Study Notes 1

8/3/2019 Study Notes 1

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STUDY NOTE-1

ARITHMETIC

RATIO AND PROPORTION

1.1 Ratio : The ratio between two quantities a and b of same kind is obtained by dividing a by b

and is denoted by a : b.

Inverse Ratio : For the ratio a : b, inverse ratio is b : a .

A ratio remains unaltered if its terms are multiplied or divided by the same number.

a: b = am : bm (multiplied by m)

)0mbydivided(mb

:ma

b:a =

Thus 2 : 3 = 22 : 32 = 4 : 6, again4 6

4 : 6 : 2 : 32 2

= =

If a = b, the ratio a : b is known as ratio of equality.If a > b, then ratio a : b is known as ratio ofgreater inequality i.e. 7 : 4 And for a < b, ratio a : b

will be the ratio of Lesser inequality i.e. 4 : 7.

For solving problems :

1. Reduce the two quantities in same unit.If a = 2 kg, b = 400 gm, then a : b =2000 : 400 = 20 : 4 = 5 : 1 (here kg is changed to gm)

2. If a quantity increases by a given ratio, multiply the quantity by the greater ratio.If price of crude oil increased by 4 : 5, which was, Rs. 20 per unit of then present

price = .unitper25.Rs45

20 =

3. If again a quantity decreases by a given ratio, then multiply the quantity by the lesser ratio.In the above example of the price of oil is decreased by 4 : 3, the present

price = .unitper15.Rs43

20 =

4. If both increase and decrease of a quantity are present is a problem, then multiply the quantity

by greater ratio for inverse and lesser ratio for decrease, to obtain the final result.

Proportion : The equality of two ratios is called the proportion thus 2 : 3 = 8 : 12 is written as

2 : 3 : : 8 : 12 and we say 2, 3, 8, 12 are in proportion.In proportion the first and fourth terms are known as extremes, while second and third terms

are known as means.

In proportion, product of means = product of two extremes As 2, 3, 8, 12 are in proportion, we

have 212 = 38 (=24)

Few Terms :

1. Continued proportions : The quantities a, b, c, d, e.. are said to be in continuedproportion of a : b = b : c = ..Thus 1, 3, 9, 27, 81, .. are in continued proportion as 1 :

3 = 3 : 9 = 9 : 27 = 27 : 81 =

MATHS 1.01


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MATHS1.02

A

RITHMETIC

Example : If 2, x and 18 are in continued proportion, find x Now 2 : x = x : 18 or,

6x,or36x,or18

x

x

2,or

18

x

x

2 2 ====

Obs. If a, b, c are in continued proportion , the aca,acb2 == .

2. Compound Proportion : If two or more ratios are multiplied together then they are

known as compounded.

Thus a1 a2 a3 : b1 b2 b3 is a compounded ratios of the ratios a1 : b1 ; a2 : b2 and a3 : b3. This method

is also known as compound rule of three.

Example : 10 men working 8 hours a day can finish a work in 12 days. In how many days can 12

men working 5 hours a day finish the same work.?

Men Hours day

Arrangement : 10 8 12

12 5 x

daysx 1612

10

5

812 == Obs : less working hour means more working days, so multiply

by greater ratio58

. Again more men means less number of days, so multiply by lesser ratio12

10.

Derived Proportion : Given quantities a, b, c, d are in proportion.

(i) Invertendo : If a : b = c : d then b : a = d : c

(ii) Alternendo : If a : b = c : d, then a : c = b : d

(iii) Componendo and DividendoIf

dcdc

baba

thendc

ba

-+-

-+=

Proof : Let ,kdc

ba == then a = bk, c = dk

L. H. S. =1k1k

)1k(b

)1k(b

bbkbbk

-+=

-+

=-+

R. H. S. =1k1k

)1k(d

)1k(d

ddkddk

-+=

-+

=-+

. Hence the result.

An Important Theorem

If ,then......f

e

d

c

b

a ==

each ratio =

n/1

nnn

nnn

......rfqdpb

......reqcpa

++

++where p, q, r,. are quantities positive or negative.

Let ........fke,dkc,bkathatso,kfe

dc

ba ======

Hence, Pan

= p(bk)n

= pnk

n, qc

n= qd

nk

n, re

n= rf

nk

n, etc.


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MATHS 1.03

.k)k(......rfqdpb

......rfqdpb(k

.....rfqdpb

.....reqcpan

nn

nnn

nnnnn

nnn

nnn

==

+++

++=

++

++\

1

11

Hence the result.

Cor. 1. Putting n = 1, we get

If ,then......fe

dc

ba == each rati o, =

.....rfqdpb

.....reqcpa

++++++

Cor. 2. Puttting p = q= r= .=1, we find

If If ........fe

dc

ba === each ratio

n1

nnn

nnn

.....fdb

......eca

++

++

Cor.3. If ........f

e

d

c

b

a === then each rat io =a c e .....

b d f ....

Putting p = .1.Corin......1r,1q,1 ==

Note. 1.x y z

a b c= = is sometimes written as x : y : z = a : b : c.

2. If x : y = a : b, it does not mean x = a, y = b. It is however to take x = ka, y = kb.

Solved problems :

1. If,a2

x3y4

b3

y3z4

c4

z3x4 -=-

=- , show that each ratio is equal toc4b3a2

zyx

++++

.

Each of the given ratioc4b3a2

zyxa2b3c4

x3y4y3z4z3x4++++=

++-+-+-=

2. Ifh

g

fe

dc

ba === show that

4 4 4 4

4 4 4 4

aceg a c e g

bdfh b d f g

+ + +=

+ + +

h

g

f

e

d

c

b

a === =k (say), so that a = bk, c = dk, e = fk, g = hk.

L. H. S. = 4kbdfg

hk.fk.dk.bk =

R. H. S. =4 4 4 4 4 4 4 4

4 4 4 4b k d k f k h k b d f h+ + ++ + + = ( )4

4444

44444

khfdb

hfdbk =+++ +++. Hence the result

3. If a1, a2, , an, be continued proportion, show that

1n

2

1

n

1

a

a

a

a-

=

We have, 31 2 n 1

2 3 4 n

aa a a..... k (say)

a a a a

-= = = = n

1

n

1n

4

3

3

2

2

11n

a

a

a

a....

a

a

a

a

a

ak == --

again,

n 1

n 1 1

2

ak

a

-

- =

sum (or difference)of numerators

sum(or difference)of deno min ators=

4444

hfdb +++((

4444 hfdb +++

4

k 4k=

denomin ators


4/39

1

2

11-

=\

n

n a

a

a

a

4.abc

xyz

cab

zxy

bca

yzxthatprove

cz

b

y

ax

-

-=

-

-=

-

-==

2

2

2

2

2

2

[I.C.W.A. (F) June 2007]

kc

z

b

y

a

x === (say); x = ak, y = bk, z = ck

( )( )

2

2

22

2

2

kbca

bcak

bca

yzx=

-

-=

-

-, Similarly

abc

xyzk

cab

zxy

-

-==

-

-2

22

2

2

(to show in detail) Hence the

result.

5. Ifba

rac

q

cb

p

-=

-=

-prove that p + q + r = 0 = pa + qb + rc [I.C.W.A. (F) Dec. 2007]

bar

ac

q

cb

p

-=

-=

-= k (say), p = k (b-c), q = k (ca), r = k (ab)

Now p + q + r = k (b c + c a + a b) = k 0 = 0And pa + qb + rc = ka (b c) + kb (c a) + kc (a b) = k (ab ac + bc ba + ca cb) = k 0

= 0. Hence the result.

6. Ifba

zac

y

cbx

+=

+=

+prove that

222222 ba

)yx(z

ac

)xz(y

cb

)zy(x

-

-=

-

-=

-

-[I.C.W.A (F) Dec. 2005]

,kba

zac

y

cbx =

+=

+=

+x = k (b+c), y = k(c+a), z = k(a+b)

22

k)cb)(cb(

)ac)(cb(k

)cb)(cb(

)baac(k).cb(k

cb

)zy(x-=

-+-+

=-+

--++=

+-

Similarly,22

2

22 ba

)yx(zk

ac

)xz(y

-

-=-=

-

-(To show in detail) Hence the result.

7. The marks obtained by four examinees are as follows :

A : B = 2 : 3, B : C = 4 : 5, C : D = 7 : 9, find the continued ratio.

A : B = 2 : 3

B : C = 4 : 5 =4

15:3

4

35:

4

34 = [for getting same number in B, we are to multiply by

4

3]

C : D = 7 : 9 = 7 28

135:

4

15

28

15 = [to same term of C, multiply by28

15]

\A : B : C : D = 2 : 3 :4

15: .135:105:84:56

28135 =

8. Two numbers are in the ratio of 3 : 5 and if 10 be subtracted from each of them, the

remainders are in the ratio of 1 : 5, find the numbers.

Let the numbers be x and y, so that )1...(y3x5,or53

yx ==

Again5

1

10y

10x =--

MATHS1.04


5/39

or, 5xy = 40 .(ii) , Solving (I) & (ii), x = 12, y =20

\ regd. Numbers are 12 and 20.

9. The ratio of annual incomes of A and B is 4 : 3 and their annual expenditure is 3 : 2 . If each

of them saves Rs. 1000 a year, find their annual income.

Let the incomes be 4x and 3x (in Rs.)

Now23

1000x31000x4 =

--

or, x = 1000 (on reduction)

\ Income of A = Rs 4000, that of B = Rs. 3000.

10. The prime cost of an article was three times the value of material used. The cost of raw

materials was increased in the ratio 3 : 4 and the productive wage was increased in the ratio

4 : 5. Find the present price cost of an article, which could formerly be made for Rs. 180.

[I.C.W.A (F) June 2007]

Prime cost = x + y, where x = productive wage, y = material used.

Now prime cost = 180 =3y or, y = 60, again x + y = 180, x = 180y = 18060 = 120

Present material cost = ,3

y4present wage = ,

4

x5

\Present prime cost = .230.Rs1508041205

3604 =+=+

SELF EXAMINATION QUESTIONS :

1. The ratio of the present age of a father to that of his son is 5 : 3. Ten years hence the ratio

would be 3 : 2. Find their present ages. [ICWA (F) 84] [Ans. 50,30]

1. The monthly salaries of two persons are in the ratio of 3 : 5. If each receives an increase ofRs. 20 in salary, the ratio is altered to 13 : 21. Find the respective salaries.

[Ans. Rs. 240, Rs. 400]

3. What must be subtracted from each of the numbers 17, 25, 31, 47 so that the remainders may be

in proportion. [Ans. 3]

4.ba

zac

y

cbx

If+

=+

=+

show that (bc)x+(ca)y + (ab)z = 0

5.a2

x3y4

b3

y3z4

c4

z3x4If

-=

-=- show that each ratio =

c4b3a2

zyx

++

++.

6. If 0)zyx(if,21

kthatprovekyx

zxz

y

zyx ++==

+=

+=

+

7. If ,21

ba

ba =+-

prove that ]2001June)F(ICWA[7391

baba

baba22

22=

+-

++

2 2 2

2 2 2

Hint s :2 a 2 b a b or, a 3 b or,a 9b.

81b 9b bL.H.S &etc.

81b 9b b

- = + = =

+ +=

- +

MATHS 1.05


6/39

8. If 2c

cbathatprove,

9c

5b

4a =++== [ICWA Dec. 2000]

===== .etc&k9c,k4a;k

9c

5b

4a

:sintH

9. (i) Ifc

ba

b

ac

a

cb +=+=+ and a + b + c 0 then show that each of these ratios is equal to 2.

Also prove that 222 cba ++ = ab + bc + ca. [ICWA (Prel.) Dec. 90]

(ii) If a : b = c : d show that xa + yb : b-a+=b-a dc:ydxcba:

10. Givenqpprrq -

g=

-b

=-a

, prove that g+b+a==g+b+a rqp0 [ICWA, July, 62]

11. In a certain test, the number of successful candidates was three times than that of unsuccessful

candidates. If there had been 16 fewer candidates and if 6 more would have been unsuccessful, the

numbers would have been as 2 to 1. Find the number of candidates. [Ans. 136]

12. (i) Monthly incomes to two persons are in ratio of 4 : 5 and their monthly expenditures are in

the ratio of 7 : 9. If each saves Rs. 50 a month, find their monthly incomes.

[Ans. Rs. 400, Rs.500]

(ii) Monthly incomes of two persons Ram and Rahim are in the ratio 5 : 7 and their monthly

expenditures are in the ratio 7 : 11. If each of them saves Rs. 60 per month. Find their monthly

income. [ICWA(F) June 2003][Ans. 200, Rs. 280]

=-

-.etc&11

7

60x7

60x5sinth

13. A certain product C is made or two ingredients A and B in the proportion of 2 : 5. The price

of A is three times that of B.

The overall cost of C is Rs. 5.20 per tonne including labour charges of 80 paise per tonne. Find

the cost A and B per tonne. [Ans. Rs. 8.40, Rs. 2.80]

14. The prime cost of an article was three times than the value of materials used. The cost of

raw materials increases in the ratio of 3 : 7 and productive wages as 4 : 9. Find the present prime

cost of an article which could formerly be made for Rs. 18. [Ans. Rs. 41]

15. There has been increment in the wages of labourers in a factory in the ratio of 22 : 25, but

there has also been a reduction in the number of labourers in the ratio of 15 : 11. Find out in what

ratio the total wage bill of the factory would be increased or decreased. [Ans. 6 : 5 decrease]

16. Three spheres of diameters 2, 3 and 4 cms. respectively formed into a single sphere. Find the

diameter of the new sphere assuming that the volume of a sphere is proportional to the cube of its

diameter. [Ans. cm993 ]

MATHS1.06

A

RITHMETIC


7/39

MATHS 1.07

OBJECTIVEQUESTIONS:

1. Find the ratio compounded of 3 :7, 21 : 25, 50 : 54 [Ans. 1 :2. What number is to be added to each term of the ratio 2 : 5 to make to equal 4 :5. [Ans. 13. Find the value of x when x is a mean proportional between : (i) x2 and x+6(ii) 2 and 32 [Ans. (I) 3 (ii) 8] [ICWA (F

4. If the mean proportional between x and 2 is 4, find x [ICWA (F) June 2007] [Ans.5. If the two numbers 20 and x + 2 are in the ratio of 2 : 3 ; find x

[ICWA(F) Dec.2006] [Ans. 2

6. Ifba

find21

ba

ba =+-

[I.C.W.A. (F) Dec. 2006] [Ans.19

7. If 3, x and 27 are in continued proportion, find x [Ans. 8. What number is to be added to each term of the ratio 2 : 5 to make it 3 : 4 ? [Ans. 7]

=+

+.etc&

4

3

x5

x2:sinth

9. If 2ba

ba =-+

find the value of22

22

baba

baba

++

+-[Ans.

13

7] [I.C.W.A. (F) Dec.200

10. Ifa2

x3y4

b3

y3z4

c4

y3x4 -=

-=

-show that each ratio is equal to

c4b3a2

zyx

++++

[I.C.W.A (F) June, 200

++

-+-+-= .etc&

a2b3c4

x3y4y3z4z3x4ratioeach:sinth

11. The ratio of the present age of mother to her daughter is 5 : 3. Ten years hence the ratiowould be 3 : 2. Find their present ages. [I.C.W.A. (F) Dec. 2004] [Ans. 50; 30 year

12. If A : B = 2 : 3, B : C = 4 : 5 and A : C [Ans. 8 :1

13. If x : y = 3 : 2, find the value of (4x2y) : (x + y) [Ans. 8 :

14. If 15 men working 10 days earn Rs. 500. How much will 12 men earn working 14 days?

[Ans. Rs. 56

15. Fill up the gaps :-

==-

=-

=22

1

1 ba

b

a[Ans. ab, ordein,ab,ba,ab 3

16. If x, 12, y and 27 are in continued proportion, find the value of x and y [Ans. 8 ; 1

17. If ,43

yx = find the value of

yx3

y4x7

+-

[Ans.135

18. What number should be subtracted from each of the numbers 17, 25, 31, 47 so that the

remainders are in proportion. [Ans

=

--=

--

3orx47x31

x25x17

:sinth

19. 10 years before, the ages of father and son was in the ratio 5 : 2; at present their total age is

90 years. Find the present age of the son. [Ans. 30 year

[Ans. 28]

[Ans. 10]

[Ans. 1 : 3]

[Ans. 8]


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MATHS1.08

A

RITHMETIC

20. The ratio of work done by (x1) men in (x+1) days to that of (x+2) men in (x1) days is

9 : 10 , find the value of x. [Ans.

=-++-

.etc&10

9

)1x()2x(

)1x()1x(:sinth

1.2 AVERAGE :

The arithmetic average or arithmetic mean or simple mean of a number of quantities is the sum of

the quantities divided by their number.

If x1, x2,.,x3, are the n numbers, then the average X is given by

n

x....xxX n21

+++=

For example, if there are 5 boys whose height are 50, 54, 52, 56 and 58 inches, then the average

(or mean) .Inches545

5856525450 =++++=

quantitiesofnumber

quantitiesofsumaverage,findweThus =

or, average number of quantities = sum of quantities.

Note : When quantities x1, x2,.,xn are all different, the average of the numbers is known as

simple average.

WEIGHTED AVERAGE :

If there are n quantities x1, x2,.,xn and w1, w2,.wn are their respective weights, then the

weighted arithmetic average is given by

n21

nn2211

w.....ww

xw.............xwxw

+++

+++

For example, a contractor pays wages to his employees at the following rates : Rs. 2.50 per man

per day and Rs. 2.25 per woman per day. If he engages 20 men and 12 women,


9/39

Let number of girls be m and that of boys be n so that total numbers of all candidates be (m + n).

Now total score of girls = 75 m

Total score of boys = 70 n

Total score of all candidates = 72 (m + n)

By question, 75m + 70 n = 72 m + 72 n

Or, 3m = 2n or,32

nm =

.3:2isratio.reqd\

3. In a class there are three divisions. The number of students and the average marks in

mathematics in the three divisions are 30, 40 30 and 40%, 30%, 50% respectively. What is the

average marks in mathematics of the class?

For division I of 30 students, total marks = 30 40 = 1200

,, ,, II ,, 40 ,, ,, ,, = 40 30 = 1200,, ,, III ,, 30,, ,, ,, = 30 50 = 1500

For the whole class of 100 students total marks = 3900

\ average mark of the class =1003900

= 39.

4. The average monthly income of 6 employees is Rs. 310. After one of them receives an

increment the average rises to Rs. 312.50. Find the amount of increment.

Total income of 6 employees = 6 310 = Rs. 1860

Total income of 6 employees after increment of one employee only = 6 312.50 = 1875.00

\ amount of increments = 18751860 = Rs. 15.

5. During the year 1932, the bank rates where as follows : 6% for 1 week, 5% for 1 week, 4%

for 1 week, 4% for 17 weeks, 3% for 3 weeks, 3 % for 5 weeks and 4% for 24 weeks. What

was the average rate during the year?

Diff. rates are 6% 5% 4% 4% 3% 3% 4%

Occurs (time) 1 1 1 17 3 5 24

\ required average rate

%245317111

244533317414151621

21

++++++

++++++=

%.5249

3%50205 ==

Again, the simple average rate6

33445621

21 +++++

= = %2

14 (as there are 6 diff. rates)

This average is different from the correct average.

MATHS 1.09


10/39

6. The average dividend paid by a company during the last five consecutive years was 10%. The

dividends for the last four years were %21

8 , %21

9 , 11% and 12%. What was the dividend for the

first year?

Total dividends for last four years = %

2

18 + %

2

19 +11+12 = 41%

Since the average dividend for 5 years was 10%, so the total dividend for 5 years = 5 10% =

50%.

\Dividend for the first year = 5041 = 9%.

7. The weight of a person during July is 85.50 kg. and every following month he reduces his

weight by 1/50 kg. than the previous month. Find out his weight in June and by how much

percent there is a shortfall from his average weight of the year.

From July to June we find 1 year (= 12 months)

His weight in August = 85.50 1.50 kg.; that in September = 85.50 21.50 = 82.50 kg.

In the same way his weight in June = 85.50 111.50 = 85.50 16.50 = 69.00 kg.

\ total weight for the whole year will be

85.50 + 84 + 82.50+81.00+79.50+78.00+76.50+75.00+73.50+72.00+70.50+69.00 = 927.00 kg.

\Average weight = .kg25.7712

927 =

Again shortfall = 77. 25 69.00 = 8.25 kg.

Av. Weight Shortfall x = .)app(%..

.6810100

2577

258 =

77.25 8.25

100 x \ reqd. percentage = 10.4 (app.)

8. The average age of a class of 17 boys at the end of a term was 14 years 3 months. Of these 5 left

of the average age of 15 years 2 months and at the beginning of the next term, 2 months later 6

new boys of average age 13 years 11 months came into the class. What was then to the nearest

month, the average age of the class?

Total age of 17 boys = 14 yrs. 3m. 17 = 242 yrs. 3m.

Total age of 5 boys left the class = 15 yrs. 2m. 5 = 75 yrs. 10 m.

Now, total age of (17 5=) 12 boys belonging to the class = 242 yrs. 3m. 75 yrs. 10 m. = 166yrs. 5 m.

Again, total age of 6 new boys joined the class = 13 yrs. 11 m. 6 = 83 yrs. 6m.

Now, total age of (12+6=) 18 boys in the class = 166 yrs. 5 m. + 83 yrs. 6 m = 249 yrs. 11 m.

\Average age = .)app.(m11.yrs1318

.m11.yrs249=

9. A school with 78 boys and 72 girls on the books meets 432 times in the year. If each boy loses

one meeting in 9, and each girl one in 8, find the average attendance of each sex for the year.

Out of 9 meetings each boy attends 8 meetings.

MATHS1.10

A

RITHMETIC


11/39

\.432... .. .. 4329

8 = 384 meeting.

78 boys meet = 384 78 times and average attendance of boys per meeting =2169

43278384 = i.e.

69 boys.

Similarly, out of 72 girl, each girl attends = 43287 meetings.

\72 girls meet = 7284327 times

\average attendance of girls per meeting = 634328

724327 =

girls.

10. A person drove his car for first 20 km and then 30 km at an average speed of 20 km and 30 km

per hour respectively. At what speed must he drive next 50 km if the average speed of the whole

distance of his driving is 40 km per hour? [(ICWA (F)

June 2006]

Time required for first 20 km =2020

= 1hr. and time for next 30 km. =3030

= 1hr. lastly time

required for next 50 km =x

50(where x = reqd. average speed)

Again time for whole journey =40

503020 ++=

40

100=

2

5

So we get, 1+1+2

5

x

50 = or,2

5

x

502 =+ or,

2

5

x

50x2 =+

Or, 5x = 4x + 100 or, x = 100.

\ reqd. average speed = 100/hr.

11. The average salary per head of all the works in an institution is Rs. 60. The average per head

of 12 officers is Rs. 400. The average salary per head of the rest is Rs. 56. Find the total number of

workers.

Let total number of workers (including officers) = n and number of others = 12

No. of workers other than officers = n12.

Now, 60n = 400 12 + 12+ 56 (n 12) or, 60n = 4800 + 56n 672,

or, 60n 56n = 4128 or, 4n = 4128 or, 1032

4

4128n ==

SELFEXAMINATIONQUESTIONS:

1. The average height of 50 students of a class is 160 cm. The average height of 30 of them is 155

cm. Find the average height of remaining 20 students. [Ans. 167.5 cm.]

2. The mean of 5 numbers is 27. If one number is excluded, the mean becomes 25. Find the

excluded number. [Ans. 35]

3. The average earning of a man for 9 days from 1st

to 9th

January is Rs. 5 per day and that for 10

days from 1st

to 10th

January is Rs. 5.10. What is his earning on the 10th

January?

[I.C.W.A (Prel.) June, 1986] [Ans. Rs.6]

MATHS 1.11


12/39

4. In a factory out of 60 workers 40 are males and 20 are females. The mean wage of a male

worker is Rs. 70. If the mean wage of a male worker is Rs. 80; find the mean wage of a female

worker.

[ICWA(Prel) Dec.1984] [Ans. Rs. 50]

5. From the following table find the value of x.Class No of students Average wt. in kgs

VIII 30 35

IX 35 40

X 50 45

XI x 50

Average weight of all students is 42.4. [ICWA(Prel)June1984] [Ans. 30]

6. A candidate obtains following percentage of marks in an interview for a technical job :

(i) Written examination 75

(ii) Practical 80

(iii) Qualification and achievements 70

(iv) Viva 60.

Find the weighted average marks obtained by the candidate of the weights allotted are 3, 2, 2, 2

respectively. [Ans. 70.5]

7. A candidate obtained the following % of marks in a job interview.

Written examination 83, Practicals 75, Viva 71, G. K. 65. Find the weighted mean in the weights

are 3, 2, 3, 2 respectively [ICWA(Prel) Dec. 1984] [Ans. 74.2]

8. A candidate obtained the following percentages of marks in an examination.

English 75, Mathematics 60, G.K. 59, Accountancy 55, Hindi 63.

Find the weighted mean if the weight are 2,1, 3, 3 and 1 respectively.

[ICWA (prel.) June 1984] [Ans. 61.5]

9. The employees of a company consist of its officers and clerks. The average wages of employees

in the company is Rs. 100. The average wages of 150 clerks is Rs. 80 and the average wages of

officers is Rs. 300. Find the number of officers in that company.

[ICWA (Prel) June 1988] [Ans. 15]

10. (I) A truck runs 25 km. at a speed of 50 km.p.h. another 50 km. at a speed of 60 km. p.h. then

due to bad condition of road travels for 12 minutes at a speed of 5 km. p.h. and finally covers the

remaining distance of 24 km. at a speed of 30 km. p.h. show that the average speed of the truck is

76

42 km. p.h.

MATHS1.12

A

RITHMETIC


13/39

(ii) Form the following data, find the value of x, if the average marks in mathematics for all the

centres together is 52.

Centre No. of student Average mark

Scored in mathematics

Bombay 30 45

Calcutta 30 55

Delhi x 50

Madras 40 60

[ICWA(Prel) June 1990] [Ans. 100]

(iii) The average marks in Elements of Mathematics of Preliminary students of 3 centres in

India is 50. The number of candidates in three centres are respectively 100, 120, 150. If the

averages of the first two centres are 70 and 40, find the average marks of the third centre.

ICWA (Prel.) Dec. 1990] [ Ans.44.67 marks]

++

++= .etc&

150120100

x150401207010050:sintH

11. A person drove his car 20 km. at an average speed of 25 km. per hour. At what average speed

for the whole distance is to be 30 km. per hour? [Ans. 37.5 km. per hr.]

12. The average score of boys is 60, that of girls is 70 and that of all the candidates is 64

appearing in mathematics of annual examination. Find the ratio of number of boys and the

number of girls there. If the total number of candidates appearing in mathematics is 150, find the

number of boys there.

[ICWA (F) Dec. 2006] [Ans. 3 : 2 ; 90]

OBJECTIVE QUESTIONS :

1. The average of 3 numbers is 15. With inclusion of a fourth number, the average becomes 17.Find the included number. [ICWA(F) June 2004] [Ans. 23]

2. Speed of a car to go up a hill us 10 km. per hour and to go down is 20 km per hour. Compuleit average speed. [ICWA(F) June, 2003] [Ans.

31

13 km. per. Hr.]

[hints : refer w.o. ex. 12]

3. The mean of 5 numbers is 27. If one number is excluded the mean becomes 25. Find theexcluded number. [Ans. 35]

4. The average height of 50 students in a class is 160 cm. The average height of 30 of them is155 cm. Find the average height of remaining 20 students. [Ans. 167.5 cm]

5. The mean of 4 numbers is 9. If one number is excluded the mean becomes 8. Find theexcluded number. [Ans. 12]

6. The average weight of 24 students is a class is 35 kg. If the weight of the teacher be included,the average rises by 400 gms. Find the weight of the teacher. [Ans. 45 kg]

MATHS 1.13


14/39

MATHS1.14

A

RITHMETIC

1. A batsman scored 87 runs in the 17th innings and thus increased his average by 3. Find hisaverage after 17

thinnings. [Ans. 39]

2. The average of 50 numbers is 38. If two numbers, say 45 and 55 are discarded, find average ofremaining numbers. [Ans. 37.5]

3. The average age of three boys is 15 years. If their ages are in the ratio 3 : 5 : 7; Find the ageof youngest boy [Ans. 9 years]

4. The average score of a Cricketer for 10 matches is 43.9 runs. If the average for the first sixmatches is 53, find the average for the last four matches. [Ans. 30.25]

5. The average salary of 20 workers in an office is Rs.1900 per month. If the managers salaryis added, the average salary becomes Rs. 2000 p/m. Find the managers annual salary.]

[Ans. 48,000]

6. The average age of four children in a family is 12 years. If the spacing between their ages is 4years, find the age of the youngest child. [Ans. 6 years]

1.3 MIXTURE :

Problems of mixture can be classified into two waysDirect and Inverse.

In case of direct problems, if the individual price of the different ingredients and the proportion in

which they are to mixed are given, we can find the price of mixture.

Formula. Price of mixture (per unit quantity) =quantitytotal

tcostotal.

In case of inverse problems, we are to find the proportions of the ingredients are given.

Example : To find proportion in which two kinds of tea costing Rs. 6 and Rs. 6.30 P per kg aremixed up to produce a muxture of costing Rs. 6.20 P per kg.

If 10 kg. of first kind are taken, then there is a gain of 10 20P = Rs. 2.

Similarly, if 20 kg. of second kind are taken, the loss is 20 10 P = Rs. 2.

So, 10 kg. of first kind are to be mixed with 20 kg. of second kind. Hence the mixture should be

10 : 20 = 1 : 2.

Rule. First kind; gain per kg. (i.e. difference) = 20

Second kind per kg. loss = 10

\First : Second = 10 : 20 = 1 : 2

i.e. First : Second = diff. of second : diff. of first.

WORKING RULE :

(i) All the prices must be expressed in the same unit.

(ii) The cost price (C.P.) of the mixture must lie in-between the cost prices of the other

ingredients.

(iii) All prices used should be cost prices. If in some cases, the selling price of the mixture at

profit or loss is given, find the corresponding cost price.


15/39

(iv) Now applying the following formula to find the required ratio

kindst1ofthatwithmixtureof.P.Cofdifference

kindnd2ofthatwithmixtureof.P.Cofdifference

kindnd2

kindst1=

Alternatively :

Working Rule

\1st kind : 2nd kind = 2nd difference : 1st difference = () : (.)

SOLVED EXAMPLES :

(On direct problems)

1. A dealer blends 75 kg. of salt @ Rs. 1.50 per kg. with 45 kg. of salt @ Rs. 1.30 per kg. what is

the lowest price at which he can sell the mixture so as to gain at least 25%?

Cost price (C.P.) of the mixture120171

.Rs7545

50.1530.145 =+

+=

C.P. S.P.

100 125 \x = .)app(78.1.Rs100120

171125 =

120171

x

the lowest price per kg = Rs. 1.78.

2. A man mixes 5 gallons of spirit at Rs. 5.50 per gallon,12 gallons at Rs. 6 per gallon and 13

gallons at Rs. 6.50 per gallon. At what price per gallon must he sell mixture so as to gain Rs. 26

on the whole?

C.P. of 5 gallons = Rs. 5.50 5 = Rs. 27.50

C.P. of 12 gallons = Rs. 72.00

C.P. of 13 gallons = Rs. 6.50 13 = Rs. 84.50

\C.P. of 30 (= 5+12+13) gallons = Rs. (27.50 + 72.00 + 84.50) = Rs. 184.

Total gain = Rs. 26

\ S. P. of 30 galls. = Rs. (184 + 26) = Rs. 210

\ S. P. per gallon = Rs.30

210= Rs. 7.

3. A merchant buys sugar at Rs. 4.10, Rs. 3.75 and Rs. 4.50 per kg. and mixes in the proportion

5 : 4 : 1. At what price must he sell the mixture so to gain 25%?

C.P. of the mixture per kg.

4.Rs10

00.40.Rs

145

150.4475.3510.4 ==++

++

Cost Price of ingredients 1st kind 2nd kind

Cost Price of mixture

Difference

(...) (...)

(...) (...)

(...)

MATHS 1.15


16/39

C.P. S.P.

100 125 4X 125 Rs.5.00100

\ = =

4 x\S. P. per kg. = Rs. 5.00.

4. There are two one-litre decanters containing liquor and water in the ratios 7: 3 and 8 : 2

respectively. The contents of the two decanters are poured into a two litre decanter, and the

mixture is sold at Re. 1 per litre, thereby realising a profit of %2

112 on the returns. Find the cost

price of a litre of liquor.

Quantity of liquor the 1st

decanter litre12

7=

Quantity of liquor the 2nd

decanter litre108=

Total quantity of liquor in the 2-litre decanter =1015

108

107 =+ litre.

Now S.P. of 2 litres = Rs. 2, gain = 0.25 P. (= %2

112 of Rs. 2)

\ C.P. = 2 0.25 = Rs. 1.75

2 litres of mixture contains1015

litre of liquor

\C.P. of1015

litre liquor = Rs. 1.75 (as water is free of cost)

C.P. of 1 litre liquor = 1.75 1510 = Rs1.16 P. (app.)

5. 30 litres of mixed paint containing 10% oil is to be made into a paste containing 50% oil. Find

the quantity of oil added. If 6 more litres are added, find the percentage of colouring matter in the

mixture.

Quantity of oil in 30 lmixed paint = 33010010 = l; so quantity of paint = 30 3 = 27 l. For 50%

oil in the mixed paint quantity of oil should be made equal to that paint. So quantity of oil = 27 l

i.e. oil to be added = 27 3 = 24l.

If again 6 loil being added more, total quantity of oil = 27 + 6 = 33 l

And total quantity of mixed paint = 27 + 27 + 6 = 60 l.

\Reqd. percentage of paint %.451006027 ==

6. In a liquid mixture 20% is water, and in another mixture water is 25%. These two mixtures are

mixed in the ratio 5 : 3. Find the percentage of water in the final mixture.

In the first pot liquid : water= 80 : 20 = 4 : 1

(as water = 20% so liquid = 80%)

In the second pot liquid : water = 3: 1

MATHS1.16

A

RITHMETIC


17/39

Let 5 litres of mixture from first pot be mixed with 3 litres of second pot In

5lmixture, water 155

15

41

1 ==+

=

In 3 lmixture, water43

341

331

1 ==+

=

In 8 lmixture, total water47

431 =+=

In 100 lmixture total water =87

218

10047 =

\ reqd. percentage =87

21

7. Two types of oil A and B are mixed in the ratio 3 : 2. After selling 25% of this mixture, type A

is mixed to the remaining mixture so as to make a new ratio 5 : 3. If this new mixture is 400 litres

by quantity, find the quantity of original mixture (before selling) and also the quantity of A mixed

later on.

In 400 lmixture, quantity of A = 40035

5 +

= 250 l.

And that of B = (400 250) = 150 l. Let the quantity of mixture (after selling 25%) = x l, quantity

of5x3

A = , quantity5x2

B = .

Now as quantity of B remains fixed, so5x2

= 150 or, x = 375 l

i.e., quantity of A = 3755

3

= 150 l

So quantity of A added = 250 225 = 25.

Let original mixture = y l.

Now 75 % of y = 375 or, 375y10075 =

or, 50075

100375y ==

\ reqd. quantity = 500 l.

8. 4 Litres are drawn from a cask full of sepecial liquid, it is then filled with water then 6 litres of

the mixture are drawn and the cask is again filled with water. If the quantity of special liquid now

in the cask be to the quantity of water in it as 1 to 2; how much does the cask hold?

Let x = capacity of cask. After taking 4lof liquid and filling with water, the ratio of liquid and

water is (x4) : 4. Now in 6 lof drawn mixture, there are ( ) 4)4x(6

644x

4x -=+-

-lof liquid.

Remaining liquid in the cask

= (x 4)

x

24x10x

x

24x6x4x

x

)4x(6 22 +-=+--=-

MATHS 1.17


18/39

remaining water in caskx

24x10xx

2 +--= =x

24x10 -

Now 2:1x

24x10:

x

24x10x2 =-+-

or, 24x1024x10x2

-+-

= 21

or 2(x2

10x + 24) = 10x 24

or, 036x15x2 =+-

or, (x 3) (x 12) = 0 or, x = 3, 12

Now x = 3 is impossible, as in the first case 4lof liquid drawn out.

\ reqd. capacity = 12l.

(On inverse problem)

9. How should salts at Rs. 2.50 and Rs. 3.35 perlb. be mixed to produce a mixture worth Rs. 3

per lb.?In the 1

stkind, profit per lb = 3 2.50 = 50 P.

In the 2nd

loss per lb = 3.35 3 = 35 P.

\ 1st : 2nd = 35 : 50 = 7 : 10.

10. A dealer mixes two varieties of grains costing Rs. 6 per quintal and Rs. 15 per quintal in such

a way that can gain 10% by selling the resultant mixture at Rs. 8.25 P. per quintal, What is the

proportion in which the gains are mixed?

On 10% profit, S.P. = 100 + 10 = 110. Now to find the C. P. of mixture.

S.P. C.P.

110 100

8.25 x x = 100 50.7110

25.8 =

So, for the 1st

variety, gain = 7.50 6 = 1.50

And for the 2nd

loss = 15 7.50 = 7.50

\ 1st variety : 2nd variety = 7.50 : 1.50 = 5 : 1

11. A dealer mixes tea costing Rs. 8 per kg, with tea costing Rs. 7 per kg. and thereafter, sells the

mixture at Rs. 8 per kg and earns profit of 7.5% on his sale price. In what proportion does he mixthem? [ICWA (F) Dec. 2003]

To find cost price of mixture. Here gain is on selling price. So

S.P. C.P.

100 92.5

8 ? ? 4.78100

5.92 == (C.P. of mixture)

32

60.040.0

40.78740.7

kindnd2

kindst1==

--=

MATHS1.18

A

RITHMETIC


19/39

12. A dealer mixed two varieties of teas having costs Rs. 1200 and Rs. 2500 each per kg in such a

way that he can gain 20% by selling the resultant mixture at Rs. 1800 per kg. Find the proportion

in which the two types of tea are mixed. [ICWA (F) Dec. 200]

We are to find cost price, here gains is on cost price So

S.P. C.P.

120 100

1800 ? ? 15001800120

100 == \Cost price of mixture = Rs. 1500

310

30001000

1200150015002500

kindnd2

kindst1==

--= \reqd proportion is 10 : 3.

13. In mixing tea, 1 kg. in every 100 kgs. is wasted. In what proportion must a dealer mix teas

which cost him Rs.24 and Rs. 18 per kg. respectively, so that the cost is Rs. 20 per kg.

Let us take 100 kg. of two kinds. Then during the time of mixing, 1 kg. is wasted, leaving 99 kg.

only.

C.P. of 99 kg. of the mixture = Rs.99 20. Again , C.P. of 100 kg. of two kinds = 100x, where x =

average price, if there would be no loss

\ 99 20 =100 x

or, x 80.19100

1980 ==

So,mixtureandkindst1of.)kgper(pricesof.diff

mixtureandkindnd2of)kgper(pricesof.diff

kindnd2

kindst1=

7

3

20.4

80.1

80.1924

1880.19 ==

-

-= \ reqd. proportion = 3 : 7.

14. A lump of gold and silver weighing 200 gm. is worth Rs. 825, but if the weights of gold and

silver be interchanged it would be worth Rs. 562.50. If the price of 10 gm. of gold be Rs. 67.50;

find the rate per 10 gm. of silver. Also calculate the weight of silver in the lump.

Worth of 200 gm. Gold and Silver = 825

Worth of 200 gm. Gold and Silver = 562.50

(When weight of gold and silver interchanged)

\Worth of 200 gm gold and 200 gm of silver = Rs. (825+562.50) = Rs. 1387.50

50.37.Rssilver.gm200ofworth,gSubtractin

00.1350.Rs)2050.67(gold.gm200ofworthAgain

===

worth of 10 gm. 88.1.Rs20

50.37 == From 1st lump,23

52550.787

825135050.37825

silver

gold==

--=

\gold : silver = 3 : 2 Weight of silver = .gm802005

2 =

15. Two vessels contain mixture of milk and water are in the ratio of 5:1 and in the other in the

ratio of 9 : 1. In what proportion the quantities from the two should be mixed together so that the

mixture thus formed may contain milk and water in the ratio 8 : 1?

Let 1 litre mixture from 1st vessel be mixed with klitres mixture of 2nd vesselMATHS 1.19


20/39

MATHS1.20

A

RITHMETIC

From 1st vessel :65

milk= l,61

water= l, taken

From 2ndvessel :10

k9milk= l,

10k

water= l, taken

Total milk60

k545010

k965 +=+ l, Total water

60k610

10k

61 +=+=

By condition, 1:860

k610:

60

k5450 =++ or,1

8

60

k5450 =+

or, k = 5 (on reduction)

\ reqd. proportion = 1 : 5.

Mixture of three or more kinds :

Problems regarding the mixture containing three or more kinds are also solved in similar way. At

first, two kinds are to be selected, one of which is at a higher price than the average price while

the other at a lower price. Similar treatment is made for the other two and so on, till all the kinds

are taken at least once.

It may be noted that such problems give infinite number of solutions.

16. The prices of 3 kinds of tea are Rs. 5.00, Rs. 6.00 and Rs. 6.80 respectively. In what

proportion they should be mixed so that the price of the mixture may be Rs. 6.50 per kg?

Let A, B and C are the three kinds.

Taking A and C, at first,

For A, profit per kg. = 6.50 5.00 = Rs. 1.50 = 150P.

For C, loss per kg. = 6.80 6.50 = Rs. 0.30 = 30 P. \A : C = 30 : 150 = 1 : 5

Taking B and C.

For B, profit per kg. = 6.50 6.00 = Re. 0.50 = 50 P

For C, loss per kg. = 6.80 6.50 = Re. 0.30 = 30 P. \ A : B : C = 1 : 3 : 10

Note. In the above example, if we take A : C = 3 : 15 and B : C = 6, 10, then A : B : C = 3 : 6 : 25

(a different result from the former)

Alternatively

C.P. of 1st

kind = Rs. 5

.2nd kind = Rs. 6

.3 rd kind = Rs. 6.80Let the proportion of mixing the three kinds be x : y : z

\ C.P. per kg. of mixture.

zyx

z5

34y6x5

.Rs++

++=

)zyx(5

z5

34y30x25

.Rs++

++=

Here, C.P. of the mixture = Rs. 6.50 = Rs.2

13per kg.

\ 2

13)zyx(5

z3430x25 =++++


21/39

Solving, 15x + 5y = 3z

If we take x = 1 and y = 3, then z = 10 (A)

Thus one possible proportion is 1 : 3 : 10

Such a problem will have infinite number of solutions.

Note. The values of x and y in (A) should be chosen such that after substituting the values of x and

y in (A) . L.H.S. does not become negative.

On withdrawal :

Rule. Original quantity of liquid in vessel= x units (say) from which if y units being withdrawn

and replaced in same quantity of water for n operations (repeated), then amount of liquid left = x

n

x

y1

- units.

Example . A vessel contains 50 litres of milk. From the vessel 5 litres of milk being withdrawn

and replaced by water by same quantity. The operation being repeated three times. Find the

quantity of milk left in the cask.

Here x = 50, y = 5 , n = 3. Now using the rule, we get quantity of milk left in vessel

33

)10.1(5050

5150 -=

-= 45.36729.50)9.0(50 3 === l

SELFEXAMINATIONQUESTIONS:

1. A milkman mixes 61 litres of water with 349 litres of milk which he buys for Rs. 123. At what

price should he sell a litre so as to save one-third of his outlay? [Ans. 40P]

2. A grocer blends 75kg. salt at Rs. 1.12 P per kg. with 45 kg. of salt at Rs. 1.31 P. per kg. What isthe price at which he should sell the mixture so as to gain 25% [Ans. Rs.1.49P]

3. Three equal jars are filled with mixtures of milk and water, the proportion of milk to water

being at 5 : 2, 4 : 3, and 8 : 3 respectively. The mixture of the jar are poured into a drum. What is

proportion of milk and water in the final mixture? [Ans. 155: 76]

4. There are two one-litre decanters containing liquor and water in the ratios 11 : 4 and 10 : 5

respectively. The contents of the two decanters are poured into a two-litre decanter and the

mixture is sold at Re . 1 per litre thereby realising a profit of %2

112 on the returns. Find the cost

price of a litre of liquor. [Ans. Rs. 1.25]

5. A dealer mixes 90 litres of wine containing 10% of water with 60 litres of wine containing 20%

of water. What is the percentage of water in the mixture. [ICWA(F) June, 2005] [Ans. 14%]

6. Two grades of motor oil A and B are mixed in the proportion 3 : 1 to make 96 litres of grade

C. When half of C has been sold, a further quantity of A is added to increase the proportion of A

to B in the resulting mixture to 7 : 2, Find the quantity of last added. [Ans. 6 litres]

7. Two grades C and D of alcohol are mixed in the proportion 3 : 2. After 25% of this has been sold

from stock, a sufficient quantity of C is mixed with the remainder to raise the proportion to

MATHS 1.21


22/39

5 : 3. If the stock is now 7200 gallons what was the quantity of the original mixture and how much

of grade C was added to make the new mixture? [Ans. 9000 gals. ; 450 gals.]

8. A dealer mixes tea costing Rs. 8. Per kg. with tea costing Rs. 7 per kg. and sells the mixture at

Rs. 8 per kg. with tea costing Rs. 7 per kg. and sells the mixture at Rs. 8 per kg. and earns a profit

of 21

7 % on his sale price. In what proportion does he mix them? [Ans. 2: 3]

9. A tea dealer mixes two qualities of tea at Rs. 2.56 P. and Rs. 5.53 P. per kg. and sells the

mixture at Rs. 3.41 P. per kg. to make a profit of 10% on his outlay. In what proportion does he

mix the two qualities? [Ans. 9 :2]

10. A dealer buys oranges of two qualities one at Rs. 12 a dozen and the other at Rs. 8 a dozen.

These were mixed up and he sells them at 15 per Rs. 12 and thereby makes 5% on his total

outlay. Find the ratio of the number of oranges of the two kinds. [Ans. 2 : 5]

11. A grocer sells one kind of tea at Rs. 2.70 P. per kg. and loses 10% and another at Rs. 4.50P.

per kg. and gains2

112 %. How the two quantities of tea should be mixed, so that the mixture may

be sold at Rs. 3.95. P. kg. at a profit of 25%. [Ans. 21 : 4]

12. A grocer mixes 4 kinds of rice which cost him Rs. 5, Rs. 4, Rs. 3 and Rs. 2 per kg.

respectively, in the proportion of 2, 3, 4, 7 respectively. Find at what rate he must sell the mixture

so as to gain 25% on the whole. [Ans. Rs. 3.75]

13. A dealer mixes varieties of grains at Rs. 6 per kg. and Rs. 15 per kg. and in such a way that he

can gain 10% by selling the resulting mixture at Rs. 8.25 per kg. What is the proportion in which

the grains are mixed? [Ans. 5 : 1]

14. (i) In mixing tea 1 kg. in every 100 kg. is wasted. In what proportion must a dealer mix teas

which cost him Rs. 42 and Rs. 32 per kg. respectively so as to gain 10% by selling the mixture at

Rs. 40 per kg. [Ans. 2 : 3]

(ii) In mixing tea 1% is wasted. In what proportion must a dealer mix two grades of tea which

cost him Rs. 4.50 and Rs. 7.00 per kg. respectively, so as to gain 8.9% by selling the mixture at

Rs. 5.50 per kg? [Ans. 4: 1]

15. What weight of tea worth Rs. 7.00 per kg. should be mixed with 60 kg. of tea worth Rs. 5.00

per kg. so that when the mixture is sold at Rs. 5.50 per kg. there may be neither gain nor loss?

[Ans. 20 kg]

16. A gold is value at Rs. 10 per gm. An alloy of gold and silver weighs 1 kg. and its value is Rs.

6080. If the weights of gold and silver are interchanged it would worth Rs. 4120. Find the

proportion of gold and silver in and alloy also the price of silver per kg. [Ans. Rs. 200; 3 : 2]

17. With grain worth Rs. 10 per kg. a trader mixes an inferior quality worth Rs. 6 per kg. In what

proportion must he mix them so that by selling the mixture at the higher price he may gain 16%?

[Ans. 19 : 20 ]

MATHS1.22

A

RITHMETIC


23/39

18. (i) Two casks A and B are filled with the two kinds of liquids; mixed in cask A in the ratio of

2 : 7 and in cask B in the ratio of 1 : 5. What quantity must be taken from each to form a mixture

which shall consist of 2 litres to one kind and 9 litres of the other? [Ans. 3 : 8]

(ii) Two vessel contain mixtures of milk and water in the proportion 2: 3 and 4 : 3 respectively. In

what proportions should the two mixtures be mixed so as to form a new mixture containing equal

quantities of milk and water? [ICWA (F) Dec, 2005] [Ans. 5 : 7]

19. Three vessels have capacities in the ratio 1 : 2 : 3. All the three glasses are filled with

mixtures of milk and water. The proportion of milk and water in the first glass is 3 : 1, in the

second the proportion is 5 : 3 and in the third it is 9 : 7. If the contents of all the three glasses are

put into a bigger single vessel what will be the proportion of milk and water in the mixture in the

bigger vessel? What is the percentage of milk in the final mixture in the bigger vessel?

[Ans. 59 : 37 : 61.46%]

20. A trader mixes 100 lbs. of tea at one price with 70 lbs. of tea at a dealer price. By selling the

mixture at Rs. 2.12 per lb.he gained 20% on his outlay. Find the value of each kind of tea, the

difference in their prices being Re. 0.36 per lb. [Ans. Rs. 1.62; Rs. 1.98]

21. Nine litres are drawn from a pot of full of milk. It is then filled with water, then nine litres of

the mixture are drawn and the pot is again filled with water. If the ratio of the quantity of milk

now in the pot to the quantity of water in it 16 : 9, how much does the pot hold?

[Ans. 45 litres]

22. Several litres of milk were drawn off a 54 litre vessel full of milk and an equal volume of

water added. Again the same volume of mixture were drawn off. As a result the mixture in thevessel contains 24 litres of pure milk. How much milk was drawn off initially? [Ans. 18 l]

=

- .etc&24

54

x154Use.Hints

2

23. Four litres are drawn from a cask full of wine. It is then filled with water. Four litres of

mixture are again drawn and the cask is again filled with water. The quantity of wine now left in

the cask is to that of wine in it is 36 : 13. How much does the cask hold? [Ans. 28 l]

OBJECTIVEQUESTIONS:

1. A mixture of 30 litres contain wine and water in the ratio 7 : 3, how much water must be added

to it., so that the ratio of wine and water may become 3 : 7 ? [Ans. 24 l]

2. 65 litres of a mixture of wine and water contain 15 litres more wine than water. Find the ratio of

wine to water. [Ans. 8 : 5]

Hints : 65 15 = 50 litres water = 255021 = l.

Total wine = 25 + 15 = 40; ratio = 40 : 25 = 8 : 5]

3. In what ratio salt worth Rs. 6 per kg. be mixed with salt worth Rs. 7.30 per kg. to make mixture

worth Rs. 6.50 per kg? [ Ans. 8 : 5 ]

MATHS 1.23


24/39

SOLVEDEXAMPLES :

1. Amit deposited Rs. 1200 to a bank at 9% interest p.a. find the total interest that he will

get at the end of 3 years.

Here P = 1200, ,09.0100

9i == n = 3, I = ?

I = P. i.n = 1200 0.09 3 = 324.

Amit will get Rs. 324 as interest.

2. Sumit borrowed Rs. 7500 at 14.5% p.a. for2

12 years. Find the amount he had to pay

after that period.

P = 7500, ,145.0100

5.14i == ?A,5.2

21

2n ===

A = P (1+ in) = 7500 (1+0.1452.5) = 7500 (1+0.3625)

= 7500 1.3625 = 10218.75

\reqd. amount = Rs. 10218.75.

3. Find the simple interest on 5600 at 12% p.a. from July 15 to September 26, 1993.Time = number of days from July 15 to Sept. 26

= 16 (July) + 31 (Aug.) + 26 (Sept.)= 73 days.

P = 5600,10012

i = =0.12, .yr36573

n = = .yr51

S.I. = P. i.n. = 5600 0.12 51

=134.4

\ reqd. S.I. = Rs. 134.40.(In counting days one of two extreme days is to be excluded, Usually the first day is

excluded).

To find Principle :

4. What sum of money will amount to Rs. 1380 in 3 years at 5% simple interest?

Here A = 1380, n = 3, ?P,05.0100

5i ===

From A = P (1 + 0.05.3) or, 1380 = P (1+0.15)

Or, 1380 = P (1.15) or, P1380

12001.15= =

\reqd. sum = Rs. 1200

5. What sum of money will yield Rs. 1407 as interest in2

11 year at 14% p.a. simple interest.

Here S.I. = 1407, n = 1.5, I = 0.14, P = ?

S.I. = P. i.n. or, 1407 = P 0.14 1.5

Or, 670021.0

14075.114.1

1407P ==

=

\ reqd. amount = Rs. 6700.MATHS 1.25


25/39

6. What principal will produce Rs. 50.50 interest in 2 years at 5% p.a. simple interest.

S.I. = 50.50, n = 2, i = 0.05, P = ?

S.I. = P. i.n. or, 50.50 = P 0.05 2 = P 0.10

or, 50510.050.50

P ==

\ reqd. principal = Rs. 505.

Problems to find rate % :

7. A sum of Rs. 1200 was lent out for 2 years at S. I. The lender got Rs. 1536 in all. Find the

rate of interest p.a.

P = 1200, A = 1536, n = 2, i = ?

A = P (I + ni) or 1536 = 1200 (1 +2I) = 1200 + 2400 I

or, 2400i = 1536 1200 = 336 or, 14.02400336

i ==

\ reqd. rate = 0.14100 = 14%.

8. At what rate percent will a sum, become double of itself in2

15 years at simple interest?

A = 2P, P = Principal,21

5n = , i = ?

A = P (1 + ni) or, 2P = P (1+2

11i)

or, 2 = 1+2

11i or,

11

2i =

or, 18.18100112

n == (app.); \reqd. rate = 18.18%.

Problems to find time :

9. In how many years will a sum be double of itself at 10% p.a. simple interest.

A = 2P, P = Principal,100

10i = = 0.10, n= ?

A = P (1 + ni) or, 2P = P [1 + n(.10)] or, 2 = 1 + n (.10)

or, n (.10) = 1 or, 1010.1

n ==

\Reqd. time = 10 years.

10. In what time Rs. 5000 will yield Rs. 1100 @2

15 %?

P = 5000, S.I. = 1100, ,055.0100

5i 2

1

== n = ?

S.I.= P. ni or, 1100 5000 n (0.055) = 275 n

or, .4275

1100n == \reqd. time = 4 years.

MATHS1.26

A

RITHMETIC


26/39

Miscellaneous worked out Examples :

11. In a certain time Rs. 1200 becomes Rs. 1560 at 10% p.a. simple interest. Find the

principal that will become Rs. 2232 at 8% p.a. in the same time.

In 1st case : P = 1200, A = 1560, i = 0.10, n = ?

1560 = 1200 [1+n (.10)] = 1200 + 120 nor, 120 n = 360 or, n = 3

In 2nd case : A = 2232, n = 3, i = 0.08, P = ?

2232 = P (1+30.08) = P (1 + 0.24) = 1.24 P

or, .180024.1

2232P ==

12. A sum of money amount to Rs. 2600 in 3 years and to Rs.2600 in 3 years and to Rs.

2900 in2

14 years at simple interest. Find the sum and rate of interest.

Amount in2

14 yrs. = 2900

Amount in 3 years = 2600

\ S. I. for2

11 yrs. = 300

S.I. for 1 yr. 20032

3001

300

21

===

and S.I. for 3 years.= 3200=600

\ Principal = 2600 600 = Rs. 2000P = 2000, A = 2600, n = 3, i = ?

2600 = 2000 (1+ 3 i) = 2000 + 6000 i

or, 6000 i = 600 or101

6000600

i == or, 1010

100r ==

\ reqd. rate = 10%.

Alternatively. 2600 = P(1+3 i).(i), 2900 = P (1+ 4.5 i).(ii)

Dividing (ii) by (I),,i31

i5.41

)i31(P

)i5.41(P

26002900

++

=+

+=

or,i31

i5.41

2629

++

= or, i = 0.10 (no reduction)

or, r = 0.10 100 = 10%

From (i) 2600 = P (1+30.10) = P (1+0.30) = P (1.30)

.200030.1

2600P ==\

MATHS 1.27


27/39

13. A person lent some amount at 12% p.a. for2

12 years and some amount at 12.5% p.a. for

2 yrs. If he had amount of Rs. 10,000 in hand and on such investment earned Rs. 2700 in all,

find the amount he invested in each case.

In 1st case let the investment be Rs. x, then in 2nd case, it would be Rs. (10000 x)

In 1st case, interest earned10

x3212

10012x ==

In 2nd case, interest earned = (10000 x)12.5 x(1000 x) 2 2500

4100= - = -

By question, 27004x

250010

x3 =-+

or, 4000xor20020x

or2004x

10x3 ===-

\ reqd. investment in 1st case = Rs. 4000.

And that in 2nd case = Rs. (10000 4000) = Rs. 6000.

14. Divide Rs. 2760 in two parts such that simple interest on one part at 12.5% p.a. for 2

years is equal to the simple interest on the other part at 12.5% p.a. for 3 years.

Investment in 1st case = Rs. x (say)

Investment in 2nd case = Rs. (2700 x)

Interest in 1st case = x 5

x2

100

10 =

Interest in 2nd case8x3

10353100

5.12)x2760( -=-=

By question,5x

=1035 8x3

or, 10358x3

5x =+

or, 103540

x15x8 =+ or, 103540

x23 = or, x = 1800

\ Investment in 2nd case = Rs. (2760 1800) = Rs. 960.

15. A person borrowed Rs. 8,000 at a certain rate of interest for 2 years and then Rs. 10,000

at 1% lower than the first. In all he paid Rs. 2500 as interest in 3 years. Find the two rates

at which he borrowed the amount.

Let the rate of interest = r, so that in the 2nd case, rate of interest will be (r1). Now

25001100

)1r(000,102

100r

8000 =-

+

or, 160r + 100 r 100 = 2500 or, r = 10

\In 1st case rate of interest = 10% and in 2nd case rate of interest = (10 1) = 9%

Calculations of interest on deposits in a bank : Banks allow interest at a fixed rate on deposits

from a fixed day of each month up to last day of the month. Again interest may also be calculated

by days.

MATHS1.28

A

RITHMETIC


28/39

16. A man deposited Rs. 5000 on 20th

April in a Co., paying interest at 2% p.a. He

withdraws Rs. 3000 on 15th

may and deposited Rs.400on 6th

June. How much interest was

due to him on 30th June following?

From 21st

April to 14th

May = (10 +14) = 24 days. For investment of Rs. 5000 for 24 days, the

corresponding interest (l2) due

,73

480.Rs= From I =

n.r.P.

100

From 15 th may to 5th

June = (17+5) = 22 days.

For the investment of Rs. (5000 3000) for 22 days,

The interest (I2)= Rs.176

73

From 6th

June to 30th

June = 25 days

For the investment of Rs. (2000 + 4000) for 25 days

The interest (I3) = Rs.

73

600

Total interest I1+I2+I3 =73

600

73

176

73

480 ++ = Rs. 17.21 (approx).

Interest on instalment basis :

For purchasing costly goods, instead of each down price, instalment payment is introduced. In that

case, the seller charges some interest, calculation of which will be clear from the following

example:

17. If I buy a watch with cash down, I pay Rs. 150. If I pay it on instalment basis, I pay

Rs.10 cash down and 10 monthly instalment of Rs. 15. What rate of interest is calculated in

the second case ?

The price of the watch = Rs. 150. Now Rs. 10 is paid in cash and the balance Rs. 140 is to

cleared up with interest in 10 monthly instalment of Rs. 15 each. So the amount of Rs. 140 for 10

months = the sum of amount of the successive instalments of Rs. 15 each, with the respective

earning of interest. Taking i = interest of Re. 1 for one month, in this case

140(1+10 i) = 15 (1 + 9 i) + 15 (1+8 i) + .+ 15 (1 + i) + 15

= 15 10 + 15 i (9 + 8+ ..+1)

= 150 + 15 i 45 = 150 + 675 i

or, 140 + 1400 i = 150 + 675i or, 725 i = 10 or,725

10i =

Now interest for Rs. 100 for 1 year2916

161210072510 ==

\ reqd. rate of interest .a.p%2916

16=

MATHS 1.29


29/39

18. A Pressure cooker is available for Rs. 350 cash or Rs. 80 down-payment and Rs. 50 per

month for 6 months.

Find (I) total amount paid (ii) rate of interest charged.

Principal left for 1st month = 350 80 = Rs. 270

2nd . = 270 50 = Rs. 220

... 3rd = 220 50 = Rs. 170

. .. 4th = 170 50 = Rs. 120

.. .. 5th. = 120 50 = Rs. 70

. .. 6th = 70 50 = Rs. 20

Total = Rs. 870

Instalment charge = amount paid in all cash prige

= (50 6 + 80) 350 = Rs. (380 350) = Rs. 30

Thus Rs. 30 is the interest on Rs. 870 for 1 month

Now reqd. interest .%38.411210087030 ==


1. What sum will amount to Rs. 5,200 in 6 years at the same rate of simple interest at which

Rs. 1,706 amount to Rs. 3,412 in 20 years? [Ans. Rs. 4

2. The simple interest on a sum of money at the end of 8 years is52

th of the sum itself. Find the

rate percent p.a. [Ans. 5

3. A sum of money becomes double in 20 years at simple interest. In how many years will it betreble? [ICWA (F) Dec. 2004] [Ans. 40

4. At what simple interest rate percent per annum a sum of money will be double of it self in 25

years? [ICWA(F) June, 2005] [Ans.

5. A certain sum of money at simple interest amounts to Rs. 560 in 3 years and to Rs. 600 in 5

years. Find the principal and the rate of interest. [Ans. Rs. 500; 4%]

6. A tradesman marks his goods with two prices, one for ready money and the other for 6 months

credit. What ratio should two prices bear to each other, allowing 5% simple interest.

[Ans. 40 : 41]

7. A man lends Rs. 1800 to two persons at the rate of 4% and2

14 % simple interest p.a.

respectively. At the end of 6 years, he receives Rs. 462 from them. How much did he lend to each

other? [Ans. Rs. 800 ; Rs. 1000]

8. A man takes a loan of Rs. 10,000 at the rate of 6% S.I. with the understanding that it will be

repaid with interest in 20 equal annual instalments, at the end of every year. How much he is to

pay in each instalment? [Assam H.S. 1986) [Ans. 700.64]

[Hints. Refer W.O. ex. 4. 10,000 (1+20 .06) = P (1+19.06) + P(1+18.06) + P]

MATHS1.30

A

RITHMETIC


30/39

MATHS 2.31

9. The price of a watch is Rs. 500 cash or it may be paid for by 5 equal monthly instalments of

Rs. 110 each, the first instalment to be paid one month after purchase. Find the rate of interest

charged. [Ans.76

42 %]

10. Divide Rs. 12,000 in two parts so that the interest on one part at 12.5% for 4 years is equal to

the interest on the second part at 10% for 3 years. [Ans. Rs. 4500; Rs. 7500]

11. Alok borrowed Rs. 7500 at a certain rate for 2 years and Rs. 6000 at 1% higher rate than the

first for 1 year. For the period he paid Rs. 2580 as interest in all. Find the two of interest.

[Ans. 12%; 13%]

12. If the simple interest on Rs. 1800 exceeds the interest on Rs. 1650 in 3 years by Rs. 45, find

the rate of interest p.a. [Ans. 10%]

1. A certain sum put out at S.I. amounts to Rs. 708 in 3 years. If the rate of interest increased byone third, it will amount to Rs. 744 in the same time. Find the sum and rate on interest.

[Ans. Rs. 600; 6%]

+=

+= )ii......(

1003r4

.31p744);i.....(

100r3

1P708.sintH

=

+

+

= .etc&Pfind),i(from%;6r,or100

r31p

100r4

1p

708

744

,or

14. A certain sum at a certain rate of interest p.a. S.I. becomes Rs. 1150 in 3 years and Rs. 1250 in

5 years. Find the rate percent p.a. [Ans. 5%]

15. Mr. X deposited a total of Rs. 95000 in two different banks which give 5% and21

7 % interest.

If the amounts repayable by the two banks at the end of 7 years are to be equal. Find the individual

amount of deposit. [Ans. Rs. 5700; Rs. 3800]

16. A man left Rs. 130000 for his two sons aged 10 years and 16 years with the direction that thesum should be divided in such a way that the two sons got the same amount when they attain the

age of 18 years. Assuming the rate of simple interest is2

112 % p.a. calculate how much the elder

son got in the beginning. [Ans. Rs. 80,000]

[Hints.n (for elder son) = 18 16 = 2, n (for younger son)

= 18 10 = 8 ; x = elder sons sum. Now

+-=

+ 8

100

5.121)x130000(2

100

5.121x or, x = 80000]

13.


31/39

MATHS1.32

A

RITHMETIC

17. What annual installment will discharge a debt of Rs. 3094 due in 4 years at 7% S.I.?

[Ans. Rs. 700]

[Hints. (amount of x for 3 yrs) + (x for 2 years) + (x for 1 yr.)

18. A dealer of radio offers radio for Rs. 2700 cash down or for Rs. 720 cash down and 24

monthly instalments of Rs. 100 each. Find the rate of simple interest charged per annum.

[ICWA (F) June, 2007] [Ans. 10.06%(opp)]

[Hints : rafer solved ex. 17]

19. A dealer offers an item for Rs. 270 cash down or Rs. 30 cash down and 18 equal monthly

instalments of Rs. 15. Find the rate of simple interest charged. [ICWA (F) June, 2004]

[Hints : refer solved problem no. 17] [Ans. %97

17 ]

OBJECTIVE QUESTION :

1. At what rate of S.I. will Rs. 1000 amount to Rs. 1200 in 2 years? [Ans. 10%]

2. In what time will Rs. 2000 amount to Rs. 2600 at 5% S.I.? [Ans. 6 yrs.]

3. At what rate per percent will S.I. on Rs. 956 amount to Rs. 119.50 in2

12 years? [Ans. 5%]

4. The S.I. on a sum of money at the end of 8 years is52

th of the sum itself. Find the rate percent

p.a. 5. To repay a sum of money borrowed 5 months earlier a man agreed to pay Rs. 529.75. Find the

amount borrowed it the rate of interest charged was2

14 % p.a. [Ans. Rs. 520]

6. What sum of money will amount to Rs. 5200 in 6 years at the same rate of interest (simple) at

which Rs. 1706 amount to Rs. 3412 in 20 years? [Ans. Rs. 4000]

7. A sum money becomes double in 20 years at S.I., in how money years will it be triple?

[Ans. 40 years]

8. A certain sum of money at S.I. amount to Rs. Rs. 560 in 3 years and to Rs. 600 in 5 years. Find

the principal and the rate of interest. [ Ans. Rs. 500; 4%]

9. In what time will be the S.I. on Rs. 900 at 6% be equal to S.I. on Rs. 540 for 8 years at 5%.

[Ans. 4 years]

10. Due to fall in rate of interest from 12% to2

110 % p.a.; a money lenders yearly income

diminishes by Rs. 90. Find the capital. [Ans. Rs. 6000]

11. A sum was put at S.I. at a certain rate for 2 years. Had it been put at 2% higher rate, it would

have fetched Rs. 100 more. Find the sum. [Ans. Rs. 10,000]

9. Complete the S. I. on Rs. 5700 for 2 years at 2.5% p.a.[ICWA(F) June, Rs. 2007] [Ans. Rs. 285]

13. What principal will be increased to Rs. 4600 after 3 years at the rate of 5% p.a. simple

interest? [ICWA (F) Dec. 2006] [Ans. Rs. 4000]

14. At what rate per annum will a sum of money double itself in 10 years with simple interest ?[ICWA (F) Dec. 2005] [Ans. 10%]

diminishes by Rs. 90. Find the capital.


32/39

15. The simple interest on Rs. 300 at the rate of 4% p.a. with that on Rs. 500 at the rate of 3% p.a.

both for the same period, is Rs. 162. Find the time period. [ICWA (F) June, 2005] [Ans. 6 years]

16. Calculate the interest on Rs. 10,000 for 10 years at 10% p.a.[ICWA(F) Dec. 2003] [Ans. Rs. 10,000]

17. A person deposited Rs. 78,000 in Post office monthly interest scheme (MIS) after retirement at

8% p.a. Calculate his monthly income. [ICWA (F) June, 2003] [Ans. Rs. 520]

[Hints : Income (monthly) = .]etc&12

1

100

878000

1.5 DISCOUNTING OF BILLS :

Few Definitions :

Present Value (P.V.) : Present value of a given sum due at the end of a given period is that sum

which together with its interest of the given period equals to the given sum i.e.

P.V. + Int. on P.V. = sum due Sum due is also known as Bill Value (B.V.)

Symbols : If A = Sum due at the end of n years, P = Present value, i = int. of Re. 1 for 1 yr.

n= unexpired period in years, then A = P+P n i == P(1+n i).(i)

or,ni1

AP

+=

True Discount (T.D) :

True discount of a given sum due at the end of a given period, is the interest on the present value

of the given sum i.e. T.D. = P n i..(ii)

T.D.= Int. of P.V. = amount due Present value i.e. T.D. = A P(iii)

Again T.D. = )iv..(..........ni1Ani

ni1A

A +=+-

Example : Find P.V. and T.D. of Rs. 327 due in 18 months hence at 6% S.I..

3 6327

Ani 2 100T.D 27,3 61 ni

12 100

= = =

+ + here A = 327, n = 18 m = 3/2 yrs. i= 6/100.

We know P.V. +Int. on PV (i.e.T.D)= sum due (i.e.B.V)

Or, P.V. = B.V. T.D. = 327 27 = Rs. 300.

SOLVED EXAMPLES :

(T.D; n; i are given, to find A)

1. The true discount on a bill due 6 months hence at 8% p.a. is Rs. 40, find the amount of the bill.

In the formula,ni1

AniD.T

+= , T.D. = 40,

2

1

12

6n == , i = 8/100 = 0.08

)08.0(21

1

)08.0.(21

.A40

+=\ , or, A = 1040 (in Rs.)

2 : (T.D.; A, i are given, to find n)

Find the time when the amount will be due if the discount on Rs. 1,060 be Rs. 60 at 6% p.a.MATHS 1.33


33/39

)06.0(n1

)06.0.(n.106060,or,

ni1

Ani.D.T

+=

+= or,

32

n = yrs.

3. : (T.D.; A; n are given, to find i)

If the discount on Rs. 11,000 due 15 months hence is Rs. 1,000, find the rate of interest,

Here, A =11,000, 4

5

12

15

n == , T.D. =1000, i = ?

So we have, ,ori.

45

1

i.45

,110001000

+= 12500 i = 1000 or i = 0.08. %8r=\

4. (If A, n; i; are given to find T.D.)

Find the T.D. on a sum of Rs. 1750 due in 18 months and 6% p.a.

06.0100

6i;yrs.

2

3.yrs

17

18n;1750AHere =====

So we get )app(50.14409.01

09.01750

06.023

1

06.0231750

.D.T =+

=+

=

.50.144.Rs.D.T.reqd =\

5. Find the present value of Rs. 1800 due in 73 days hence at 7.5% p.a. (take 1 year = 365 days)

Here : A = 1800,5

1

365

73n == years; 075.0

100

5.7i ==

60.26015.1

27

015.01

015.01800

)075.0(511

)075.0.(51

.1800.D.T ==

+

=+

=

We know P.V. = A T.D. \ P.V. = 1800 26.60 = 1773.40

6. The difference between interest and true discount on a sum due in 5 years at 5% per annum is

Rs. 50. Find the sum. [ICWA (F) Dec. 2005]

Let sum = Rs. 100, Interest = 10050.05 = Rs. 25, 05.0100

5i ==

Again ;20Re05.1

2505.01

05.05100ni1

Ani.D.T ==

+=

+= So difference = 25 20 = 5

Diff sum ? = ,1000505

100 = .1000.Rssum.reqd =\

5 10050 ?7. If the interest on Rs. 800 is equal to the true discount on Rs. 848 at 4% When the later amount

be due?

T.D. = A P.V. = 848 800 = Rs. 48, here A = 848, P.V. 800

Again T.D. = P. n.i or, 48 = 800 n 0.04, or,

2

11n = yrs.

MATHS1.34

A

RITHMETIC


34/39

MATHS 1.35


1. The true discount on a bill due21

1 years hence %21

4 p.a. is Rs. 54. Find the amt. of the bill?

[Ans. Rs. 854]

2. The true discount on a bill due 146 days hence at %2

14 p.a. is Rs. 17. Find the amount of the

bill (take 1 year = 365 days) [Ans. Rs. 1017]

3. When the sum will be due if the present worth on Rs. 1662.25 at 6%p.a. amount to Rs. 1,525.

[Ans.21

1 yrs.]

4. Find the time that sum will be due if the true discount on Rs. 185.40 at 5% p.a. be Rs. 5.40

(taking 1 year = 365 days) [Ans. 219 days]

5. If the true discount on Rs. 1770 due2

12 years hence, be Rs. 170, find the rate percent.[Ans.

2

14 % p

6. If the present value of a bill of Rs. 1495.62 due4

11 years hence, be Rs. 1424.40; find the rate

percent. [An

7. Find the percent value of Rs. 1265 due2

12 years at 4% p.a. [Ans. Rs.1150]

8. The difference between interest and due discount on a sum due 73 days at 5% p.a. is Re.1. Find

the sum. [Ans. Rs. 10,000]

9. The difference between interest and true discount on a sum due

2

12 years at 4% p.a. is Rs.

18.20. Find the sum. [Ans. 20,000]

10. If the interest on Rs. 1200 in equal to the true discount on Rs. 1254 at 6%, when will the later

amount be due ? [ Ans. 9 months]

BILL OF EXCHANGE :

This is a written undertaking (or document) by the debtor to a creditor for paying certain sum of

money on a specified future date.

A bill thus contains (i) the drawer (ii) the drawee (iii) the payee. A specimen of bill is as follows

Stamp Address of drawer

percent.

C.K. Roy

1/1 K.K. Bose Rd.

Kolkata 700 012A.

P. Dhar

(drawer)


35/39

Bill of Exchange is two kinds

(i) Bill of exchange after date, in which the date of maturity is counted from the date ofdrawing the bill.

(ii) Bill of exchange after sight, in which the date of maturity is counted from the date ofaccepting the bill.

The date on which a bill becomes due is called nominal due date. If now three days, added with

this nominal due date, the bill becomes legally due. Thus three days are known as days of grace.

Bankers Discount (B.D.) & Bankers Gain (B.G.):

Bankers discount (B.D.) is the interest on B.V. and difference between B.D. and T.D. is B.G.

i.e. B.D. = int. on B.V. = Ani (v)

B.G. = B.D. T.D. and B.G. = interest on T.D.

)vi....(..........ni1

)ni(A.G.B

2

+=

B.V. B.D. = Discounted value of the bill(vii)

SOLVED EXAMPLES :

Problems regarding T.D. and B.D.

8. A bill for Rs. 1224 is due in 6 months. Find the difference between true discount and bankers

discount, the rate of interest being 4% p.a.

.2402.148.24

)04.0.(21

1

)04.0.(21

.1224

ni1Ani

D.T ==+

=+

= B.D. = Ani = 1224.2

1.(0.04) = 24.48;

B.D. T.D. = 24.48 24= 0.08; \reqd. difference = Rs. 0.48 [ This difference is B.G. (0.48)

Again Int. on 24 (i.e., T.D.) ,48.0)04.0.(2

1.24 == i.e, B.G. = Int. on T.D.]

9. If the difference between T.D. and B.D. on a sum due in 4 months at 3% p.a. is Rs. 10, find the

amount of the bills.

B.G. = B.D. T.D. = 10;31

12/4n == yrs.; 03.0100

3i == ; A = ?

B.D. = A ni, T.D. = Pni, B.G. = B.D. T.D.= A ni Pni = (A P) ni

Now, (A P) ni = 10 or (A P)31 . (0.03) = 10,

or, (AP) 10001.0

10 ==

or, T.D. = 1000 (as T.D. = A P.) Again T.D. = P ni,

or,2

1.P (0.03) = 1000 or,

01.01000

P =

\P = 1,00, 000. Now, A = P.V. + T.D. = 1,00,000 + 1000 = 1,01,000

\ reqd. amount of the bill = Rs. 1,01,000.

MATHS1.36

A

RITHMETIC


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Aliter :ni1

)ni(A.G.B

2

+= ,

03.31

1

03.31

.A

10

2

+

= , A = 1,01,000.

10 : The T.D. and B.G. on a certain bill of exchange due after a certain time is respectively Rs. 50

and Re. 0.50. Find the face value of the bill.

We know B.G. = Int. on T.D. or, 0,50 = 50 n i or, 01.05050.0ni ==

Now, B.D. = T.D + B.G. or, B.D. = 50 + 0.50 = 50.50

Again B.D. int. on B.V. (i.e., A)

or, 50.50 = A. ni or, 50.50 = A.. (0.01) or, 505001.050.50

A ==

\reqd. face value of the bill is Rs. 5050.

Problems regarding discounting of bills.

11 : A bill of Exchange drawn on 4.1.81 at 5 months was discounted on 26-3-1981. If the bankers

discount at 3% be Rs. 603.60; find the value of the bill and also B.G.

Unexpired days from 26 March to 7 June, 1981.

(M) (A) (M) (J) drawn on 4. 1. 81

5+30+31+7 = 73 period 5

(excluding 26 3 = 81) normally of grace 4. 6. 81

days of grace 3

leqully due on 7. 6. 81

Let the face value of the bill = Rs. 100, here100

3i = = 0.03,

51

36573

n == yr.

B.D. = int. on B.V. = 100. 1/5. (0.03) = 0.60

B.D. B.V

0.60 100 600,00,160.60360.0

100x ==

603.60 x Hence, reqd. face value = Rs. 1,00,600

Now,03.0

511

3.051

100600

nii

)ni(A.G.B

2

2

+

=+

= 6.30060.1

000036.0100600 ==

Hence, B.G. = Rs. 3.60

[Observation : In this case, the discounted value of the bill = 1,00,600 603.60 = Rs.99,996.40

(i.e., this amount the holder of the bill will receive.]

12 : A bill of exchange drawn on 5.1.1983 for Rs. 2,000 payable at 3 months was accepted on the

same date and discounted on 14.1.83, at 4% p.a. Find out amount of discount.

[ICWA (F) June, 2004]

Unexpired number of days from 14 Jan to 8 April = 17 (J) + 28 (F) + 31 (M) + 8 (A) = 84

(excluding 14.1.83)

MATHS 1.37


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1983 is not a leap year, Feb. is of 28 days. 41.18100

436584

2000.D.B == (after reduction)

Hence, reqd. discount = Rs. 18.41.

Drawn on period 5.1.83

Period 3

Nominally due on 5.4.83Days of grace 3

Legally due 8.4.83


Problems regarding T.D. B.D. B.G.

1. At the rate of 4% p.a. find the B.D., T.D. and B.G. on a bill of exchange for Rs. 650 due 4

months hence. [Ans. Rs. 8.67; Rs. 8.55; Re. 0.12]

2. Find the difference between T.D. and B.D. on Rs. 2020 for 3 months at 4% p.a. Show that thedifference is equal to the interest on the T.D. for three months at 4%. [Ans. Re. 0.20]

1. Find the T.D. and B.D. on a bill of Rs. 6100 due 6 months hence, at 4% p.a.[Ans. Rs. 119.61;Rs.122]

4. Find out the T.D. on a bill for 2550 due in 4 months at 6% p.a. Show also the bankers gain in

this case. [Ans. Rs. 50, Re.1]

5. Find the T.D. and B.G. on a bill for Rs. 1550 due 3 months hence at 6% p.a.

[Ans. Rs. 22.9]; Re.0.34]

6. Calculate the B.G. on Rs. 2500 due in 6 months at 5% p.a.[ICWA(F) June, 2003] [Ans. Rs. 1.54]

7. If the difference between T.D. and B.D. on a bill to mature 2 months after date be Re. 0.25 at

3% p.a.; find. (i) T.D. (ii) B.D. (iii) amount of the bill. [Ans. Rs. 50; Rs. 50.25; Rs.10,050]

8. If the difference between T.D. and B.D.(i.e. B.G.) on a sum of due in 6 months at 4% is Rs. 100

find the amount of the bill. [ICWA(F) June, 2007] [Ans. Rs. 2,55,000]

[hints : refer solved problem no 9]

9. If the difference between T.D. and B.D. of a bill due legally after 73 days at 5% p.a. is Rs.10,

find the amount of the bill. [Ans.1,01,000]

10. If the bankers gain on a bill due in four months at the rate of 6% p.a. be Rs. 200, find the bill

value, B.D. and T.D. of the bill. [Ans. (F) Dec. 2006][Ans.5,10,000; Rs. 10,200; Rs.10,000]

11. A bill for Rs. 750 was drawn on 6th

March payable at 6 months after date, the rate of discount

being 4.5% p.a. It was discounted on 28th

June. What did the banker pay to the holder of the bill?

[Ans. Rs. 743.62]

12. A bill of exchange for Rs. 846.50 at 4 months after sight was drawn on 12.1.1956 and

accepted on 16th

January and discounted at 3.5% on 8 th Feb.1956. Find the B.D. and the

discounted value of the bill. [Ans. Rs. 8.18; Rs. 838.32]

MATHS1.38

A

RITHMETIC


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MATHS 1.39

13. A bill of exchange for Rs. 12,500 was payable 120 days after sight. The bill was accepted on

2nd Feb.1962 and was discounted on 20 th Feb. 1962 at 4%. Find the discounted value of the bill.

[Ans. 12,356.17]

14. A bill for Rs. 3,225 was drawn on 3rd

Feb. at 6 months and discounted on 13th

March at 8%

p.a. For what sum was the bill discounted and how much did the banker gain in this?

[Ans. Rs. 3121.80; Rs. 3.20]

Problems regarding rate of interest :

15. What is the actual rate of interest which a banker gets for the money when he discounts a bill

legally due in 6 months at 4% p.a. [Ans. 4.08% app]

16. What is the actual rate of interest which a banker gets for the money when he discounts a bill

legally due in 6 months at 5%. [Ans. %9

35 ]

17. If the true discounted of a bill of Rs. 2613.75 due in 5 months be Rs. 63.75; find rate of

interest. [Ans. 6%]

BOOK FOR REFERENCE :

1. Basic Mathematics and Statistics N. K. Nag2. A Text Book of ICSE Mathematics O. P. Sinhal


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NOTES

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