, x4+x2+1 and x6–1

b. a2–6ab+8b2, a2–16b2 and a3–a2–2a+8

c. (a+b)2– 4ab, a3b3– b6 and a2+ab– 2b2

d. 5a3–20a, a3–3a2–10a and a3–a2–2a+8

e. x3–27y3, x4+9x2y2+81y4 and x3+3x2y+9xy2

f. 3x2–8x+4, 2x2–5x+2 and x4– 8x

g. a2– 4, a3+8 and a2+5a+6

h. x2– 9, x3– 27 and x2+x– 12

i. x3– 64y3, x2– 6xy+8y2 and x2– 16y2

j. 4x4+16x3-20x2, 3x3+14x2– 5x and x4+125x

k. x2– 18x– 19+20y– y2,x2+x– y2+y and x2– y2+2y– 1

l. 8x3+27y3, 16x4+36x2y2+81y4 and 4x3– 6x2y+9xy2

m. a2+2ab+b2– c2,b2+2bc+c2– a2 and c2+2ca+a2– b2

n. 9x2– 4y2– 8yz– 4z2, 4z2– 4y2– 9x2– 12xy and 9x2+12xz+4z2– 4y2

o. e2+f2– g2+2ef,e2– f2+g2+2ge and g2+f2+2fg – e2

p. x3y+y4, x4+x2y2+y4 and 2ax3– 2ax2y+2axy2

q. m3–m2– m+1, 2m4– 2m and 3m2+3m– 6

Ans

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1)

1.a. highest common factor of all given algebraic expressions.b. when there is no common factor c. x2 d. 1 e. 2x2 2. a. (a+b) b. 3x(x2-25) c. 2x (x+2) d. 2xy (x-1) 3. a.(x+3) b. (x+2) c. (x+y) d. 4(x+5) e. (a2+a+1) f. x (x+2) g. (a-b-1) h. (p+2) i. 4x2+2x+1 j. (a-1) k. (4a2-2ab+b2) l. (m2-mn+n2) 4.a. (x2+x+1) b. 1 c. (a-b) d. (a+2) e. (x2+3xy+9y2) f. (x-2) g. (a+2) h. (x-3) i. (x-4y) j. x(x+5) k. (x-y+1) l. (4x2-6xy+9y2) m. (a+b+c) n. (3x+2y+2z) o. (e+f+g) p.(x2-xy+y2) q. (m-1)


4.2 Lowest Common Multiple (L.C.M.)Let us consider two monomial expressions 4x3y2 and 6x2y3

Here, all the possible factors of 4x3y2 = 2 × 2 × x × x × x × y × yAlso, the possible factors of 6x2y3 = 2 × 3 × x × x × y × y × yAll the common factors of the given expression = 2 × x × x × y × yThe remaining factors of 4x3y2 = 2xThe remaining factors of 6x2y3 = 3yThe product of common factors and remaining factors = 2x2y2 .2x.3y = 12x3y3

Here, 12x3y3 is the lowest expression which is multiple of both the expression. Therefore, 12x3y3 is said to be the Lowest Common Multiple (L.C.M.) of the given expression.

Similarly, let us consider the algebraic expression x2-9 and 3x2 + 81

On factorization of the above expression. we get,

First expression = (x2-9)

= (x2-32) [∵ a2-b2 = (a+b)(a-b)]

= (x+3)(x-3)

Second expression = 3x3+81

= 3(x3+27)

= 3(x3+33) [∵ a3+b3 = (a+b)(a2-ab+b2)]

= 3(x+3)(x2-3x+32)

= 3(x+3)(x2-3x+9)

The common factor of the given expressions = (x+3)

The remaining factors of x2-9 = (x-3)

The remaining factors of 3x3+81 = 3(x2-3x+9)

The product of common factors and all the remaining factors = 3(x+3)(x-3)(x2-3x+9) = 3(x2-9)(x2-3x+9)

Therefore, the Lowest Common Multiple (L.C.M.) of the given expressions is 3(x2-9)(x2-3x+9).

Steps for finding the L.C.M. of two or more than two algebric expression:

From the above discussion we can easily find the L.C.M. of two or more than two algebric expression by following the given steps:


Example 1 : Find the L.C.M. of 4x4+4x2y2+4y4 and 6x3-6y3.Solution : Here,First expression = 4x4+4x2y2+4y4

= 4(x4+x2y2+y4)

= 4(x4+y4+x2y2)

= 4{(x2+y2)2-2x2y2+x2y2}

= 4{(x2+y2)2-x2y2}

= 4{(x2+y2)2-(xy)2}

= 4{(x2+y2+xy)(x2+y2-xy)}

= 4{(x2+xy +y2)(x2-xy +y2)}

= 2×2{(x2+xy +y2)(x2-xy +y2)}

Example 2 : Find the L.C.M. of x3-2x2y+2xy2-y3,x3+y3 and x4-y4

Solution: Here,First expression = x3-2x2y+2xy2-y3

= x3-y3-2x2y+2xy2

= (x-y)(x2+xy+y2)-2xy(x-y)

= (x-y)(x2+xy+y2-2xy)

= (x-y)(x2-xy+y2)

Second expression = x3+y3

= (x+y)(x2-xy+y2)

Worked Out Examples

1. Factorize each of the expression

2. Take the common factors of each of the expression.

3. If there are more than two expressions, take the factors which are common to any two or more expressions.

4. Take the remaining factors of each of the expression.

5. Express all the common factors and remaining factors in the product form.

6. Thus, L.C.M. = common factors × remaining factors.

Second expression = 6x3-6y3

= 6(x3-y3)

= 6(x-y)(x2+xy+y2)

= 2×3(x-y)(x2+xy+y2)

L.C.M. = 2(x2+xy+y2).2(x2-xy +y2).3(x-y)

= 12(x-y)(x2+xy +y2)(x2-xy +y2)


Third expression = x4-y4

= (x2)2-(y2)2

= (x2+y2)(x2-y2)

= (x2+y2)(x+y)(x-y)

∴ L.C.M. = (x+y)(x-y)(x2-xy+y2)(x2+y2)

= (x2-y2)(x2+y2)(x2-xy+y2)

= (x4-y4)(x2-xy+y2)

Example 3 : Find the L.C.M. of a2-18a-19+20b-b2,a2+a-b2+b and a2-b2+2b-1

Solution: Here,First expression = a2-18a-19+20b-b2

= a2-18a+81-100+20b-b2

= a2-18a+92-100+20b-b2

= (a-9)2-(100-20b+b2)

= (a-9)2-(102-20b+b2)

= (a-9)2-(10-b)2

= {(a-9)+(10-b)}{(a-9)-(10-b)}

= (a-9+10-b)(a-9-10+b)

= (a-b+1)(a+b-19)

Second expression = a2+a-b2+b

= a2 -b2+a +b

= (a-b)(a+b)+1(a+b)

= (a+b)(a-b+1)

∴ L.C.M. = (a+b) (a+b-1)(a-b+1)(a+b-19)

Example 4 : Find the L.C.M. and H.C.F. of: x2+5x+6 and x3+4x2+4x+3Solution :

First expression = x2+5x+6

= x2+(3+2)x+6

= x2+3x+2x+6

= x(x+3)+2(x+3)

Third expression = a2-b2+2b-1

= a2-(b2-2b+1)

= a2-(b-1)2

= {a-(b-1)}{a+(b-1)}

= (a-b+1)(a+b-1)


= (x+3)(x+2)

Thus, if there is a common factor, it is either (x+3) or (x+2). But, (x+2) cannot be the factor second expression as the last term is 3. So, let us try with ( x +3)

Second expression = x3+4x2+4x+3

[Balancing the expression x3+4x2+4x+3 by adding or subtracting the required term]

= x2(x+3)+x2+4x+3

= x2(x+3)+x(x+3)+x+3

= x2(x+3)+x(x+3)+1(x+3)

= (x+3)(x2+x+1)

∴ L.C.M. = (x+3)(x+2)(x2+x+1)Example 5 : Find the L.C.M. of : a4 + a2 + 1 , a3 – 1 and a3 – a2 + aSolution : Here,First expression = a4 + a2 + 1

= { (a2)2 + (1)2 + a2 } = { ( a2 + 1 )2 – 2 a2 + a2 } = { ( a2 + 1 )2 – a2 } = ( a2 + a +1 ) ( a2 – a +1 )

L.C.M. = Common factor × Remaining factor = a ( a – 1 ) ( a2 + a + 1 ) ( a2 – a + 1 )Thus, the required L.C.M. is a (a – 1) (a2 + a + 1) (a2 – a + 1)

1. a. What is L.C.M. of algebraic expression.?

b. Establish the relation between H.C.F. and L.C.M.

c. What is the L.C.M. of a2 and a3 ?

d. What is the L.C.M. of a2 and b3 ?

2. Find the L.C.M. of the following expressions:

a. 3x(x+1)(x–1) and 2x2(x–1) (x+3) b. 4x3(x–3)(x+2) and 6x2(x+2)(x+3)

c. 8a2b(a–b)(a2+ab+b2) and 12ab2(a–b)(a+b) d. (x+2)(x+3),(x+3)(x–2) and (x–2)(x–3)

e. (a–3)(a–4),(a–4)(a–5) and (a–5)(a–3) f. x4+x2y2+y4 and x3–y3

Exercise (4.2)

Second expression = a3 – 1 = (a)3 – 13

= (a – 1) ( a2 + a + 1)Third expression = a3 – a2 + a = a ( a2 – a + 1 )


g. a4+a2b2+b4 and a3+b3 h. 2x2+5x–3 and 8x3–4x2–2x+13. Find the L.C.M. of the following expressions: a. 3x2+6x, 2x3+4x2 b. ax2+ax, –ax2–a c. 4x5y4+2x4y5, 10x4y3+5x3y4 d. x2– xy, x3y– xy3

e. 4x2– 2x,8x3– 2x f. x2+5x+6, x2– 4 g. x2– 9,3x3– 81 h. a4b– ab4,a4b2– a2b4 4. Find the L.C.M. of the following expressions: a. x3– 9x, x4– 2x3– 3x2 and x3– 27 b. a3– 4a, a4– a3– 2a2 and a3– 8 c. x2y+2xy– 3x– 6, x3– 2x2+8– 4x and x2– 4 d. x3+y3, x4+x2y2+y4 and x3– x2y+xy2

e. a3b+b4, a4+a2b2+b4, 2xa3– 2xa2b+2xab2 f. a2– 4, a3– 8 and (a+2)2

g. (a– 3)2, a2– 9 and a3+27 h. a3– 2a2+a and a3+a2– 2a+a3– 4a i. x4– y4,x2– y2 and x3– y3 j. 4x3– 10x2+4x, 3x4– 8x3+4x2 and x4– 8x k. a2+2ab+b2– c2,b2+2bc+c2– a2 and c2+2ca+a2– b2

l. x2+2xy+y2– z2,y2+2yz+z2– x2 and z2+2zx+x2– y2 m. x3– 2x2y+2xy2– y3, x4– y4 and x3+y3 n. x2+3x+2, x2+5x+6 and x2+4x+35. Find H.C.F and L.C.M. of the following expressions. a. x4 + ( 2b2 – a2 )x2 + b4 and x4 + 2ax3 + a2x2 – b4

b. x2 + 2x – 15, x2 + 9x + 20 and x2 + 11x + 28 c. 1+4m+4m2- 16m4, 1+2m-8m3-16m4

d. x3y + y4, x4 + x2y2 + y4 and 2ax3 – 2ax2y + 2axy2

Ans

wer

s (4.

2)

1.a. Lowest common multipleb. L.C.M × H.C.F = Product of two expressions. c. a3 d. a2 b3

2. a.6x2 (x+1) (x-1) (x+3) b. 12x3 (x+2) (x2-9) c. 24a2b2 (a+b) (a3-b3) d. (x2-4) (x2-9) e. (a-3) (a-4) (a-5) f. (x-y) (x4 + x2y2 + y4) g. (a+b) (a4 + a2b2 + b4) h. (2x+1) (2x-1)2 (x+3)3. a. 6x2 (x + 2) b. – ax(x+1) (x2+1) c. 10x4y4(2x+y) d. xy (x2-y2)

e. 2x (4x2-1) f. (x+2) (x-2) (x+3) g. 3(x2-9) (x2+3x+9) h. a2 × b2 (a+b) (a3-b3) 4. a. x2 (x+1) (x2-9) (x2+3x+9) b. a2 (a+1) (a+2) (a3-8) c.(x+2) (x-2)2 (xy-3) d. x(x+y) (x4+x2y2+y4) e. 2xab (a+b) (a4+a2b2+b4) f. (a+2)2 (a3-8) g. (a-3)2 (a3+27) h. a(a-1)2 (a+2) (2a-3) i. (x4-y4) (x2+xy+y2) j. 2x2(x-2) (3x-2) (2x - 1) (x2+2x+4) k.(a+b+c) (a+b-c) (b+c-a) (c+a-b) l. (x+y+z) (x+y-z) (y+z-x) (z+x-y) m. (x4-y4) (x2-xy+y2) n. (x+1) (x+2) (x+3) 5. a.(x2+ax+b2) and (x2+ax+b2) (x2+ax-b2) (x2-ax+b2) b. 1 and (x-3) (x+4) (x+5) (x+7) c. (4m2+2m+1) and (2m+1) (4m2+2m+1) (1-2m) (1+2m-4m2) d. (x2-xy+y2) and 2axy (x+y) (x4+x2y2+y4)


4.3 Simplification of Algebraic ExpressionRational expressionRational algebraic expressions are those expression which can be expressed in the

form pq

where q ≠ 0.

The expression y2 ,

5x3bc5 ,

a2 - b2

a+b ,

3x2y + 5x2 + 6x-y are few examples of rational

expression.

Simplification of rational expression

To simplify rational algebraic fraction we have to follow some process as numeric fraction i.e. rational number. The process of addition and subtraction of two or more algebraic function consists of the following steps:

1. Factorize the denominator and numerator if required.6

a2 - 9 +

13 - a

- 12(a+3)2

= 6

a2 - 32 +

13 - a

-

12(a+3) (a+3)

= 6

(a+3) (a-3) +

13 - a

-

12(a+3) (a+3)

2. Change the sign of the denominator of fraction (or fraction as a whole) in order to arrange the denominator in ascending order of variable.

= 6

(a+3) (a-3) -

1(a-3)

-

12(a+3) (a+3)

So, here we have to change the sign of the denominator of the second fraction and write the expression as:

= 6

(a+3) (a-3) -

1(a-3)

-

12(a+3) (a+3)

3. Find the L.C.M. of denominator of each fraction:

L.C.M. of (a+3)(a-3),(a-3) and 2(a+3)(a+3) is;

L.C.M. 2(a+3)(a-3)(a+3)

4. Divide the L.C.M. of the denominator of first fraction and multiply the quo-tient with numerator of the fraction. Follow same process for other fraction.

6 × 2 (a +3) - 2(a+3)2 - (a - 3)

2(a+3) (a+3) (a -3)


5. Simplify the numerator and reduce the last fraction into lowest term.

=12 (a+3) -2 (a+3)2 - (a-3)

2(a+3) (a+3) (a-3)

= 12 (a+3) -2 (a+3)2 - (a-3)2(a+3) (a+3) (a-3)

= 12a + 36 - 2 (a2+6a+9) - (a-3)2(a+3)2 (a-3)

= 12a + 36 - 2a2-12a -18-a+32(a+3)2 (a-3)

= 21 - 2a2 - a2(a+3)2 (a-3)

= 21 - a - 2a2

2(a+3)2 (a-3)

6. If two or more algebraic functions with common denominator are to be added or subtracted, we should add or subtract the numerators. The denominator of the resulting fraction is the common denominator of the given fractions.

For example:x

x2 - y2 - y

x2 - y2

= x-y

x2 - y2

= (x-y)

(x+y) (x-y)

= 1

(x + y)Sign rules

1. The value of fraction is not changed if we change the signs of both nu-merator and denominator.

Example: +a+b

= - a- b

- a- b

= + a+ b

= 21 - (7-6) a-2a2

2(a+3)2 (a-3)

= 21 - 7a+6a-2a2

2(a+3)2 (a-3)

= 7(3-a) + 2a(3-a)2(a+3)2 (a-3)

= (3-a) (7 + 2a)2(a+3)2 (a-3)

= - (a -3) (7 + 2a)2(a+3)2 (a-3)

= - (7 + 2a)2 (a+3)2


- a+b

= + a- b

+a- b

= - a+ b

2. The signs of a fraction may be changed by changing the sign of every term in the numerator or denominator.

Example: aa - b

=

- a- a + b

=

- ab - a

- pp - q

=

p-p + q

=

pq - p

Worked Out Examples

Example 1 : Simplify : x x + 2y - x

2y - x - 8y2

x2 - 4y2

Solution : Here,

x x + 2y - x

2y - x - 8y2

x2 - 4y2

= x x + 2y + x

x - 2y - 8y2

x2 - 4y2

= x x + 2y + x

x - 2y - 8y2

(x+2y) (x-2y)

= x(x-2y) + x(x+2y) - 8y 2 (x+2y) (x-2y)

= x2 - 2xy+x2+2yx -8y2

(x+2y) (x-2y)

= 2x2 - 8y 2

(x+2y) (x-2y)

= 2(x2 - 4y2)(x+2y) (x-2y)

= 2{x2 - (2y)2}(x+2y) (x-2y)

= 2(x+2y) (x-2y)(x+2y) (x-2y)

= 2


Example 2 : Simplify : 1(x-2) (x - 3) + 2

(x -1) (3 - x) + 3

(1-x) (2 - x)

Solution : Here,

= 1(x-2) (x - 3) + 2

(x -1) (3 - x) + 3

(1-x) (2 - x)Writing the factors in descending power of x,

(3-x) = -(x-3), (1-x) = -(x-1), (2-x) = -(x-2)

∴ 1(x-2) (x - 3) + 2

(x -1) (3 - x) + 3

(1-x) (2 - x)

= 1(x-1) - 2(x-2) + 3(x-3)(x -1) (x-2) (x-3)

= x-1-2x + 4 + 3x -9(x -1) (x-2) (x-3)

= 2x - 6(x -1) (x-2) (x-3)

= 2 (x - 3)(x -1) (x-2) (x-3)

= 2(x - 1) (x - 2)

Example 3 : Simplify : 1x - 1 - 2

2x+1 + 1x + 1 -

22x - 1

Solution: Here,

1

x - 1 - 22x+1 + 1

x + 1 - 2

2x - 1It is easier to combine the fractions in suitable pairs instead of taking the L.C.M. of all denominator at once.

1

x - 1 + 1x+1 - 2

2x + 1 - 2

2x - 1

= 1x - 1 + 1

x+1 - [ 22x + 1 +

22x - 1 ]

= x+1+x-1(x-1) (x+1) - [2(2x-1) + 2(2x +1)

(2x+1) (2x-1) ]= 2x

x2 - 1 - [ 4x - 2 + 4x + 2(2x)2 - 1 ]

= 2xx2 - 1 -

8x4x2 - 1


= 2x (4x2 -1) - 8x (x2 -1)

(x2-1) (4x2 - 1)

= 8x3 - 2x - 8x3 + 8x

(x2-1) (4x2 - 1)

= 6x

(x2-1) (4x2 - 1)

Example 4 : Simplify : 1

x2 - 5x + 6 - 2

x2 - 4x + 3 + 1

x2 - 3x + 2

Solution : Here,

1

x2 - 5x + 6 - 2

x2 - 4x + 3 + 1

x2 - 3x + 2

= 1

x2-(3+2)x+6 - 2

x2-(3x+1)x+3 + 1

x2-(2+1)x+2

= 1

x2-3x-2x+6 - 2

x2-3x-x+3 + 1

x2-2x-x+2

= 1

x(x-3)-2(x-3) - 2

x(x-3)-1(x-3) + 1

x(x-2)-1(x-2)

= 1

(x-3) (x-2) - 2

(x-3) (x-1) + 1

(x-2) (x-1)

= (x-1) -2(x-2) +1 (x-3)

(x-1) (x-2) (x-3)

= x-1-2x+4+x-3

(x-1) (x-2) (x - 3)= 0

Example 5 : Simplify: 1

1-x+x2 - 1

1+x+x2 - 2x

1+x2+x4

Solution : Here,

1

1-x+x2 - 1

1+x+x2 - 2x

1+x2+x4

= (1+x+x2) - (1-x+x2)(1-x+x2) (1+x+x2) -

2x1+x4+x2

= 1+x+x2 - 1+x-x2

(1-x+x2) (1+x+x2) - 2x

1 + (x2)2 + x2


= 2x

(1-x+x2) (1+x+x2) - 2x

(1+x2)2 - 2.1x2 +x2

= 2x

(1-x+x2) (1+x+x2) - 2x

(1+x2)2 - x2

= 2x

(1-x+x2) (1+x+x2) - 2x

(1+x2+x) (1+x2-x)

= 2x

(1-x+x2) (1+x+x2) - 2x

(1-x+x2) (1+x+x2)

= 2x-2x

(1-x+x2) (1+x+x2)

= 0

(1-x+x2) (1+x+x2)= 0

Example 6 : Simplify 2

1+a + 1

a-1 + 3a

1-a2 + a

1+a3

Solution : Here,

2

1+a + 1

a-1 + 3a

1-a2 + a

1+a3

= 2

1+a -1

1-a + 3a

1-a2 + a

1+a3

= 2(1-a) - 1(1+a)

(1+a)(1-a) + 3a

(1+a)(1-a) + a

(1+a)(1-a+a2)

= 2-2a-1-a

(1+a)(1-a) + 3a

(1+a)(1-a) +a

(1+a)(1-a+a2)

= 1-3a+3a

(1+a)(1-a) + a

(1+a)(1-a+a2)

= 1

(1+a)(1-a) +a

(1+a)(1-a+a2)

= 1-a+a2+a(1-a)

(1+a)(1-a)(1-a+a2)

= 1-a+a2+a-a2

(1+a)(1-a)(1-a+a2)

= 1(1-a2) (1-a+a2)


Example 7 : Simplify: x2 - (y-z)2

(x+z)2 -y2 + y2-(x-z)2

(x+y)2-z2 + z2 - (x-y)2

(y+z)2 - x2 Solution : Here,

x2 - (y-z)2

(x+z)2 -y2 + y2-(x-z)2

(x+y)2-z2 + z2 - (x-y)2

(y+z)2 - x2

= (x-y+z) (x+y-z)(x+z+y) (x+z-y) +

(y+x-z) (y-x+z)(x+y+z) (x+y-z) +

(z+x-y) (z-x+y)(y+z+x) (y+z-x)

= x+y-zx+z+y +

y-x+zx+y+z +

z+x-yy+z+x

= x+y-z+y-x+z+z+x-y

x+z+y

= x+z+yx+z+y

= 1

Example 8 : Simplify: 1

x+1 + 1

x2+1 + 4

x4+1 + 8

x8-1Solution : Here,

Here we will first take the last two fractions, then preceding them and so on. The work will be complicated if we start with the first fraction.

Now,

1

x+1 + 1

x2+1 + 4

x4+1 + 8

x8-1

= 1

x+1 + 2

x2+1 + 4

x4+1 + 8

(x4 ) 2-12

= 1

x+1 + 2

x2+1 + 4

x4+1 + 4 (x4 - 1) + 8(x4+1) (x4-1)

= 1

x+1 + 2

x2+1 + 4 (x4 - 1) + 8(x4+1) (x4-1)

= 1

x+1 + 2

x2+1 + 4x4 - 4+8

(x4+1) (x4-1)

= 1

x+1 + 2

x2+1 + 4x4+4

(x4+1) (x4-1)


= 1

x+1 + 2

x2+1 + 4(x4+1)

(x4+1) (x4-1)

= 1

x+1 + 2

x2+1 + 4

(x4-1)

= 1

x+1 + 2

x2+1 + 4

(x2)2-12

= 1

x+1 + 2

x2+1 + 4

(x2+1) (x2-1)

= 1

x+1 + 2(x2-1) + 4

(x2+1) (x2-1)

= 1

x+1 + 2x2 - 2+4

(x2+1) (x2-1)

= 1

x+1 + 2x2 + 2

(x2+1) (x2-1)

Exercise (4.3)

= 1

x+1 + 2(x2+1)

(x2+1) (x2-1)

= 1

x+1 + 2

(x2-1)

= 1

x+1 + 2

(x+1) (x-1)

= x-1+2

(x+1) (x-1)

= x-1+2(x+1) (x-1)

= x + 1(x+1) (x-1)

= 1

x-1

1. Simplify:

a. 1

x-2 - 1

x2-4 b. 1

a-2 - 1

a2-4 c. 1

2x-3y - x+y

4x2-9y2

d. 1

a-b -2b

a2-b2 e. 1

x+2 +1

(x+1)2 f. x2

y(x-y)+y2

x (y-x)

2. Simplify:

a. a+1a-1 +

a2+1a+1 b.

a+4a+2 -

a-1a-2 c.

4aa2-1 -

a+1a-1 d.

a-4a+6 -

a-6a+4

e. a2+ab+b2

a+b - a2-ab+b2

a-b f. x-2x2-1 -

x+1x2-2x+1 g.

a+2a2+a-2 +

3a2-1

h. x

x2+2xy+y2 + y

x2 - y2 i. 1

(x-y)(x-z) + 1

(x-z) (y-z) j.1

(a-b)(b-c) + 1

(c-b)(a-c)


3. Simplify:

a. 2

(x-2)(x-3) + 2

(x-1)(3-x) + 1

(1-x)(2-x) b. 1

(x-3)(x+2) + 3

(x-2)(4-x) + 2

(x-3)(x-4)

c. 1

(x-2)(x-3) + 1

(x-2)(x-1) + 2

(x-3)(x-1) d. 2(a-3)

(a-4)(a-5) + a-1

(3-a)(a-4) + a-2

(5-a)(a-3)

e. x-1

(2x-1)(x+2) + 3

(x+2)(x-1) - 1

(1-x)(1-2x) f. 1

(x-3)(x+2) + 3

(x+2)(4-x) + 2

(x-3)(x-4)

4. Simplify:

a. 1

x2-5x+6 - 2

x2-4x+3 - 1

x2-3x+2 b. 1

a2-5a+6 - 2

a2-4a+3 - 1

a2-3a+2

c. x-1

x2-3x+2 +x-2

x2-5x+6 + x-5

x2-8x+15 d. 2a-6

a2-9a+20 - a-1

a2-7a+12 - a-2

a2-8a+15

5. Simplify:

a. x

x+2 + x

x-2 - 4x

x2-4 b. 2

a+b - 2

a-b + 4a

a2-b2 c. x+yx-y -

x-yx+y +

4xyx2+y2

d. a+ba-b -

a-ba+b -

2aba2-b2 e.

x+1x-1 +

x-1x+1 -

2x2

x2-1 f. x+yx-y +

x-yx+y +

2(x2-y2)x2+y2

6. Simplify :

a. 11 - x

-

11+ x

+ 1 - x

x

b. 2x + y - 1

x - y - 3

x-yx y-

c. 18(1 - )a

- 18(1 + )a

+ 28(1 - a)

a

d. 18( )x -1

+ 18( )x +1

+ 28(x - 1)

x

7. Simplify:

a. x+2

1+x+x2 - x-2

1-x+x2 - 2x2

1+x2+x4 b. 1

1+x+x2 - 2

1-x+x2 + 2x

1+x2+x4

c. 2x-y

4x2-2xy+y2 - 2x+y

4x2+2xy+y2 + 16x3

16x4+4x2y2+y4

d. 2a+b

4a2+2ab+b2 + 2a-b

4a2-2ab+b 2 - 2b3

16a4+4a2b2+b4

e. m-n

m2-mn+n2 + m+n

m2+mn+n2 - 2n3

m4+m2n2+n4


f. x-y

x2-xy+y2 + x+y

x2+xy+y2 - 2y3

x4+ x2y2+y4

g. 3x - 1

9x2-3x+1 - 3x + 1

9x2+3x+1 + 54x3

81x4+ 9x2 + 1

8. Simplify:

a. a2 - (b-c)2

(a+c)2-b2 + b2-(a-c)2

(a+b)2-c2 + c2-(a-b)2

(b+c)2-a2 b. (a-b)2-c2

a2-(b+c)2 +(b-c)2-a 2b2-(c+a)2 +

(c-a)2-b2

c2- (a+b)2

c. (x-y)2-z2

x2-(y+z)2 + (y-z)2-x2

y2-(z+x)2 + (z-x)2-y2

z2-(x+y)2 d. (a+2b)2-b2

(a+b)2-4b2 +(a-b)2-4b2

(a+2b)2-b2 + (2a+3b)2-b2

(2a+b)2-9b2

9. Simplify:

a. x3

x-1 - x2

x+1 - x

x-1 + 1

x+1 b. 1

x-a - 2

2x+a + 1

x+a - 2

2x-a

c. 1

x-5 - 1

x-3 + 1

x+5 - 1

x+3 d. 1a-3

- 1

a-1 + 1

a+3 - 1

a+1

e. a3

a-1 + a3

a+1 - 1

a-1 + 1

a +1 10. Simplify:

a. 1

1-x + 1

1+x - 2

1+x2 + 4

1-x4 b. 2

x-1 - 1

1+x - 2

x+1 - 3

x2-1

c. 3

x-1 - x

x2+1 - 2

x+1 - 5

x2-1 d. 1

1+a + 2

1+a2 + 4

1+a4 + 8

1+a8 - 16

1-a16

11. Simplify:

a. 4

x-1 - 2x

x2+1 - 2

x+1 + 6

1-x2 b. 1+b

a-b + 2ab

a2+b2 - a

a+b + 4a3ba4+b4

c.

1x+1 +

2x2+1 +

4x4+1 +

8x8-1 d. 1

8( )x-1 + 18( )x+1 + 8( )x+1

x2

d. x-y

(a+x)(a+y) + y-z

(a+y)(a+z) + z-x

(a+z)(a+x)

e. 1+ a

(x-a) + bx

(x-a) (x-b) + cx2

(x-a) (x-b)(x-c) + dx3

(x-a)(x-b)(x-c)(x-d)

f. 12 .

x-5x2-7x+10 +

12 .

x-6x2-9x+18 g.

1+p1-p +

4p1+p2 +

8p1-p4 -

1-p1+p


h. 1 + n

m-n + 2mn

m2+n2 – m

m+n + 4m3nm4+n4

12. Simplify :

a. 1

(a-b)(a-c) + 1

(b-a)(b-c) + 1

(c-a) (c-b) b. x+a

(a-b) (a-c) + x+b

(b-c) (b-a) + x+c

(c-a) (c-b)

c. 1

a(a-b)(a-c) + 1

b(b-a)(b-c) + 1

c(c-a)(c-b) d. a3

(a-b)(a-c) + b3

(b-c) (b-a) + c3

(c-a) (c-b)

Ans

wer

s (4

.3)

1. a. x+1x2-4

b. a+1a2-4 c. x + 2y

(2x+3y) (2x-3y) d. 1a+b e.

x2 + 3x + 3(x+2) (x+1)2 f. x2 + xy + y2

xy

2. a. a2 + 3aa2-1 b. a - 6

a2-4 c. 1 - a1 +a d. 20

(a+6) (a+4) e. -2b3

a2-b2 f. 1-5x(x-1)2 (x+1)

g. a +4a2-1 h. x2 + y2

(x-y) (x+y)2 i. 1(x-y) (y-z) j. 1

(a-b) (a-c)

3. a. 1(x-2) (x-3)

b. - (x+6)(x-3) (x+2) (x-2) (x-4)

c. 4(x-1) (x-3)

d. 5(a - 3) (a - 4) (a - 5)

e.

x + 4(2x-1) (x + 2)

f.

9

(x + 2) (x - 3) (x - 4)

4. a.

2(1-x) (x-2)

b. 2(1-a) (a-2)

c. 3x-7(x-2) (x-3)

d. 5(a - 3) (a - 4) (a - 5)

5. a.

2xx+2

b. 4a+b

c. 8x3yx4-y4

d. 2aba2-b2

e. 2x2-1

f. 4 (x4+y4)x4-y4

6. a. (1-x)3 x

b.

(y-x)9 x

c.

4(1-a)9 a

d.

4(x-1)9 x

7. a. 41+x2+x4

b. -11-x+x2

c. 2 (2x-y)4x2-2xy + y2

d. 2 (2a-b)4a2-2ab + b2

e. 2 (m-n)m2-mn + n2

f. 2 (x-y)x2-xy + y2

g. 2 (3x-1)

9x2-3x+1

8. a. 1 b. 1 c. 1 d. 3a2+4ab+9b2

(a-b) (a+3b)

9. a. x2 + 1 b. 6a2x(x2-a2) (4x2-a2)

c. 32x(x2-9) (x2-25)

d. 16 a(a2-9) (a2-1)

e. 2 (a2+1)

10. a. 41-x2

b. 2-xx2-1

c. 2xx4-1

d. 1a-1

11. a. 4xx4-1

b. 8a7ba8-b8

c. 1x-1

d. 0 e. x4

(x-a) (x-b) (x-b) (x-d) f. 2x-5

2 (x-2) (x-3)

g. 16 p1 -p4

h. 8m7nm8-n8

12. a. 0 b. 0 c. 1abc

d. a + b + c


4.4 IndicesMeaning of indicesBy general multiplication we know that,

a.a = a2

a.a.a = a3

a.a.a.a = a4

Similarly, a.a.a.a..........to m = am, where m is a positive integer. = am is read as "a to the power m". Here, 'a' is called the base and 'm' is called the index or power or exponent. Indices is the plural form of index.Laws of IndicesLaws of indices are certain verified rules which are used in the operation of Indices of algebric term. The verification of these laws have been already discussed in our earlier grades, however we are just reviewing these laws to solve the problems related to indices. These laws can be summarized in the table given below.

Law Operation1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

am×an = am+n, where m and n are positive integers

am ÷an = am-n, where m>n

am ÷an = 1

an-m , where m<n

(am)n = (an)m = am×n

(ab)m = ambm

ab( )m

= am

bm

a-m =1

am

a0= 1

amn=

man

a × b = ab

am × bm

= abm

x3× x2 = x3+2 = x5

x3 ÷x2 = x3-2 = x

x2 ÷x3 = x2-3 = x-1 = 1x

(x2)3= (x3)2 = x2×3 = x6

(xy)3 = x3y3

xy( )2

= x2

y2

x-2 = 1x2

x0= 1, (1000)0 = 1, (xy)0 = 1

x = x12

2 × 3 = 6

a3 × b3

= ab3


Example 1 : Evaluate :

a. 32243

( )-25 b. 127 × 286

217 × 166

a. Solution : Here, b. Solution : Here,

32243

( )-25

127 × 286 217 × 166

=

25

35( )-25

=

(4×3)7 × (7×4)6

(7×3)7 × (4×4)6

=

23

( )5 ×-25

=

47 × 37 × 7 6 × 46

77 × 37 × 412

= 2

3( )-2 = 413 - 12

77 - 6

= 94 =

47

Example 2 : Simplify : (27)3x + 1 (243)9x+5. 33x-7

-4x5

Solution :

= (33)3x + 1 (35)(32)x+5. 33x-7

-4x5

= 39x + 3. 3-4x

32x + 10. 33x - 7

= 39x + 3 -4x

32x + 10 + 3x - 7

= 35 x + 3

35x + 3

= 35x + 3 - 5x - 3

= 30 = 1

Example 4 : Simplify: a. 1

1 - ax - y + 1

1 - ay-x b. (125x3 ÷ 27y-3)-23

c. 27a33 × 625a4×b44

Worked Out Examples

Example 3 : Simplify: 49x. 7x-1 - 7x

72x+1 . 7x-2 - 7x Solution : Here,

= 49x. 7x-1 - 7x

72x+1. 7x-2 - 7x

= 72x. 7x-1 - 7x

72 x +1 . 7x-2 - 7x

= 73x - 1 - 7x

73x - 1 - 7x

= 1


a. Solution :

= 1

1 - ax - y + 1

1 - ay-x

= 11- ax

ay

+ 11- ay

ax

= 1ay - ax

ay

+

1ax - ay

ax

=ay

ay - ax + ax

ax - ay

=ay

ay - ax -ax

ay - ax

= ay - ax

ay - ax

= 1

Example 5 : Find the value of : 6n+2 + 7×6n

6n+1 × 8-5 × 6n Solution :

= 6n+2 + 7×6n

6n+1 × 8-5 × 6n

= 6n. 62 + 7 × 6n

6n .61 × 8 - 5× 6n

= 6n(62 + 7)

6n (6× 8 - 5)

= 36 + 748 - 5

= 4343

= 1Example 6 : Simplify:

1xb( )xa.

a2+ab+b2 . 1

xc( )xb.b2+bc+c2

. 1xa( )xc.

c2 + ca + a2

Solution: Here,

= 1xb( )xa.

a2+ab+b2 . 1

xc( )xb.b2+bc+c2

. 1xa( )xc.

c2 + ca + a2

= (xa-b)a2+ab+b2. (xb-c)b2+bc+c2 . (xc-a)c2+ca+a2

b. Solution : Here,

= (125x3 ÷ 27y-3)-23

= 125x3y3

27( )

23

-

= 27

125x3y3( )23

= 35xy

( )2

=9

25x2y2

c. Solution : Here,

27a33 × 625a4×b44

= 33a33 × 54a4 × b44

= (3a)33 × (5ab)44

= 3a × 5ab

= 15a2b


= x(a-b) (a2 + ab + b2) .x(b-c) (b2 + bc + c2) . x(c-a) (c2 + ca + a2) = xa3-b3

. xb3-c3 . xc3-a3

= xa3-b3 + b3 - c3 + c3-a3

= x0 = 1Example 7 : Simplify: xa2-b2a+b

. xb2-c2b+c .

xc2- a2c+a

Solution:

= xa2-b2a+b . xb2-c2b+c

.

xc2- a2c+a

= xa2 - b2

a + b . x b2 - c2

b + c . x c2 - a2

c + a

= x(a+b)(a-b)

(a+b) . x (b+c)(b-c)

(b+c) . x(c+a)(c-a)

(c+a)

= xa-b.xb-c.xc-a

= xa-b+b-c+c-a

= x0

= 1Example 8 : Simplify: ( )1 -m+(mn2)

13 +(m2n)

13

m - n× n

13

m13Solution:

= ( )1 -m+(mn2)13 +(m2n)

13

m - n× n

13

m13

= m13 - n

13

m13

m + m13 .n

23 + m

23 n

13

m - n×

= m13 - n

13

m13

m13 . m

23 + m

13 n

23 + m

23 n

13

m - n×

= m13 - n

13

m13

m13 ( m

23 +n

23 + m

13 n

13)

m - n×

= 2+ m

13 n

13 +[( )m

13 ( )2

n13

(m13 - n

13 )

m - n

=

3-( )m

13 ( )3

n13

m - n

Alternative process

Let, m13 = a, m = a3, m

23 = a2

n13 = b, n = b3, n

23 = b2

( )1 -m+(mn2)13 +(m2n)

13

m - n× n

13

m13

= m13 - n

13

m13

m + m13 .n

23 + m

23 n

13

m - n×

= a3 + ab2 + a2 ba3 - b3 ( )a - b

a

= a(a2 + b2 + ab)

(a - b) (a2+ab+b2) (a - b)

a

= 1


= m - nm - n

= 1 Example 9 : Simplify:

11+xa-b + xc-b +

11+xb-c + xa-c +

11+xc-a + xb-a

Solution: Here,

= 1

1+xa-b + xc-b + 1

1+xb-c + xa-c + 1

1+xc-a + xb-a

= 11 + xa

xb + xc

xb

+

11 + xb

xc + xa

xc

+ 11 + xc

xa + xb

xa

= xb+xa+ xc

xb

1

+

xc+xb+ xa

xc

1

+ xa+xc+ xb

xa

1

= xb

xb + xa + xc + xc

xc + xb + xa + xa

xa + xc + xb

= xb+xc + xa

xa + xb + xc

= xa+ xb + xc

xa + xb + xc

= 1 Example 10 : If x = 3

13 + 3

23 , prove that x3-9x-12=0

Solution: Here, x = 313 + 3

23

Cubing on both sides, we get

x3 = (313 + 3

23 )3

or, x3 = (313 )3 +3. 3

13 . 3

23 (3

13 + 3

23 ) + (3

23 )3 [(a+b)3=a3+3ab(a+b)+b3]

or, x3 = 3 + 3.31+23 . x + 32

or, x3 = 3 +3.3.x +9

or, x3 = 3 + 9x + 9

or, x3 = 9x + 12

or, x3 - 9x - 12 = 0 Proved.

Example 11 : Simplify p2

(p-y)y -

2p(p-y)y-1

+

1(p-y)y-2


Solution : Here, p2

(p-y)y -

2p(p-y)y-1

+

1(p-y)y-2

= p2

(p-y)y - 2p

(p-y)y(p-y)-1 +

1(p-y)y(p-y)-2

= p2

(p-y)y - 2p(p-y)(p-y)y

+ (p-y)2

(p-y)y

= p2-2p(p-y)+(p-y)2 (p-y)y

= {p-(p-y)}2

(p-y)y

= y2

(p-y)y

Example 12 : If xyz +1 =0, prove that 11-x-y-1

+

11-y-z-1

+

11-z-x-1 = 1

Solution: Here, xyz +1 = 0

or, xyz = -1

or, x = -1yz

and 1x

= -yz i.e. x-1 = -yz

Now,

L.H.S. = 11-x-y-1

+

11-y-z-1

+

11-z-x-1

= 11 + 1

yz - 1y

+ 11 - y -

1z

+

11 - z+yz

= 1yz+1-z

yz

+ 1z-yz-1

z + 1

1-z+yz

= yzyz+1-z

+

zz-yz-1

+

11-z+yz

= yzyz+1-z

-

zyz+1-z

+

1yz+1-z

= yz-z+1yz+1-z

= yz+1-zyz+1-z

= 1 = R.H.S. Proved.


Exercise (4.4)

1. a. What index of x will be 1x4 ?

b. What should be the index of x so that its value will be equal to 1 ?

c. What should be the index of x so that its value will be equal to 1x3 ?

d. What is the rational value of ( 5-1 )2 ?

2. Find the value of :

a. (27)-13 b. (125)

-23 c.

162401[( )

-38 ]

43

d. (0.5)-2

e. x15 ÷(x4 × x3) f. x9

x3 g. 125×285

216×167 h. 35×274×94

815×33×9-3

i. 1632 - 3(4)0 j. 16

32 - 3(4)0 + 2 (4)1

2

3. Evaluate:

a. 12564[( )- 1

3 ]2

b. -125

27( )13

÷

-259( )

12

c. 12564[( )- 1

3 ]2

÷

-254( )

12

d.

-827( )

13 ×

-8116( )

14

÷

-32243( )

15

e. -27

8( )13

8116[ ]1

4 ÷ ( ) -425( )

12 f.

-2516( )

12

12564[ ]1

3 ÷ ( ) -827( )

13

g. 729

64[ ]( )3

-1

4. Simplify:

a. (64x3 ÷ 27a-3)-23 b. (125a3 ÷ 27b-3)

-23

c. (81a4 ÷ 16b-4)-34 d. a6b-2c4 ÷ a4b-4c84

e. 27a12b33 ÷ 16a8b44

f. 9x2y73 × 3x5y-23

g. 3x7y11z-13 × 72x-1yz43 h. (x+y)-83

. (x+y)23


i. (x+y)-83 . (x+y)

13 j. (a + b)-35

÷ (a+b)25

k. (x-3 × y23 × z

76 )

-12 ÷ (x

-92 × y × z

-54 )

-13

5. Simplify:

a. 4x+2 - 4x

4x b. 3a+2+ 3a

5.3a c. 2x+4 - 2x

5.2x d. 5x+2 - 5x

5x+1 + 5x

e. 2x ×3-2x

2x+2-2x+1 f. 5

m + 5m+1

5m+1 +5m g. 4m+4m+1

4m+2 - 4m h. 6n+2-6n

6n+1+6n

i. 2x+4 - 2x

5.2x j. 5n+2-2.5n

23.5n k. 5n+2 - 16×5n

3 × 5n l. 33a+2 - 33a+1

6 ×27a

m. 132x+1 - 5 × 169x

9 × 169x n. 112x+1 - 6 × 121x

5 × 121x o. 7a+1 + 9 × 7a

7a+2 - 45 × 7a p. 9n+2 + 10 × 9n

9a+1 × 11 - 8x9n

q. 11n+2 - 55.11n

11n × 112 r. 273n-1 (243)-

4n5

9n+1 × 33n-5s. 7n+2 - 35.7n-1

7n × 11 t.

6n+2-36.6n-1

6n × 30

u. 9a+2 - 45.9a-1

9n × 76

6. Simplify:

a. 25p2q5r-33 × 40p-5qr93 b. 9a-2 b53 × 24a5b43

c. 9x-2y73 × 3x5y-13 d. 8a2b34 × 162a2b4

e. 54a8y53 ÷ 2a-1y23 f. 243a-6x-54 ÷ 3a2x-94

g. (a+b)-1 × (a+b) (a - b)2 h. (x+y)-1 × (x+y) (x - y)2

7. Simplify :

a. xb-c × xc-a × xa-b b. (ax+y)x-y × (ay+z) y-z × (az-x)z+x

c. xa+b × xa-b × xc-3a

xc-a d. ya

za( )b. zb

xb( )a. xa

ya( )b

e. x-1y-1+y-1z-1+z-1x-1

x+y+z f. a2 + b2 +c2

a-2b-2+b-2c-2+c-2a-2

8. Simplify:

a. xa

x-b( )a-b× xb

x-c( )b-c× xc

x-a( )c-a

b. (xm2 -n2)

1m+n × (xn2 -p2

)1

n+p × (xp2 -m2)

1p+m


c. [(ax. ay)x-y ay

xz( )y+z ] × az

ax( )z+x

d. (ax ÷ ay)x2 + xy + y2. (ay ÷ az)y2+yz+z2. (az ÷ ax)z2 + zx + x2

e. (xa ÷ xb)a2+ab+b2 × (xb ÷ xc)b2+bc+c2 × (xc ÷ xa)c2+ca+a2

f. 1xb( )xa ×

a2 + ab + b2

× 1xc( )xb ×

b2 +bc + c2

× 1xa( )xc ×

c2 + ca + a2

g. xa+b

xc( )a-b× xc+a

xb( )c-a×

xb+c

xa( )b-c

h. xl

xm( ) × xm

xn( ) × xn

xl( ) i. (ax+y)2 . (ay+z)2 . (az+x)2

(ax.ay.az)4

j. xa2+b2

xab( )a-b× xb2+bc

x-c2( )b-c×

xc2+ac

x-a2( )c-a

k. xb

xc( )b+c-a× xc

xa( )c+a-b× xa

xb( )a+b-c

l. xb+c

xc-a( )a-b× xc+a

xa-b( )b-c×

xa+b

xb-c( )c-a

m. xa

x.x-b( )a-b× xb

x.x-c( )b-c× xc

x.x-a( )c-a

n. xb

xc( )1bc

× xc

xa( )1ca

× xa

xb( )1

ab

o. x

bc

xcb

1bc

× x

ca

xac

1ca

× x

ab

xba

1ab

p. 1

x-y( )a1

x-z ×

1y-z( )a

1y-x

× 1

z-x( )a1

z-y

q.

13

p-qp + (pq2) + (p2q)

13

× q

13

p13

1-


r.

13

m-nm+(mn 2) (m2n)

13

× n13

m13

1-

9. Simplify :

a. a

az2

y2

y2

z2

y2z2 × a

ax2

z2

z2

x2

z2x2 × a

ay2

x2

x2

y2

x2y2

b. xa2-b2a+b × xb2-c2b+c × xc2-a2c+a c. ax

ayxy

× ay

azyz

× az

axzx

d. a

azy

yz

yz × a

axz

zx

zx × a

ayz

xy

xy

e. a

a1b

1a

1ab

× a

a1c

1b

1bc

× a

a1a

1c

1ca

10. Simplify :

a.

mn( )1 + n

m( )1 -m

m-n

mm-n

nm-n

nm-n

nm( )+ 1 m

n( )- 1

b.

xy( )1 + y

x( )1 -x

x-y

xx-y

yx-y

yx-y

yx( )+ 1 x

y( )- 1

c.

1y2( )xx2- 1

y( )y-xx -

1x( )x-yy +1

x2( )yy2-

d.

1b2( )aa2- 1

b( )b-aa -

1a( )a-bb -1

a2( )bb2-

e.

1x( )x+yy + 1

y( )x+yx -

1x2( )yy2-1

y2( )xx2- f.

1b( )a+ba + 1

a( )a+bb -

1b2( )aa2-1

a2( )bb2-

g.

1b( )x a + 1

b( )y- a

1a( )y - b1

a( )x b + h.

1y( )a -x 1

y( )ax +

1x( )a- y1

x( )a y +

11. a. a2

(a-b)y -

2a(a-b)y-1

+ 1(a-b)y-2

b.

p2

(p-y)y -

2p(p-y)y-1

+ 1(p-y)y-2


12. Simplify:

a. 11+xa-b+xc-b

+ 11+xb-c+xa-c

+

11+ xc-a+xb-a

b. 11+xa-b+xa-c

+ 11+xb-c+xb-a

+

11+ xc-a+xc-b

c. 11+ xp-q +xr-q

+ 11+xq-r+xp-r

+

11+xr-p+xq-p

d. 11+ax-y+ax-z

+ 1a+ay-x+ay-z

+

11+az-x+az-y

13. a. If a+b+c=0, prove that:

11+xa+x-b

+ 11 + xb + x-c

+

11+ xc+ x-a = 1

b. If x3+y3+z3=1, Show that:

ax

a-y( )x2-xy+y2. ay

a-z( )y2-yz+z2. az

a-x( )z2-zx+x2 = a2

c. If x2+y2+z2=xy+yz+zx, Show that:

ax

ay( )(x-y). ay

az( )(y-z). az

ax( )(z-x) = 1

d. If pqr=1, Show that:

1

1 + p + q-1 + 1

1 + q + r-1 +

11 + r + p-1 = 1

14. a. If x = p13 - p

-13 then prove that x3 + 3x = p - p-1

b. If x = a13 - a

-13 then prove that x3 + 3x = a- a-1.

c. If x = 213 - 2

23 then prove that x3 - 6x = 6

d. If x2 - 2 = 223 + 2

-23 then prove that 2x3 - 6x = 5

e. If 2x = 3y = 12z then prove that 1z

- 1y

= 2x


Ans

wer

sExercise 4.4

1. a. -4 b. 0 c.-3 d. 125

2. a. 13

b. 125

c. 494

d. 10.25

e. x8

f. x6 g. 15376

h. 6561 i. 61 j. 65

3. a. 1625

b.1 c. 45

d. 23

e. 25

f. 23

g. 23

4. a. 916a2x2

b. 925a 2b2

c. 827a3b3

d. a2 e. 32

a2

f. 3x73 y

53 g. 6x2 y4 z h.

1x + y i. 1

3(x+y)7 j. 1

a+b k. 1

z5. a. 15 b. 2 c. 3 d. 4 e. 1 f. 4 g. 1

3 h. 5 i. 3 j. 1

k. 3 l. 1 m. 89

n.1 o.4 p. 1 q. 12

r. 1 s. 4 t. 1 u.1

6. a. 10q2r2

p b. 6ab3 c. 3xy2 d. 6ab e. 3a3y f. 3x

a2 g.a-b h. x-y

7. a. 1 b.1 c.1 d.1 e. 1xyz

f. (abc)2

8. a. 1 b. 1 c. a2 (z2-y2)

d.1 e. 1 f. 1 g. 1 h. 1 i. 1 j. 1 k.1 l. 1 m. 1 n. 1 o. 1 p. 1 q. 1 r.1 9. a.1 b.1 c.1 d.1 e.110. a. m

n b. x

y c. x

y( )x + y d. a

b( )a + b e. x

y( )y - x

f. ab( )b-a

g. ab( )x + y

h. xy( )2a

11. a. b2

(a-b)y b. y2

(p-y)y

12. a. 1 b. 1 c. 1 d. 1


Example 1. Solve : 52x+1 = 252x-1

Solution : Here, 52x+1 = 252x-1

or, 52x+1 = (52)2x-1

or, 52x+1 = (5)4x-2

Since the bases are same,

∴2x+1 = 4x-2

or, 4x-2x = 3

or, 2x = 3

∴ x = 32

Example 3 : Solve : 2x+3.3x+4 = 18Solution : Here,

2x+3.3x+4 = 18

or, 2x.23.3x.34 = 18

or, 2x.3x.8.81 = 18

or, (2×3)x = 188×81

or, 6x = 136

or, 6x = 162

Worked Out Examples

4.5 Exponential Equation

The equation of the form, ax = b, where 'a' is base and 'x' is the index or exponent is called exponential equation.

22x+1 = 16, 92-x = 81 are few examples of exponential equations.

Solving exponential equations

Exponential equations can be solved by using the axioms.

“If ax = ab, then x must be equal to b”. Therefore, we can convinently solve exponential equation if we follow the steps given below:

a. Firstly, simplify the given equation so that both the sides of the equation have the same base.

b. Use the above axiom (ax = ab), to equate their exponents and powers.

Example 2 : Solve : 3x+2+3x+1 = 1 13

Solution :

3x+2+3x+1 = 1 13

or, 3x.32+3x.31 = 43

or, 3x(32+3) = 43

or, 3x(9+3) = 43

or, 3x.12 = 43

or, 3x = 43

× 112

or, 3x = 13 × 1

3

or, 3x = 19

or, 3x = 3-2

∴ x = – 2


or, 6x = 6-2


∴ x = –2Example 4 : Solve : ( 2 )x+4 = ( 2

3)2x+5

Solution : Here, ( 2 )x+4 = ( 23 )2x+5

or, 2x+4

2 = 22x+5

3

Since the bases are same,x+4

2 = 2x+5

3

or, 3(x+4) = 2(2x+5)

or, 3x+12 = 4x+10

or, 4x –3x + 10 - 12 = 0

or, x - 2 = 0

∴ x = 2

Example 6 : Solve : 4x – 6.2x+1+32 = 0Solution: Here,

or, 4x – 6.2x+1+32 = 0

or, (22)x – 6.2x.21+32 = 0

or, (2x)2 – 12.2x+32 = 0

Let, 2x = a

∴ a2 – 12a+32 = 0

or, a2– (8+4)a+32 = 0

or, a2–8a–4a+32 = 0

or, a(a–8) –4(a–8) = 0

or, (a–8)(a– 4) = 0

Either, a–8 = 0

∴a = 8

or, a–4 = 0

∴a = 4

The required values of x are 2 and 3.

Example 5 : Solve : 23x-5.ax-2 = 2x-2.a1-x

Solution : Here,

or, 23x-5.ax-2 = 2x-2.a1-x

or, 23x-5.ax-2

2x-2a1-x = 1

or, 23x-5-x+2 .ax-2-1+x = 1

or, 22x-3 . a2x-3 = 1

or, (2a)2x-3 = (2a)0


or, 2x-3 = 0

or, 2x = 3

∴ x = 32

Now, Putting the value of a,

a = 8

or, 2x = 8

or, 2x = 23


∴ x = 3

Again,

a = 4

or, 2x = 4

or, 2x = 22


∴ x = 2


Example 7 : Solve : 5x + 15x = 25 1

25

Solution: Here, 5x + 1

5x = 25 125

or, 5x + 15x = 626

25

Let, 5x = a

∴ a + 1a

= 62625

or, a2+1a

= 62625

or, 25(a2+1) = 626a

or, 25a2– 626a +25 = 0

or, 25a2– (625+1)a +25 = 0

or, 25a2– 625a-a +25 = 0

or, 25a(a–25) –1(a – 25) = 0

or, (a–25)(25a –1) = 0The required values of x are 2 and –2.

Either, a–25 = 0

∴ a = 25

or, 25a–1 = 0

∴ a = 125

Now, restoring the value of a,

a = 25

or, 5x = 25or, 5x = 52

Since the bases are same,∴ x = 2Again,

a = 125

or, 5x = 5-2

Since the bases are same,∴x = –2

1. a. If 3y = 1 , what is the value of y ? b. If 53x = 1, what is the value of x ?

c. If 52x = 125

, what is the value of x ?

d. If 3b = 127-1 , what is the value of b ?

e. If 6a = 1(4×9)-3 , what is the value of a ?

2. a. If ( 5p ÷ 52 ) 53 = 55, what is the value of p ? b. If( ap ÷ aq ) ar = ax, then express x in terms of p,q and r ? c. If a-l ÷ am = an × a-p, then express p in terms of l,m and n ?3. Solve a. 2x-4 = 4x-6 b. 9x-1 = 3x+1 c. 42x-1 = 2x+1 d. 32x+1 = 92x-1 e. 4x+1 = 1

8x f. 4x-1 = ( 2 )x

Exercise (4.5)


g. ( 2 )3x-1 = ( 4 )x-2 h. ( 9 )x-3 = ( 3 )x+2 i. 3x+1 - 3x = 54

j. 25x+3 = 10.04

k. 103y-3 = 10.01

l. (0.5)x2 = 0.25

m. 3x+2 + 3x = 109

n. 32x+3– 2.9x+1 = 19

o. 5x+5x+1+5x+2 =155 4. Solve:

a. 2x-3× 3x-4 = 3-1 b. 23x-5ax-2 = 2x+2a5-x c. 5x-3× 32x-8 = 225

d. 2x-5× 5x-4 = 5 e. 72x+1× 52x-1 = 75

5. a. If ax=b,by=c and cz = a, prove that xyz = 1 b. If x=yz,y=zx and z = xy, prove that xyz =1 c. If ax=by and b=a2, show that x-2y = 0 d. If a=10x, b=10y and aybx=100, show that xy =1 e. If xa.xb=(xa)b, prove that a

b + b

a= – 1

f. If xa=yb=zc and y3 = xz, show that 3b

= 1a

+

1c

g. If ap = bq = cr and b2= ac, prove that q = 2prp+r

h. If 2x = 3y= 12z, show that 1z

= 1y

+

2x

i. If ax

= by

= cz and abc = 1, prove that x+y+z=0

6. Solve: a. 2x + 2x+2 = 5 b. 2x+1 – 2x = 8 c. 2x+3 + 2x = 36

d. 2x+2 + 2x+3

2= 1 e. 3x+2 + 3x+1 = 1 1

3 f. 7x + 343

7x = 56

g. 5.4x+1–16x = 64 h. 4.3x+1–9x = 27 i. 4x + 14x = 16 1

16

j. 3x+3 + 13x = 28 k. 3x+2 + 1

3x-2 = 30 l. 3x + 13x = 9 1

9 m. 2x + 1

2x = 4 n. 4x + 14x = 257

16 o. 5a + 1

5a = 25 1

25

p. 7x + 7-x = 7 17

q. 4x+2 + 4x+1 = 1 14

r. 4x+128 = 6.2x+2 s. 3.2x+1– 4x = 8 t. 2x-2+23-x = 3 u. 22x+3.2x– 4 = 0 v. 4x-6.2x-1- 32 = 0 w. 3×2y+1– 4y = 8 x. 5x+1 = 26 – 51-x

y. 5x+1+52-x = 126 7. Show that the solution of the equation 4 × 3x+1 – 9x = 27 also satisfy the equation

32x – 4 × 3x+1 + 813

= 0


Ans

wer

s (4.

5)1.a. o b. o c. -1 d. 3 e. 6 2.a. 4 b. p-q-r = x c. l+m+n = p 3.a. 8 b. 3 c. 1 d. 3

2

e. -25

f. 43

g. -3 h. 8 i. 3 j. -2 k. 53

l. 4 m. -2 n. -2 o. 1 4.a. 3 b. 72

c. 5

d. 5 e. 0 6.a. 0 b. 3 c. 2 d. -3 e. -2 f. 1,2 g. 1,2 h. 1,2 i. 2,-2 j. 0,-3 k. -1,1 l. 2,-2m.2,-2 n. 2,-2 o. 2,-2 p. 1,-1 q. -2 r. 3,4 s. 1,2 t. 2,3 u. 1 v. 6 w. 1,2 x. 1,-1 y. 2, -1

4.6 Rational Number

A rational number is a number that can be expressed as a fraction or ratio in the form pa

, where q ≠ 0 and p and q both are integers. When the fraction is divided out, it becomes a terminating or recurring decimal.

The examples of rational numbers are:

6 or 61

can be written as 6.0

-2 or can be written as -2.0

12

can be written as 0.5 (Terminating)

- 54

can be written as -1.25 (Terminating)

23

can be written as 0.66666...

= 0.6 (Non Terminating but recurring)

2155

can be written as 0.381818...

= 0.318 (Non Terminating but recurring)

5383

can be written as 0.62855421687...(Non Terminating but recurring)

Irrational numbers

Irrational number is a part of rational number that cannot be expressed as a fraction. Irrational numbers cannot be represented as terminating or repeating decimals. Irrational numbers are non-terminating or non repeating decimals.

Examples of Irrational numbers are

p = 3.141592654......


2 = 1.414213562......and 0.12122122212......Here, the major confusion may arise with some students whether p is the terminating decimal 3.14 but it is not, However while solving certain Mathematical problem it is customary to round off the value of p to 3.14 to make calculation convenient. p actually is a non-terminating decimal and is irrational number.Surds

A surd is a root of rational number of which the exact value cannot be determined.

Example: 3 , 73 , 1025 etc.

But, 9 , 83

are not surd because in 9 and 83 the radical sign (√) can be removed

and exact value 3 and 2 can be obtained.

In a surd, ba , a is called order, b is called radicand and √ is called radical sign.

Entire(Pure) or Mixed Surd

A surd is said to be entire or pure surd if it does not contain rational factor or term other than 1 and -1. For example, 5

3 , 18 , 3 , - 3 .

A surd is said to be mixed surd which has a rational factor other than 1 or -1. 3 5 , 23 , 3 7 - 5 8 are some example of mixed surd.

Simple and Compound Surds : A Surd consisting of a single term is called a simple surd. For example: 15 , 18

3 etc. On the other hand a surd consisting of the algebric

sum of two or more term is called a compound surd. For example, 2 + 3 , 5 + 2 7 etc.

Like and Unlike Surds

Surds having the same order and equal radicands such as 15 35 , 14 3

5 , - 2 35 etc.

are like surds. On the other hand, the surds which are not like are unlike surds. For example: 8 12

3 , 3 8 3 , -4 25, 28 etc. are unlike surds.

Operation of surdsAddition and subtraction of surds

We can add or subtract only like surds. To add or subtract the like surds, their rational factors or coefficient should be added or subtracted as in the case of algebraic terms with the same base and powers.

Examples

i. Add the following surds: 3 5 , 7 5 and 5 .


Solution : Here,

= 3 5 + 7 5 + 5

= (3 + 7+1) 5

= 11 5

iii. Simplify: 5 8 + 6 81 3 - 2 50 - 3 243

Solution: Here,

5 8 + 6 81 3 - 2 50 - 3 243

= 5 4×2 + 6 27×33 - 2 2×25 - 3 3 × 8

3

= 5×2 2 + 6×3 33 - 2×5 2 - 3×2 3

3

= 10 2 + 18 33 - 10 2 - 6 3

3

= 0 + 12 33

= 12 33

Multiplication and Division of Surds

Two surds can be multiplied or divided only when they have the same order. To multiply or divide surds, we should multiply and divide the rational factors or coefficient as well as radicand.

Examples:

a. Multiply i. a × b × c ii. 5 2 × 7 3

Solution:

i. a × b × c = abc ii. 5 2 × 7 3 = 5×7 2×3 = 35 6

b. Divide:

i. xy by x ii. 904

by 304

Solution:

i. xyx

= xyx

= y ii. = 904

304

= 9030

4= 3

4

ii. Subtract 3 12 from 4 27Solution: Here,

4 27 - 3 12 = 4 9×3 - 3 4×3 = 4×3 3 - 3×2 3 = 12 3 - 6 3 = (12-6) 3

= 6 3

Worked Out Examples Example 2 : Subtract : 3 27 - 12 Solution : Here,

= 3 9×3 - 4×3

= 3 × 3 3 - 2 3

= 9 3 - 2 3

= 7 3

Example 1 : Add 3 3 + 7 3 + 3 Solution : Here,

3 3 + 7 3 + 3

= 11 3


Example 3 : Multiply : 4 2 × 3 3 Solution : Here,

4 2 × 3 3

= 4 × 3 2×3

= 12 6

Example 5 : Simplify: 403 - 135

3 + 2 3203

Solution: Here,

403 - 135

3 + 2 3203

= 8×53 - 27×53 + 2 64×53

= 2 53 - 3 5

3 + 2×4 53

= 2 53 + 8 5

3 - 3 53

= 10 53 - 3 5

3

= 7 53

Example 6 : a. Evaluate: 5 12

- 6 18

+ 10 150 b. (a-b)114

. (a-b)- 3

4

a. Solution : Here,

5 12

- 6 18

+ 10 12

= 5 12

- 6 14×2

+ 10 125×2

= 5 12

- 62

12

+ 105

12

= 5 12

- 3 12

+ 2 12

= 7 12

- 3 12

= 4 12

Example 4: Divide : 903

303

Solution : Here,

= 903

303

= 9030

3

= 33

b. Solution: Here,

(a-b)114 . (a - b)

-34

= (a - b)114 . (a - b)

-34

= (a - b)114

-34

= (a - b)11 -3

4

= (a - b)84

= (a - b)2.

Exercise (4.6)

1. a. Are the square roots of all positive integers irrational? If no give three examples.

b. Are the cube roots of all positive integers irrational? If no give three examples.


c. What type of number is 3 ?

2. Write the order of following surds.

a. 13 b. 752 c. 87

3 d. 8154 e. 421

6

3. Express the following surds into mixed surds.

a. 18 b. 1502 c. 54

3 d. 5124 e. 128

6

4. Express the following into pure surds.

a. 3 3 b. 3 72 c. 4 36

3 d. 5 184 e. 3 2

3

5. Express the following surds in the same order of the surds.

a. 2 73 and 6

4 b. 23 , 3

4 and 76

6. Add :

a. 9 2 + 3 2 b. 8 5 + 6 5 c. 3 18 + 3 32 d. 3 33 + 3

3 e. 4 74 + 5 7

4

7. Subtract :

a. 10 3 - 3 3 b. 9 5 - 6 5 c. 9 18 - 3 32 d. 9 33 - 3

3 e. 14 74 -5 7

4

8. Multiply :

a. 11 4 × 3 b. 9 5 ×6 4 c. 9 18 × 3 32 d. 9 33 × 4 4

3 e. 14 74 × 5

9. Divide :

a. 15 4 ÷ 3 4 b. 20 5 ÷ 6 25 c. 9 18 ÷ 3 32 d. 9 543 ÷ 3 27

3

10. Simplify :

a. 27 + 75 - 6 3 b. 4 27 + 3 75 - 4 3 c. 9 1283

+ 4 543

- 4 2503

11. Simplify :

a. ( 3 + 2 ) ( 3 - 2 ) b. ( 7 + 5 ) ( 7 - 5 ) c. (3 7 + 5 ) (3 7 - 5 )

d. ( 3 + 2 )2 e. (3 3 - 2 2 )2 f. ( 2+x + 2-x )2

g. ( 3+y + 3-y )2

12. Simplify :

a. a2-b2

a+b b. a2-36

a-6 c.

25-x23

5+x3 d.

x + yx + y

x - yx - y

+-

e. a + ba + b

a - ba - b

+-

13. Simplify :

a. 322

2+ b. 1473

27+ c. 54 - 96 + 5 216

1283

6

52 d. 81

316

3

233

-


Ans

wer

s (4.

6)1.a. consult to your teacher b. consult to your teacher c. irrational 2 a. 2 b.2 c.3 d.4 e.6 3 a. 3 2 b. 5 6

2 c. 3 23 d. 2

4 e. 2 26

4.a. 27 b. 63 c. 23043 d. 11250

4 e. 543

5a. 2612 , 7412 , 6312 b. 2412 , 3312 , 7212

6a. 12 2 b. 14 5 c. 21 2 d. 4 33 e. 9 7

4

7a. 7 3 b. 3 5 c. 15 2 d. 8 33 e. 9 7

4 8a. 22 3 b. 108 5 c. 648 d. 36 2

3 e. 70 74

9a. 5 b. 43

c. 94

d. 3 23

10a. 2 3 b. 23 3 c. 28 23

11.a. 1 b. 2 c. 58 d. 5+2 6 e. 35-12 6

f. 2(2+ 4 -x2) g. 2(3+ 9 -y2)

12a. a - b b. a + 6 c. 5 - x3 d. x2 - y2x +

y e. a2 - b2a +

b

13a. 5 b. 10 c. 54 + 24 68 63

d. 33

23

- 23

4.7. Rationalization of Surds:

Let us consider the surd . Here, is an irritional number. But when we multiply 2 by 2 then we get 2 × 2 =( 2 )2 = 2 as a rational number. This process of changing a

surd into a rational number is called rationalization. Look at the example given below carefully to have a clear concept of rationalization.

Examples: Rationalize the following:

i. 3 7

ii. 5 2x + 5

Solution : Here,

i. 3 7 = 3 7 × 7 = 3×7 =21

ii. 5 2x + 5 = 5 2x + 5 × 2x + 5 = 5 (2x+5)

In the above example, 3 7 and 2x + 5 when multiplied with given surds gives a rational number. Thus, 7 and 2x + 5 are called rationalizing factor of the given problem.


Example 1 : Rationalize the denominator of : 4 + 3 2

3 - 2 2

Solution:

4 + 3 2

3 - 2 2

= 4 + 3 2

3 - 2 2×

3 + 2 2

3 + 2 2[Multiplying both numerator and denominator

by 3 + 2 2 ]

= (4 + 3 2 ) (3+2 2 )

32 - (2 2 )2

= 4(3+ 2 2 )+3 2 (3 + 2 2 )9 -4 × 2

= 12 + 8 2 + 9 2 + 6 ×29 - 8

= 12 + 17 2 + 121

= 24 + 17 2

Worked Out Examples

Conjugate

Conjugate of a+b is a-b and conjugate of a-b is a+b, thus a+b and a-b are conjugate of each other. This concept of conjugate is very important in rationalization as multiplying with the conjugate of given surd gives the rational number.

For example:

In order to rationalize 3 2 + 2 3 we can multiply by its conjugate i.e. 3 2 - 2 3 as:

(3 2 + 2 3 ) (3 2 - 2 3 )

= (3 2 )2 - (2 3 )2

= 9 × 2 - 4 × 3

= 18-12

= 6 Where 6 is a rational number.

Therefore to rationalize a monomial surd we have to multiply by itself and to rationalize the binomial we can multiply by its conjugate.


Example 2 : Rationalize the denominator of: a + xa - x

Solution : Here,

a + xa - x

= a + xa - x

×a + xa + x

{Multiplying both numerator and denominator by a + x }

= ( a + x )2

( a )2 - ( x )2

= ( a )2 + 2 a x + ( x )2

a-x

= a + 2 ax + x

a -x

= a + x + 2 ax

a -x

Example 3 : Simplify:1

3 + 2+

12 + 1

- 2

3 + 1 Solution: Here,

1

3 + 2+

12 + 1

- 2

3 + 1

= 1

3 + 2× 3 - 2

3 - 2+

12 + 1

× 2 - 12 -1

-2

3 + 1× 3 - 1

3 -1

= 3 - 2

( 3 )2 - ( 2 )2+ 2 - 1

( 2 )2 - (1)2- 2( 3 - 1)

( 3 )2 - (1)2

= 3 - 23 - 2

+ 2 - 12 - 1

- 2( 3 - 1)3 - 1

= 3 - 21

+ 2 - 11

- 2( 3 - 1)2

= 3 - 2 + 2 - 1 - 3 + 1

= 0


Example 4 : Simplify: 1 + 485 3 + 4 2 - 72 - 108 + 8 + 2

Solution: Here,

= 1 + 485 3 + 4 2 - 72 - 108 + 8 + 2

= 1 + 16 × 3

5 3 + 4 2 - 36 × 2 - 36 × 3 + 4 × 2 + 2

= 1 + 4 35 3 + 4 2 - 6 2 - 6 3 + 2 2 + 2

= 1 + 4 32 - 3

= 1 + 4 32 - 3

× 2 + 32 + 3

Exercise (4.7)

1. Write the rationalizing factor of the following expressions:

a. 7 b. -2 5 c. 3 +1 d. 2 3 +1 e. 7 - 3 5

2. Write the conjugate of the following

a. 2 – 2 b. 5 + 3 c. 3 - 2 d. 1 + 3 e. 3 + 6

3. Express the following with rational denominator:

a. 6 + 4 36 - 4 3

b. 2 + 32 - 3

c. 2 3 + 3 25 - 2 6

d. b2

a - a2- b2

e. a+b - a-ba+b + a-b

f. x2+y2 + x2-y2

x2+y2 - x2-y2

4. Simplify:

a. 2 + 12 + 1

+ 2 - 12 - 1

b. 7 + 3 53 + 5

+

7 - 3 53 - 5

= (1 + 4 3 ) (2+ 3 )22 - ( 3 )2

= 1(2+ 3 ) +4 3 (2+ 3 )4 - 3

= 2 + 3 + 8 3 + 4 ×31

= 2 + 3 + 8 3 + 12 = 14 + 9 3


c. 3 + 7 + 23 + 7

d. 32 + 3

+

2 22 - 2

- 23 -2

e. 13 - 2

+ 33 - 2

- 2 23 - 1

f. 3 23 + 6

+ 62 + 3

- 4 36 + 2

g. 1x+ x2 - 1

+ 1x+ x2 - 1

h. x + x2 - 1x- x2 - 1

- x - x2- 1x+ x2- 1

5. Simplify:

a. 6 215 28 - 16 7 + 6 21

b. 3 + 65 3 - 2 12 - 32 + 50

6. Find the value of a and b in each of the following cases.

a. 5 + 35 - 3

= a + b 15 b. 3 + 13 - 1 = a + b 3

c. 4 + 3 54 - 3 5

= a + b 5 d. 2 + 3

3 2 - 2 3 = a – b 6

7. a. If x = 3 +1, find the value of (x + 2x )2

b. If x = 6 - 35 + 1 , find the value of x 2 + 1x2

c. If a = 3 + 23 - 2

and b = 3 - 23 + 2

, find the value of ( a + b )2

Ans

wer

s (4.

7)

1a. 7 b. 5 c. 3 -1 d. 2 3 -1 e. 7+3 52a. 2+ 2 b. 5- 3 c. 3 + 2 d. 1- 3 e. 3 - 6

3a. -(7+4 3 ) b. 7+4 3 c. 10 3 + 39 2 d.a+ a2 - b2 e.a+ a2-b2

bf. x

2+ x4-y4

y2 4a. 6 b. 3 c.6 d. 4 3+2 2+3 e. 3+ 3

f. 0 g. x- x2-1 h. 4x x2-1 5a. 3 + 32

b. 3

6a. 4,1 b. 2,1 c. -6129

- 2429

d. 2, - 56

7.a. 12 b.142 c.100


4.8 Equations involving surds(Radicals)

The equations having variables inside the radical sign such as x+5 = 3, 2x-13 = 5, 4 - 3x = 3x+4 are called radical equations. In order to solve the radical equations we should be careful in handling the radical signs especially while removing it. The following steps can help to solve radical equations.

i. Transpose the terms of equation (if required) surd alone in the left hand side (L.H.S.) and rest of the terms are in the R.H.S.

ii. Take the nth power ( according to the order of the radical) on both sides of the equa-tion, so that radical is removed.

iii. Continue this process until all the radical signs are removed.

iv. Use simple algebraic process for solving equation to get the values of variables.

v. Substitute the values obtained in the given equation in order to check whether it sat-isfies or not.

vi. The values which satisfy the given equation are required solution of the given prob-lem.

For clear understanding of solving radical equation, consider the following examples.

1. Solve: 5x -1 = 3

Solution: Here,

5x -1 = 3

Squaring both sides, we get

( 5x -1 )2 = (3)2

or, 5x-1 = 9

or, 5x = 9+1

or, 5x = 10

or, x = 105

∴ x = 2

2. Solve : x + 9 -1 = x

Solution: Here,

x + 9 -1 = x

or, x + 9 = x + 1 [By transposing -1 to the R.H.S.]

Squaring both sides, we get

( x + 9 )2 = ( x + 1)2

Verification

Substituting x = 2 in original equation.

5x -1 = 3

or, 5 × 2 - 1 = 3

or, 10- 1 = 3

or, 9 = 3

∴ 3 = 3, which is true.

Therefore, the required value of x is 2.


Verification

Substitute the value of x = 16, in original equation. x + 9 -1 = xor, 16 +9 -1 = 16or, 25-1 = 4or, 5 -1 = 4

or, 4 = 4 which is true.

Therefore, the required solution of the given

equation is x = 16.

Example 1 : Solve: 4 - 3y + 43 = 0

Solution: Here,

4 - 3y + 43 = 0

or, 4 - 3y + 43

Cubing both sides,

(4)3 = ( 3y + 43 )3

or, 64 = 3y + 4

or, 3y = 64 - 4

or, 3y = 60

or, y = 603

∴ y = 20.

Example 2 : Solve: 7x + 4 + 7x-12 = 8

Solution : Here,

7x + 4 + 7x-12 = 8

or, 7x + 4 = 8 - 7x-12

Squaring on both sides,

( 7x + 4)2 = (8 - 7x-12 )2

Worked Out Examples

Verification

4 - 3y + 43 = 0

or, 4 - 3 × 20 + 43 = 0

or, 4 - 643 = 0

or, 4 - 4 = 0


∴ y = 20

or, x + 9 = ( x ) 2 + 2. x.1+12

or, x + 9 = x + 2 x + 1

or, 9 - 1 = 2 x

or, 2 x = 8

or, x = 82

or, x = 4


( x )2 = (4)2

∴ x = 16

or, 7x + 4 = 82 - 2.8. 7x-12 + ( 7x-12 )2

or, 7x+4 = 64 -16 7x-12 + 7x -12

or, 4 = 52-16 7x-12

or, 4 - 52 = -16 7x-12

or, - 48 = - 16 7x-12

or, 7x-12 = -48-16

or, 7x-12 = 3



( 7x-12 )2 = (3)2

or, 7x - 12 = 9

or, 7x = 9 + 12

or, 7x = 21

or, x = 217

∴ x = 3

Example 3 : Simplify : x + ax - a

+ x - ax + a

= 4

Solution : Here,

x + ax - a

+ x - ax + a

= 4

or, ( x + a )2 + ( x - a )2 ( x - a ) ( x + a )

= 4

or, ( x )2 + 2 x a + ( a )2 + ( x )2 - 2 x a +( a )2 ( x )2 - ( a )2

= 4

or, x+2 ax + a + x - 2 ax+ ax - a

= 4

or, 2x + 2a = 4(x-a) or, 2x + 2a = 4x-4a or, 4x - 2x = 2a + 4a or, 2x = 6a

or, x = 6a2

∴ x = 3a. So, the required value of x is 3a.

Example 4 : Solve: x - 1x + 1

= 2 + x + 13

Solution : Here,

x - 1x + 1

= 2 + x + 13

or, ( x )2 - 12

x + 1 = 6+ x + 13

or, ( x +1) ( x -1)x + 1 = 7 + x

3

Verification

7x + 4 + 7x-12 = 8

or, 7×3 + 4 + 7×3-12 = 8

or, 25 + 9 = 8

or, 5 + 3 = 8


∴ x = 3.

Verification

3a + a3a - a

+ 3a - a3a + a

= 4

or, aa

33

((

+ 1)- 1)

+ aa

33

((

_ 1)+ 1)

= 4

or, 3 3

3 3( (( (

+ 1)2 - 1)2+- 1) + 1)

= 4

or, 3 33 + 2 1 + 3 -2

3 -1+ + 1

= 4

or, 82 = 4

or, 4 = 4 (True)

∴ x = 3a.


or, 3( x -1) = 7 + x

or, 3 x - 3 = 7 + x

or, 3 x - x = 7 + 3

or, x = 102

or, x = 5


( x )2 = 52

or, x = 25

∴ x = 25

Hence, the required value of is 25.

Verification

x - 1x + 1

= 2 +x + 1

3

or, 25 - 125 + 1

= 2 +25 + 1

3

or, 245 + 1 = 2 + 5 + 1

3

or, 246 = 2 + 6

3

or, 4 = 2 + 2


∴n = 25

Exercise (4.8)

1. Solve :

a. 5x - 9 = 4 b. 4x-11 = 3 c. 3x + 13 = 5 d. x + 3 - 2 =0

e. 4x+ 5 = 5 f. 2x +13 = x3 g. 3x+124 = 5

2. Solve

a. 3x + 4 = 2x + 28 b. 2x + 3 = x + 33 c. x - x + 8 = - 2

d. x + 2 = x + 16 e. x - 4 + x = 2

3. Solve

a. 3x = 4 - 9x2 -8x b. x2 + 5 -x = 1 c. x - x2 - 4x = 3

d. z2 + 2 + z = 2 e. 9x2 - 20 = 3x-2

4. Solve

a. 3a2 +33 = 6 b. 8 + 4x - 73 = 13 c. 12 3x - 1

= 1

d. 2t - 13 + 4 = 7 e. 4 - 3y + 43 = 0

5. Solve

a. x - 9x +3 = 1 b. 3y-1

3y +1 = 2 c. 3x - 1

3x +1 = 4


6. Solve :

a. 5x -95x + 3

= 1 + 5x - 32 b. x - 1

x + 1 = 4 + x - 1

2

c. 3x + 42 + 3x

- 3x - 22

= 4 d. 6x - 93 + 6x

= 1 + 6x - 32

e. x + 5 + x = 5 + x15

f. 3 x - 4 x + 2 =

15 + 3 x x + 40

g. x + a x - a

= 4 2 + x - a x + a

h. x + x - 1 - x = 1

7. Solve:

a. 5 + 5-x 5 - 5-x

= 5 b. x+1 + x-1 x+1 - x-1

= 2

c. 1+x2 + 1-x2

1+x2 - 1-x2 = 3 d. x2 - 2x - 4 - x2 - 3x - 3 = 1

e. x2 + 11x+20 - x2 + 5x - 1 = 3

f. 2 x - 4x - 3 = 4

4x - 3

Ans

wer

s (4.

8)

1a. 5 b. 5 c. 4 d. 1 e. 5 f. -1 g. 6133

2a. 24 b. 24 c. 1 d. 9 e. 43.a. 1 b. 2 c. 9

2 d. 12 e. 2

4.a. 71 b. 33 c. 512 d. 14 e. 20

5.a. 16 b. 3 c. 253

6.a. 5 b. 81 c. 1003 d. 25

6 e. 4 f.4

g. 2a, a2 h. 16

25

7.a. 259 b. 5

4 c. 35 d.4 e. 5 f. -1

20


4.9 Simultaneous Equations

Simultaneous Equations

Consider a equation x + y = 9, Here we can see that,

When, x = 1, y = 8

When, x = 2, y = 7

When, x = 3, y = 6

When, x = 4, y = 5

When, x = 5, y = 4 and so on.

In this way there is a value of y for every value of x and the equation has an infinite number of solutions.

Again, consider the equation, x - y = 3.

Here, also When, x = 7, y = 4

When, x = 6, y = 3

When, x = 5, y = 2

When, x = 4, y = 1 and so on.

Therefore, equation x-y=3 also has infinite number of solutions as x = 7, y = 4, x = 6, y = 3, x = 5, y = 2 etc.

Thus, x and y above have infinite number of values when two equations are taken separately. But if we take both the equations simultaneously (together), we see that they are satisfied by one and only one pair of values x = 6 and y = 3.

Therefore, this type of two equations which are satisfied by the some values of the unknown (variables) are called simultaneous equation.

Methods of solving simultaneous equations

We have already discussed in our previous grade that there are various methods to solve a pair of simultaneous equations as

1. Elimination method 2. Substitution method 3. Graphical method

Here, we are briefly going to review elimination and substitution method.

Elimination method to solve a pair of simultaneous equations

Eliminating any one of the variables either by addition or subtracting after making same coefficient of the given pair of equations is known as elimination method.


Worked Out Examples

Example 1 : Solve: 2x+3y =1

3x+2y = -6

Solution: Here,

2x+3y =1 ......................... (i)

3x+2y = -6 .......................(ii)

Multiplying equation (i) by 2 and equation (ii) by 3 and subtracting equation (ii) from (i), we get,

4x + 6y = 29x + 6y = - 18

- 5x = 20( - ) ( - ) ( + )

or, x = 20-5

or, x = - 4.

Now, substituting the value of x in equation (i), we get

2x+3y =1

or, 2(-4) +3y = 1

or, -8 +3y = 1

or, 3y = 1+8

or, 3y = 9

or, y = 93

∴ y = 3.

Therefore, required value of x = - 4 and y = - 3

Substitution Method to solve Simultaneous Equations

Expressing any one equation in term of one variable with respect to another and substituting the value in next equation to get the solution is substitution method. Consider the following example for clear understanding of this method.

Example 2 : Solve: 2x + 3y = 1 3x + 2y = - 6

Solution:

2x + 3y = 1 ............................ (i)


3x+2y = -6 ......................... (ii)From, equation (i) 2x+3y=1 or, 3y = 1 - 2x

or, y = 1-2x3 .......................... (iii)

Now, putting y = 1-2x3 in equation (ii)

[Remember put the value of equation (iii) in equation other than that from where (iii) is obtained. In this case, put the value of y in equation (ii) ]

3x+2 1-2x3( ) = - 6

or, 9x + 2(1 - 2x) = - 18 or, 9x + 2 - 4x = - 18 or, 5x = -18 -2

or, x = - 205

∴ x = - 4Now, putting x = -4 in equation (iii)

y = 1-2x(-4)3

or, y = 1 + 83

or, y = 93

∴ y = 3The required value of x = - 4 and y = 3.

Verbal problem leading to Simultaneous Equations

If the relation between two variables such as numbers, ages of person, dimension of plane figures, costs, wages of person etc. are given in the form of statements are known as verbal problems. In order to solve these types of problem we follow the following steps:

i. Choose two letters (preferably x and y) to represent two unknown number (variables) given in the problem.

ii. Express the condition explained in the problem in mathematical statement using algebraic symbols x and y, so that two equations are formed.

iii. Solve the equation obtained from step (ii) by either elimination or substitution method as per the convenience.

iv. Verify the solutions obtained in case of doubts.


Example 3 : If a number is added to twice of another number, the sum is 20. If the second number is added to three times the first number, the sum is 30. Find the numbers.

Solution: Here,

Let, the first number be x and second number be y.

According to the first equation,

x+2y = 20 .......................... (i)

According to the second equation,

3x+y = 30 .......................... (ii)

Multiplying equation (i) by 3 and subtracting equation (ii) from (i)

3x + 6y = 603x + y = 30

5y = 30( - ) ( - ) ( - )

or, y = 305

or, y = 6.

Putting the value of y in equation (i)

x + 2y = 20

or, x + 2×6 = 20

or, x + 12 = 20

or, x = 20 - 12

or, x = 8

The required number are 8 and 6.

Example 4 : The perimeter of a rectangular field is 120 m .If length is 10 m longer than its breadth, find length and breadth of the rectangular field.

Solution : Here,

Perimeter of the rectangular field (P) = 120 m

Let breadth of the rectangular field (b)= x m

By question, length of the rectangular field (l)= (x + 10) m

Now, we know


Perimeter of the rectangular field (P) = 120

Or, 2 (l + b) = 120

Or, 2 (x + 10 + x) = 120

Or, 2 (2x + 10) = 120

Or, 4x + 20 = 120

Or, 4x = 120 – 20

Or, x = 1004 = 25

∴ Breadth (b) = 25 m and length (l) = x + 10 = 25 + 10 = 35 m

Thus, the length and breadth of the rectangular field are respectively 35m and 25m.

Example 5 : A certain fraction becomes 23 when 1 is added to both the numerator

and denominator. When three is subtracted from the numerator and 2 is added to the denominator, the fraction becomes 1

5 . What is the fraction?

Solution:

Let the required fraction be xy , where x is numerator and y is denominator.

According to the first condition,

x+1y+1 = 2

3

or, 3(x+1) = 2(y+1)

or, 3x +3 =2y+2

or, 3x-2y= 2-3

or, 3x-2y = -1.............................(i)

Solving equation (i) and (ii) by substitution method.

3x-2y = -1

or, 3x-2(5x-17) = -1

or, 3x -10x +34 = -1

or, -7x = -1-34

or, -7x = -35

or, x = - 35- 7

∴ x = 5

According to the second condition,

x-3y+2 = 1

5

or, 5(x-3) = 1(y+2)

or, 5x -15 = y + 2

or, y = 5x - 17 ............................(ii)

Substituting the value of x = 5 in equation (ii)

y = 5x-17

or, y = 5×5 - 17

or, y = 25-17

∴ y = 8Therefore, the required fraction is 5

8 .


Verbal problem related to two digit number

If a problem requires a two digit number as a solution, the digits in tens place and digit in unit place should be denoted by two separate variables as x and y respectively. Then the number of two digits can be written as 10x+y (since in algebra two digit number with x and y as digits cannot be written as xy as xy means x times y in algebra unlike in arithmetic as 34, 3 and 4 being two digits)

In two digit number 10x+y, x is in tens place and y is in unit place. While digits are reversed, y comes to tens place and x to the unit place. So, new number becomes 10y+x. Similarly sum and difference of two digits means x+y and x - y respectively. and product and quotient of two digits mean xy and x

y respectively.

Example 6 : The sum of digits of a two digits number is 11. If the place of the digits are interchanged, the result after subtracting 7 so formed from the new number is twice the given number. Find the original number.

Solution : Here,

Let the number be 10x + y

By question ,

x + y = 11 ………..(i)

Also, when the place of the digits are interchanged the new number will be 10y + x

Again, by question

10 y + x – 7 = 2 ( 10x + y )

Or, 2 ( 10x + y ) = 10 y + x – 7

Or, 20 x + 2y – 10y – x = – 7

Or, 19x – 8y = –7 ………….(ii)

From equation (i) x + y = 11

Or, x = 11 – y ……..(iii)

Putting the value of x in equation (ii), we get

19x – 8y = –7

Or, 19(11 – y) – 8y = – 7

Or, 209 – 19y – 8y = – 7


Or, 209 – 27 y = –7

Or, – 27y = –7 – 209

Or, –27y = – 216

Or, y = -216-27

∴ y = 8

Thus, the required number is 10x+y = 10 × 3 + 8 = 38.

Solving Verbal problems related with age of persons

If a problem requires age as the solution, the current or present ages must be supposed by ‘x’ years and ‘y’ years separately.

Here, ago and before means subtract (-) and after and hence means add (+).

Example 7 : The sum of the present age of two teachers is 54 years. If the age of one teacher is 4 years more than the other, find their present ages.Solution : Here,Let the present age of first teacher is x years and that of second teacher is y years.By question,According to first case, x + y = 54or, x + y = 54 …………(i)According to second case, x – 4 = y …………..(ii)Putting the value of y in equation (i) we get x + x – 4 = 54 or, 2x = 58 ∴ x = 29 years

Thus, the present age of the teachers are 29 years and 25 years.

Example 8 : Two years hence, a father will be six times as old as his son. Three years ago the father was five times as old as his son will be two years hence. What is their present ages?

Solution : Here,

Let the present age of father be x years and that of son be y years.


Now, putting the value of y in equation (i), we get

x + y = 11

or, x + 8 = 11

or, x = 11 – 8

∴ x = 3

Putting the value of x in equation (i), we get

x + y = 54

or, 29 + y = 54

or, y = 25 years.


x+2 = 6(y+2)

or, x +2 = 6y +12

or, x = 6y + 12 -2

∴ x = 6y + 10 ........................(i)

Solving equation (i) and (ii)

x - 5y = 13

or, 6y + 10 -5y =13

or, y = 13 -10

or, y = 3

∴ The present age of father is 28 years and present age of son is 3 years.

Example 9 : The sum of the ages of father and son is 36 years. If they both live on until the son becomes as old as the father is now, the sum of their ages will be 84 years. Find the present ages.

Solution : Here,

Let the present age of father be x years and present age of son be y years.


x + y = 36 ........................ (i)


y+(x-y) +x+(x-y) = 84 or, y+x-y +x+x-y = 84 ∴ 3x - y = 84 ...................... (ii)Adding equation (i) and (ii) we get

x + y = 36(+) 3x - y = 84

4x = 120

or, x = 1204

∴ x =30

Therefore the present age of father is 30 years and present age of son is 6 years.

Example 10 : Four times breadth of a room is equal to 3 times the length. If the breadth had been 1 meter more and length 1 meter less, the room would have been a square. Find the dimension of the room.

Solution : Here,

Substituting y = 3 in equation (i) we get, x = 6y +10 = 6×3 +10 = 18+10 = 28

Substituting the value of x=30 in equation (i)

x + y =36 or, 30 + y = 36 or, y = 36-30 ∴ y = 6


x-3 = 5(y+2)

or, x-3 = 5y +10

or, x -5y = 10 +3

∴ x - 5y = 13 .......................... (ii)


Let, the length (l) of the room be x meter and the breadth (b) be y meters.


4y = 3x

∴ y = 34 x ................................. (i)


y+1 = x -1 [Since length = breadth in a square]

∴ y - x = -2 .................................... (ii)

Solving equation (i) and (ii) we get

y - x = -2

or, 3x4 - x = -2

or, 3x - 4x = -8

or, -x = -8

∴ x = 8

Therefore, length of the room (l) is 8m and breadth of the room (b) is 6m.

Putting the value of x in equation (i)

y = 3x4

= 3 × 84

= 244

= 6

1. Form the equation of the following cases. a. Sum of a number and 5 is 8. b. More than 6 of a number is 20. c. A number is decreased by 7 to get 25. d. 12 is three times a number. e. 4 times a number is 64. f. Twice a number is greater than 8 of the same number. g. Seven more than thrice of a number is 57. h. Eight times a number is 48. i. When 24 is divided by a number, the quotient is 6.2. Establish the relation of two variables in the form of equation. a. The sum of two number is 13. b. The difference of two number is 20. c. A number is greater than another by 5.

Exercise (4.9)


d. A number is 7 times than other. e. The sum of two and two digit number is 31. f. The reversed of two digit number is 45.3. Use the method of substitution to solve each other of the pair of simultaneous equations: a. x + y = 15 x - y = 3 b. x + y = 0 x - y = 2 c. 2x - y = 3 4x + y = 3 d. 2x - 9y = 1 5x + 2y = 274. Solve each other pair of equation given below using elimination method: a. x + 2y = -4 3x - 5y = -1 b. 4x + 9y = 5 -5x + 3y = 8 c. 9x - 6y = 12 4x + 6y = 14 d. x + y = 6 x - y = 45. Solve the following simultaneous equations: a. 3a + 4b = 43 -2a + 3b = 11 b. 4x - 3y = 23 3x + 4y = 116. a. The sum of a number and two- third of it is 50. Find the number. b. If 5 is subtracted from 7 times a number, then the number becomes 44. Find the

number. c. The sum and difference of two positive numbers are 15 and 5 respectively. Find the

number. d. If the sum of two number is 10 and their difference is 2, find the numbers. e. Four times the sum of two numbers is 32. If five times their difference is 10, find the

number. f. A number is twice the other. If their sum is 30, find the numbers. g. If the sum of two consecutive natural number is 25. Find the numbers. h. If the sum of three consecutive number is 36. Find the numbers. i. The sum of the present age of two teacher is 54 years. If the age of one teacher is 4

years more than the other, find their present ages. j. From the age of a person if two- fifth of his age is subtracted, the result becomes 24

years. Find the age of the person. k. Divide 45 into two parts so that twice the greater is equal to three times the lesser. l. If a number and its 10% are added, it will be 11 , then find the number.7. a. The sum of digits at ten’s place and unit’s place of two digits number is 11. If the digits

are interchanged the number so formed is 45 more than the original number, find the number.

b. A number consists of two digits if the numbers formed by reversing its digits is added to it the sum is 121 and same number is subtracted from it, the remainder is 27. Find the number.

8. a. A number consisting two digits exceeds 4 times the sum of the digits by 3. If 27 be added to the number, the places of the digits of the number, are interchanged, find the number.


b. A number of two digits is 3 times the sum of its digit. If 9 is added to the number by reversing its digits, the new number will be 3 times the previous one. Find the numbers.

9. a. A number consists of two digits, the sum of its digit is 16. If 18 is subtracted from the number, the digit interchange their place. Find the number.

b. The sum of the digits of a two digit number is 12. If 36 is subtracted from the number, the number so formed is a number obtained by reversing its digits. Find the original number.

c. There is a number between 10 and 100. The number is 8 times the sum of the its digit. If 45 is subtracted from it, the result is number formed by interchanging its digits. Find the numbers

10. a. The sum of the ages of father and son is 44 years. After 8 years, the age of the father will be twice as old as the age of the son. Find their present ages.

b. A year hence a father will be 5 times as old as his son, two years ago the father was 3 times as old as his son will be 4 years hence. Find their present ages.

c. A year later a mother’s age will be five times of her daughter’s age. Two years ago if the mother’s age was three times of her daughter age of four years later. Find their present ages.

11. a. A mother says to her daughter “ 5 years ago I was 5 times as old as you were but 10 years hence I shall be only twice as old as you will”. Find the present age of mother and daughter.

b. A mother says to her daughter “ 6 years ago I was 3 times as old as you were but 11 years hence I shall be only twice as old as you will”. Find the present age of mother and daughter.

12. a. The sum of present ages of a father and a son is 50 years. If they live until the son becomes as old as the father is now, the sum of their ages will be 110 years. Find their ages.

b. The sum of present ages of a father and a son is 60 years. If they live until the son becomes as old as the father is now, the sum of their ages will be 120 years. Find their ages.

13. a. 6 years ago a man’s age was 6 times to the age of the son. 4 years hence four times of his age will be equal to 8 times his son’s age. Find their present ages.

b. 8 years ago a man’s age was 5 times to the age of the son. 5 years hence four times of his age will be equal to 7 times his son’s age. Find their present ages.

14. a. Three years ago the ratio of the ages of A and B was 4:3. Three years hence, the ratio of their ages will be 11:9. Find the present ages of A and B.

b. The ages of two girls are in the ratio of 2:3. 6 years hence the ratio of their ages will be 11:15. Find their present ages.

15. a. In a fraction, the numerator is 1 less than the denominator. If 1 is added to the numerator and 5 to denominator, the fraction becomes 1

2 . Find the original fraction. b. If 3 is added to the numerator of the fraction. The value of fraction becomes 1. When


3 is added to the denominator of the fraction the value of the fraction becomes 14 .

What is the value of the fraction? Find it.

c. If 2 is added to the numerator and denominator it becomes 910 and if 3 is subtracted

from the numerator and denominator it become 45 . Find the fractions.

16. a. The combined price of 1 pen and 3 copies is Rs. 210. The combined price of 3 pen and 5 copies of same quality is Rs 430. What is the combined price of 1 pen and 1 copy? Find it.

b. I buy 3 packets of snacks and 5 packets of biscuits for Rs 30. I again buy 4 packets of snacks and 8 packets of biscuits at the same price for Rs 44. Find the price of one packet each.

c. 3 tables and 2 chairs cost Rs 1900 and 2 tables and 4 chairs cost Rs. 1800. Find the cost of table and a chair.

17. a. Eight times breadth of a room is equal to 6 times the length. If the breadth had been 1 meter more and length 1 meter less, the room would have been a square. Find the dimension of the room.

b. Perimeter of a rectangular field is 32 m , keeping the perimeter same, if length is reduced by 2 m, area will increase by 12 m2. Find the original length and breadth of the rectangle.

18. a. Two buses run from two cities of 840 km apart. One of them runs 6 km per hour faster than other. They meet exactly after five hours of their journey. Find each of their speeds, if they started at the same time.

b. Two cars start at the same time from two stations 480 km apart in opposite direction. The speed of the first car is greater than that of the second by 6 km/hr. If they meet each other after 6 hours from the start, find their speed.

19. If Ram gives one of the marbles from what he possesses to sita then they will have equal number of marbles. If sita gave one of the marbles from what she possesses to Ram, then Ram will have double of the marbles with what sita is left with. Find the number of marbles possessed by each initially.

20. 6 men and 8 boys can do a piece of work in 14 days. 4 men and 6 boys can do the same work in 20 days. How many days will a man or a boy take to do it alone?

Ans

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1.a. x+5=8 b. x+6=20 c. x-7=25 d. 3x-12=0 e. x-16=0 f. x-8=0 g. 3x-50=0 h. x-6=0 i. x-4=0 2.a. x+y=13 b. x-y=20 c. x-y=5 d. x-7y=0 e. 10x + y-29=0

f. 10y + x - 45 = 0 3.a. x=9, y=6 b. x=1, y=-1 c. x=1, y=-1

d. x = 5, y = 1 4.a. x=-2 , y=-1 b. x=-1, y=1 c. x=2, y=1 d. x= 137 , y = -6

355.a. a=5, b=7 b. x=5, y=-1


Ans

wer

s (4.

9)6.a. 30 b. 7 c. 10 , 5 d. 6 , 4 e. 5 , 3 f. 10 , 20 g. 12 , 13 h. 11,12,13 i. 29 , 25 j. 40 k. 27 , 18 l. 107.a. 38 b.74 8.a. 47 b. 27 9. a. 97 b. 84c. 72 10.a. 32 , 12 b. 14 , 2 c. 14 , 2 11.a. 30, 10 b. 23 , 57 12. a. 40 , 10 b. 45, 15 13. a. 21 , 17/2 b. 23, 11 14.a. 19, 15 b. 16 , 24

15.a. 45 b. 2

5 c. 78 16. a. 60 , 50 b. 5 , 3 c. 500 , 200

17.a. l=8 , b=6 b. l=12 , b=4 18 a.81 , 87 b. 43 , 37 19. Ram=7 , Sita=5 20. A men=140 days, a boy=280 days

4.10 Quadratic Equations

Introduction

The name quadratic comes from the word ‘quad’ means square, because the variable get squared (like x2).

An equation with one variable, in which the highest power of the variable is two, is called quadratic equation.

For examples

i. x2 = 9 ii. x2-4=0 iii. 2x2 + 3x – 5 = 0.In these three equations the highest power of x is 2. Such equations of the second degree is called quadratic equations.In (ii) there is a term in x2 and no term in x. Such an equation is called a pure quadratic equation.In (iii) there is a term in x2 and x as well. Such an equation is called adfected quadratic equation. Remember

i. The standard form of quadratic equation is ax2 + bx +c = 0 , where a and b are real numbers and a ≠ 0 .

ii. Every quadratic equation gives two values of the unknown variables which are called roots of the equation.

iii. Whenever the product of two expressions is zero, at least one of the expression is zero.

Method of solving pure quadratic Equation:

1. Remove brackets or fractions, if any. 2. Solve the equation for in the same way as you solve a simple equation in x. 3. Then take the square root of both sides to get the two values of x (one positive

and the other negative)


Worked Out Examples

Example 1: Solve: x2 = 36

Solution : Here,

Or, x2 = 36 or, x2 - 36=0 or, (x + 6)(x - 6) = 0So, one of x + 6 and x - 6 must be zero From x + 6 = 0, we get x = -6 From x - 6 = 0, we get x = 6

Thus, the required solutions are x = ± 6

Example 3 : Solve: 2+x2-x + 2-x

2+x = 4 14

Solution: 2+x

2-x + 2-x2+x = 4 1

4

or, (2+x)2 + (2-x)2

(2-x) (2+x) = 174

or, 4 + 4x + x2 + 4 - 4x + x2

4 - x2 = 174

or, 8 + 2x2

4-x2 = 174

or, 32 + 8x2 = 68 - 17x2 (By cross multiplication)

or, 25x2 = 36

or, x2 = 3625

or, x = ± 3625

or, x = ± 65

∴ x = ± 1 15

Example 5 : If 39 is subtracted from the square of a number, the result is 130. Find the positive number.

Solution : Here,

Let the positive number be x.

Then according to the question,

Example 2 : Solve: 5x2 =125

Solution : Here,

5x2 =125

or, x2 = 1255

or, x2 = 25

Taking the square root of both sides,

x = ± 25

or, x = ± 5 (i.e. +5 or -5)

Thus, required value of x is either 5 or – 5

Example 4 : Solve: x2-11x+30=0

Solution : Here,

x2-11x+30=0

or, x2- (6 + 5 ) x + 30 = 0

or, x2- 6 x - 5 x + 30 = 0

or, x (x – 6 ) – 5 ( x – 6 ) = 0

or, ( x – 6 ) ( x – 5 ) = 0

Either ( x – 6 ) = 0 Or, ( x-5 )=0 ∴x = 6 ∴x = 5

Thus, the required value of x is either 6 or 5.


x2-39 = 130

Or, x2 = 130+39

Or, x2 = 169

Or, x = ± 169

∴ x = ± 13

Thus, the required positive number is 13.

Example 6 : If the sum of a positive number and its square is equal to 56, Find the number.

Solution: Here,

Let the positive number be x.

According to the question

x + x2 = 56

Or, x2 + x -56 = 0

Or, x2+(8-7)x -56 = 0

Or, x2 + 8x-7x -56 = 0

Or, x(x+8) -7(x+8) = 0

Or, (x+8) (x-7) =0

Either, or,

x +8=0 x-7= 0

x = -8 (impossible) x = 7

Thus, the required positive number is 7.

Example 7 : The product of two consecutive numbers is 156, find the number.

Solution: Here,

Let, the consecutive numbers are x and x+1.

According to the question,

x(x+1)= 156

or, x2+x=156

or, x2+x-156=0

or, x2+ (13-12)x-156 =0


or, x2+ 13x -12x -156 =0

or, x(x+13) -12(x+13) =0

or, (x+13)(x-12)=0

Either, or,

x+13 =0 x-12=0

x=-13(impossible) x= 12

so, x+1 = 12+1=13

Thus, the required consecutive numbers are 12 and 13.

Example 8 : Two numbers are x and x+2. If the sum of their reciprocals is 512 , Find

the numbers.

Solution: Here,

The two numbers are x and x+2.

According to the question,

1x

+

1x + 2

= 512

Or, x+2+xx(x+2)

= 512

Or, 2x + 2x(x+2)

= 512

Or, 12(2x +2) =5x(x+2)

Or, 24x + 24 = 5x2 +10x

Or, 0= 5x2 -24x-24+10x

Or, 5 x2 – 14x – 24 = 0

Or, 5x2 -20x + 6x -24 =0

Or, 5x(x-4) +6(x-4)=0

Or, (x-4) (5x+6) =0

Either, or,

x - 4 = 0 5x + 6 = 0

x = 4 x = - 65

When, x = 4, then x + 2 = 4 + 2 = 6


When, x = -65 then,

-65 +2 = - 6 +10

5 = 45

Thus, the number are either 4,6 or -65 , 4

5 .

Example 9 : The sum of two numbers is 11 and their product is 28. Find the numbers.

Solution : Here,

Let, two numbers be x and y

By question,

x + y = 11 ……(i)

x × y = 28 ......(ii)

Now, from (i), x + y = 11

x = 11 – y ….(iii)

Putting the value of x in equation (ii) we get

x × y = 28

(11 – y) y = 28

Or, 11 y – y2 = 28

Or, y2 – 11 y + 28 = 0

Or, y2 – (7 + 4) y + 28 = 0

Or, y2 –7 y – 4 y + 28 = 0

Or, y ( y – 7 ) – 4 ( y – 7 ) = 0

Or, ( y – 7 ) ( y – 4 ) = 0

Either ( y – 7 ) = 0 or ( y – 4 ) = 0

∴ y = 7 ∴ y = 4

Substituting the value of y in equation (iii) , we get

11 – 7 = 4 or 11 – 4 = 7

Thus, the required numbers are 4 and 7.

Example 10 : The length of a rectangle is greater than its breadth by 3m. If its area be 10 sq. m, find the perimeter.

Solution: Here,

Suppose, the breadth of the rectangle = x m.


Therefore, length of the rectangle = (x + 3) m.

So, area = (x + 3)x sq. m

Hence, by the condition of the problem

(x + 3)x = 10

Or, x2+ 3x - 10 = 0

Or, x2 + 5x – 2x - 10 = 0

Or, x( x + 5 ) – 2 ( x + 5 ) = 0

Or, (x + 5) (x - 2) = 0

Either ( x + 5) =0 or ( x - 2 ) = 0

∴ x = - 5 or ∴ x = 2

So, x = -5,2

But x = - 5 is not acceptable, since breadth cannot be negative.

Therefore x = 2

Hence, breadth = 2 m and length = 5 m

Therefore, Perimeter = 2(2 + 5) m = 14 m.

x = - 5 does not satisfy the conditions of the problem length or breadth can never be negative. Such a root is called an extraneous root. In solving a problem, each root of the quadratic equation is to be verified whether it satisfies the conditions of the given problem. An extraneous root is to be rejected.

Example 11: The product of the digits of a two-digit number is 12. If 36 is added to the number, a number is obtained which is the same as the number obtained by reversing the digits of the original number.

Solution: Here,

Let the digit at the units place be y and that at the tens place be x.

Then, the number = 10 x + y.

The number obtained by reversing the digits = 10 y + x

Case I : x × y = 12 ................... (i)

Case II : 10x + y + 36 = 10y + x

Or, 9x – 9y + 36 = 0

Or, x - y + 4 = 0Or, y = x + 4 .................................. (ii)


Putting y = x + 4 in (i), x × y = 12Or, x (x + 4) = 12Or, x2 + 4x – 12 = 0Or, x2 + 6x - 2x – 12 = 0Or, x(x + 6) - 2(x + 6) = 0Or, (x + 6)(x - 2) = 0Either x + 6 = 0 Or, x - 2 = 0

∴ x = - 6 or x = 2But a digit in a number cannot be negative. So, x ≠ -6.Therefore, x = 2.Therefore, from (iii), y = x + 4 = 2 + 4 = 6.Thus, the original number 10 x + y = 10 × 2 + 6 = 20 + 6 = 26.

Example 12 : After completing a journey of 84 km. A cyclist noticed that he would take 5 hours less, if he could travel at a speed which is 5 km/hour more. What was the speed of cyclist in km/hour?

Solution:

Suppose, the cyclist has travelled with a speed of x km/hour

Therefore, by the condition

84x

-

84x+5

= 5

Or, 84x + 420 - 84xx(x+5)

= 5

Or, 420x2+5x = 5

Or, 5 x2 + 25x = 420

Or, 5(x2 + 5x) = 420

Or, x2+ 5x - 84 = 0

Or, x2+ 12 x – 7 x - 84 = 0

Or, x ( x + 12) – 7 ( x +12 ) = 0

Or, (x + 12) (x - 7) = 0Either ( x + 12) = 0 or ( x - 7 ) = 0 ∴ x = - 12 ∴ x = 7


Therefore, x = -12, 7But x ≠- 12, because speed cannot be negative, so, x = 7Therefore, the cyclist has travelled with a speed of 7 km/hour.Example 13 : An association has a fund of Rs 195. In addition that, each member of the association contributes money equal to the number of members. The to-tal money is divided equally among the members. If each of the members gets Rs 28, find the number of members in the association.

Solution:

Let the number of members be x.Total contributions from them = Rs x2

and the association has a fund of Rs 195.

According to the problem,

x2 + 195 = 28x or, x2 - 28x + 195 = 0 or, x2 - 15x - 13x + 195 = 0 or, x(x - 15) - 13(x - 15) = 0 or, (x - 15)(x - 13) = 0Either ( x - 15) =0 or ( x - 13 ) = 0 ∴ x = 15 or ∴ x = 13Therefore, x = 15 or 13There are 15 or 13 members in the association.

Exercise (4.10)

1. a. What is quadratic equation ?

b. How many values are given by a quadratic equation ?

c. What type of equations can be solved using quadratic formula?

2. Form the equation of the following cases. a. 4 and square of a number is 20.

b. Five times of a square number is 80.

c. 2 less than square of number is 14.

d. 5 less than square of number is 31.


3. Solve each of the following quadratic equation.

a. ( x – 4 ) ( x + 4 ) = 0 b. ( x – 7 ) ( x + 8 ) = 0 c. x2 = 36 d. a2 = 49

e. ( 2x – 4 ) ( 3x + 4 ) = 0 f. 3x2 – 36 = 0 g. 4x2 – 36 = 0 h. 6x2 – 36 = 0

4. Solve the following equations.

a. x2+3x+10 = 0 b. 4x2+20x+25=0 c. x2 + 6x + 9 = (2x+1)2 d. (x-2)(x-3) = (x-4)(x+2)

e. x+2 - 6x+2

= 1 f. 32x2

- 35x2

= 1 19

g. x2

+ 2x

= x8

+ 8x

h. x+4x-4 +

x-4x+4 = 10

35. Solve the following equations by completing the square.

a. x2-5x-36=0 b. 6x2+13x =5 c. 4x2 -3x=6

6. Solve the following equations by using formula.

a. x2 - 6x = 27 b. x2 - 10x + 21 = 0

c. (3x + 4)2 - 3(x + 2) = 0 d. x+1x+3

= x+72x+8

7. a. If 3 is added to the square of a number the result is 52 . Find the number.

b. If 6 is added to the square of a positive number, the sum is 31. Find the number.

c. If 15 is subtracted from half of the square number , the result is 35. Find the num-ber.

d. If 17 is subtracted from the square number , the result is 152. Find the number.

e. Find a number such that the difference between it and 30 times its reciprocal is unity.

f. If the sum of number and 25 times its reciprocal is 10, find the number.

8. a. Find two consecutive natural numbers whose product is 20.

b. Find two consecutive even numbers if sum of their reciprocal is 940

.

c. Divide 11 into two parts so that their product will be 24.

d. If 5 times of a number is equal to the square of the number, find the number.

e. The difference of two numbers is 4 and the sum of their square is 208. Find the numbers.

f. If the sum of two numbers is 9 and their product is 18, find the numbers.

9. a. The difference of the ages of two brothers is 4 years and product of their ages is 221. Find their present age.

b. The present age of the mother and the son are 37 and 8 years respectively. How


many years ago was the product of their ages 96?

10. a. The area of a room is 128 cm2 . If its length is twice the breadth, find its breadth.

b. A rectangular meadow has an area 28 cm2 and perimeter is 22 m . Find length and breadth of the meadow.

11. a. The sum of digits of two digit number is 7 and their product is 12 . Find the num-bers.

b. A two digit number is four times the sum and three times the product of its digits. Find the number.

c. In a two digits natural number, the digit at unit place is 4 more than that of the digit at ten’s place. If the product of the digits is 21, find the number.

d. In a positive number of two digit, the ten’s place digit exceeds the unit place digit by 5. If the product of the digits 36, find the number.

e. The product of the digit in a two digits number is 15.The number formed by inter-changing the place of digits is 18 more than the original number find the original number.

f. The product of the digits of a two digit number is 20. If 9 is subtracted from the number, the number formed with the digits reversed. Find the number.

g. The product of digits in a two digit number is 18. The number formed by inter-changing the digits of that number will be 27 more than the original number. Find the original number.

12. a. The difference of the present age of sister and her younger brother is 6 years. The product of ages of numerical values is equal to 4 times the sum of their ages, find the present age of the sister and her brother.

b. The difference of the present age of brother and his younger sister is 5 years. The product of ages of numerical values is equal to 6 times the sum of their ages, find the present ages of the brothers and his sister.

c. The present ages of a father and his son are 40 years and 8 years respectively. How many years ago the product of their ages was 105? Find it.

d. The present age of the father is thrice as age of the daughter 5 years ago. If two years ago the age of the father was twice as old as the daughter 8 years hence. Find their present ages.

e. The present ages of father and his son are 35 years and 12 years respectively. Find how many years ago the product of their ages was 210?

f. The sum of the ages of a mother and her daughter is 34 years and the product of their ages is 225 years. find their present ages.


13. a. If the area of rectangular courtyard is 63 m2 and perimeter is 32 m, find the length and breadth of the courtyard.

b. If the perimeter of a rectangular plot is 104 m and area is 612 m2, find length and breadth.

c. The area of rectangular room is 45 square meter. If the length had been 3 m more and breadth is 1 m less it would have been a square. Find the length and breadth of the room.

d. The length of a rectangle is 3 cm more than its breadth. If the area of the rectangle is 88 cm2, find its perimeter.

e. The area of a rectangular room is 45 sq. m. If the length had been 3m less and breadth 1m more it would have been a square. Find the length and breadth of the room.

f. The height of a tank is 2m and its length is 3m more than its breadth. If 80 cu. m of water can be stored in the tank, find the length and breadth of the tank.

14. a. The side of a right angled triangle containing the right angle are less than its hypotenuse by 5 cm and 10cm respectively. Find the length of the sides of the triangle.

b. The hypotenuse of a right-angled triangle exceeds its base by 2cm and twice its altitude by 1cm. Find the lengths of each side of the triangle.

c. The sum of the length of the two sides containing the right angle in a right angled triangle is 22cm and the area of the triangle is 60 sq cm. Find the length of sides of right angled triangle.

d. The hypotenuse of right angled triangle is 6cm more than twice the shortest side, the longer side is 2cm shorter than the hypotenuse. Find the length of each side of the triangle.

15. 48 dolls are equally divided among a certain number of children. If 4 children are less, each would receive 2 more dolls. Find the number of children and the num-ber of dolls received by each child.

16. A journey of 192km between two cities takes 2 hours less by a fast train than by a slow train. If the average speed of the slow train is 16km/hr less than that of fast train. Find the speed of each of them.

17. The speed of an ordinary bus is x km per hour and that of an express bus is (x + 25) km per hour.

i. Find the time taken by each bus to cover 300 km.


Ans

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1.a. consult to your teacher b. 2 c. ax2 + bx + c = 0 2a. x2-16=0 b. x2-16=0 c. x2-16 = 0 d. x2-36=03a. + 4 b. 7, -8 c. + 6 d. + 7 e. 2, -4

3 f. + 2 3

g. + 3 h. + 64. a -17, 14 b. -5

2 c. -4

3 ,2 d. -4 ,12 e.1, -4 f. + 9

10g. 4 , -4 + 4 h. + 8

5.a. -4 ,9 b. 13

- 52

c. 18 ,-15

6. a. -3, 9 b. 3, 7 c. - 23

, - 53

d. + 3

7.a. + 7 b. + 5 c. + 10 d. + 13 e. -5 , 6 f. 5 8.a. 4,5 b. 8,10 c. 3,8 d. 0,5 e. 8,12 f. 3,6 g. 3,6 9 a. 13,17 b. 5 yrs 10.a. 8 cm b.4,7 11 a. 34 b. 24 c. 37 d. 94 e. 35 f. 54 g. 3612 a. bro 6 yrs , sis 12 yrs b. bro 15 yrs, sis 10 yrs c. 5 yrs d. 84,33 e. 5 yrs f.25,9 13a. 9,7 b. l=18m , b=34m c. l=5cm, b =9cm d. 38me. l=9m, b=5m f. l=5m, b=8m 14a. 25cm, 20 cm and 15 cm b. p=8cm, b=15cm, h=17cm c. l=10cm, b=12cm, h=15.6cm d. 10, 24, 26 cm15a. total children =12, Each children gets 4 dolls 16. 48 , 32 17. i. 300

x, 300x+25

hrs ii. 75 km/h

18 . 12, 18 or, 13,17 19. 300 or 26

ii. If the ordinary bus takes 2 hrs more than the express bus, calculate speed of the express bus.

18. In a class test, the sum of Ram’s marks in Mathematics and English is 30. Had he got 2 mark more in Mathematics and 3 marks less in English, the product of his marks would have been 210. Find his marks in two subjects.

19. An association has a fund of Rs 780. In addition that, each member of the asso-ciation contributes money equal to the number of members. The total money is divided equally among the members. If each of the members gets Rs 56, find the number of members in the association.


Model set [F.M. 24]Group – A ( 1 × 1 = 1 )

1. Simplify : 1xa-b

× 1xb-a

(1)

Group – B ( 5 × 2 = 10 )

2. a. Simplify : 5x - 2.5x-1

3.5x( 1

5) b. Solve : 3x + 3x+2 = 10

3(-1)

c. Simplify : x + 5x - 5

-

x - 5x + 5

( 4 5x x - 5

)

d. Solve : x - 3 = x + 2 ( 4916

) e. Simplify : 803 - 2 20 (8 5 )

Group – C ( 4 × 2 = 8 )

3. a. Find the L.C.M. of : ( x – y )2 + 4xy, ( x + y )3 - 3xy ( x + y) and x2 + 2xy + y2 (x + y) ( x3+y3)

b. The present age of two brothers are 15 years and 22 years respectively. After how many years the product of their ages will be 408 years?

Group – D ( 5 × 1 = 5 )

4. Simplify : 1x+1

+

2x2+1

+

4x4+1

+

8x8-1

( 1x-1

)

Test yourself F.M. 251. a. Solve : x - 3 = 4 (13)

b. Establish the relation between H.C.F. and L.C.M.

2. a. x4 + x2 y2 + y4 and x3 + x2 y + x2 y + x y2

b. Simplify : 4 + 3k + 3k-1

13 × 3k c. Simplify : 1

x+y +

1x-y

+

2xx2+y2

d. Simplify : 45 + 245 - 500 + 625

e. 15

of the age of father is the age of his son. If sum of their ages is 42 years,

determine the age of the father.

3. a. Find the L.C.M. of : x2 – 10 x + 24 + 6y – 9 y2 and x2 + 6 x y + 9 y2 – 36

b. Find the two numbers such that the sum of the smaller to 7 is double of larger and sum of larger to 4 is three times of smaller.

4. Prove that the roots of 7m + 3437m

= 56 also satisfy the equation 5.4x+1 – 16x = 64

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