Chapter 2

Rule 1 of Simplification:

(T) In simplifying an expression, first of all vinculum or bar must be removed. For example: we know that -8-

10 = -18but, - 8 - 1 0 = - ( - 2 ) = 2 ( i ) After removing the bar, the brackets must be removed,

- ." ; in the order ( ) , {} and [] . i f After removing the brackets, we must use the follow-

ing operations strictly in the order given below, (a) of (b) division (c) multiplication (d) addition and (e) sub-traction Note: The rule is also known as the rule of \"BODMAS' where V, B, O, D, M , A and S stand for Vinculum, Bracket, Of, Division, Multiplication, Ad-dition and Subtraction respectively.

Illnstrative Example Ec Simplify:

I * - 0 f (6 + 8x3-2)+ 1^_7__I3 _8_ 5 ' 25 1 7 + 14

1-=-- of (6 + 8 x l ) +

= 1*- of (6 + 8)+

= l * ~ of 14 +

_7__14 5 ' 25 14

1 25 X 1 5 7

= 1+6+

Exercise

1 2 7-12 5 6 7 ~ 42 ~ ~42

1 1. Simplify: 10 2

a) I b)2

8 l + J 5 - ( 7 - 6 - 4 |

c ) l d)3

2.

S i m p l i f i c a t i o n

Simplify: i 0 - | < 5 - ( 7 - ( 6 - 8 - 5 ) a)8 b)6 c)7 d) 10

Simplify: 18 + 10-4 + 32+(4 + 1 0 + 2 - l ) a) 5 b)9 c)8 d)7

c 1 3 4. Simplify: -> of

45 2 5 . 2 +of

24 3 6 5 a) 13 b) 15 c)14 d) 16

Simplify: 240 -s-10(2 + {7 x 3 + 2(75 - 4 x 13+12+6

a) 24 53

53 C)Y4

Solve: 4 - [ 6 - { l 2 - ( 5 - 4 - 3 ) } J

a)5 b)4 c)6

d)

d)8

Solve: 2 I " 7 3 4 + x 8 4 6 4 8

29 29 28 a) T b) c) d)

14 8 ' 9 ' 9

8. Simplify: 2-[2-{2-2^2}| a) 3 b)2 c)0 d ) l

9. The value of 4.5 - [4.5 + (9.0-4.5 + x)\\= 0 , the value of x is a) 9 b)0 c)4.5 d) none

10. The value of 4 - [s - 6^ - (5 - 4^3)}] is a) 4 b ) l c)0 d)5

11. I f x = 4, y = 3, then the value of x+(y + x-l) is

a) b ) l - c) d)

12. Simplify: 1 2 : 8 I + J 5 _ ( 7 _ 4 - 2 |

16 PRACTICE BOOK ON QUICKER MATHS

a) 3 b)4 c)6 d)8

13. 2 + J2 + 2J2-(2-3^2)+ 3-(2-3)}-{3(2-3^2)}]=? a) 3 b)-3 c)8 d)-4

14. l - [ 8 + {l5-(6-2-20)})=? a)0 b)4 c)2 d)5

15. 6 - [ 6 - { 6 - ( 6 - 6 ^ 3 ) } ] = ? a) 3 b ) l c)6 d) 10

16. 20

a)0

17. r

1 1

5 u 5 1 1 + -6 3 2

c) 10 d) 20

-

Simplification 17

1.43 x 1.43+ 0.43x0.43-0.86 x 1.43=? a)0 b ) l c)2 d)4

6. 1.44x0.04-0.8x1.2 = ? a ) l b)2 c)0.24 d)0.4

7. Find the value of

0.47x0.47 + 0.35x0.35-2x0.47x0.35 0.12

a)0.1 b)0.2 c)0.3 d)0.12

0.538x0.538-0.462x0.462 8. Simplify:

a)0 b ) l c)2 d) 1.5

R- A.U , , 75983x75983-45983x45983 9. Find the value of

75983 + 45983 a) 40000 b) 49830 c) 30000 d) 35983

(Clerk's Grade 1991) 10. Find the value of

0.527 x 0.527 - 2 x 0.527 x 0.495 + (0.495)2 0.032

a) 0.032 b) 0.023 c) 0.052 d) 0.042

Answers l . d 2.b 3.d 4.c 5.b 6.a 7.d 8. b; Hint: 1 - 0.924 = 0.076 = 0.538 - 0.462 9. c lO.a

Rule 4 Application of the formula [a + bf +(a- bf = 2(a2 + b1)

Illustrative Example Ex: Simplify the following

2[1.25x 1.25 + 0.25x0.25] Soln: Applying the above formula, we have

l\i.25)2 +(0.25) 2]=(l.25 + 0.25)2 +(l .25-0.25) 2

= 2.25+1=3.25

Exercise (63 + 36)2 +(63-36) 2 _ n

1 63 2 +36 2 ' a) l b)2 c)3 d)6

(ITIExam,83)

2 Find the value of (0.46 + 0.64) 2+(0.18) 2

(0.64) 2+(0.46) 2

a) l b)2 c)4 d)8

3. Find the value of (5.3 + 3.5)2 +(5.3-3.5) 2

a) 40.34 b) 80.68 c)40 d)80

4. (400 + 50) 2 +(400-50) 2 =? a) 162500 b) 16250 c)325000 d)32500

8xj(l64 + 64) 2 + ( l 6 4 - 6 4 ) 2 } _ 0

5 - (l64) 2 +(64) :

a) 16 b) 4 c) 8 d) Can't be determined

Answers l . b 2.b 3.a 4.c 5.a

Rule 5 Application of the formula, (a + bf - (a - bf = 4ab

Illustrative Example Ex: Simplify the following

(l4.5 + 6.23) 2 -( l4 .5-6.23) 2 (14.5 x 6.23)

Soln: Appliying the above formula, we have a = 14.5, and b = 6.23

4x14.5x6.23 , . Expression= 1 4 _ 5 x 6 - 2 3 " 4

Exercise (l 125+-143)2 - ( l 125-143) 2 _ 0

' 4x1125x143 a) 4 b) 1 c) 0 d) Can't be determined

(0.576 + 0.324)2 - (0.576 - 0.324)2 _ 0 4x0.162x0.288

a ) l b)4 c)3 d)2

3. Find the value of (l 00 + 25) 2 - (l 00 - 25)2

a) 1000 b)100 c)40000 d) 10000

(250 + 50)2 - (250-50) 2 100x125

a)4 b)2 c ) l d)8

(6.25 + 5.25)2 +(6.25-5-0.25) 2 5. Find the value of ^

6.25x21

a ) l b)4 c)2 d)8

Answers l . b 2.b 3.d 4.a 5.a

Rule 6 Application of the formula, (a + b) (a - b) = a2 -b2

Illustrative Example Ex: Simplify (5O2 - 40 2 )= ?x 45

Soln: Suppose a = 50 and b = 40 and required number = x Applying the above formula,

2.



x = (50 + 40X50-40)_ 90x10

45 45 Required answer = 20

= 20

Exercise

1. Find the value of (l.73) 2 -(0.27) 2

1.73-0.27 a) l b)2 c)4 d)0.2

2. (4.44)2 -(0.56) 2

a)3.28 b)5 c)3.88 d)4

3. Find the value of (l 75)2 - (75)2

a) 25000 b) 20000 c) 40000 d) 50000 4. Find the value of

(2.35)2 -(0.35) 2 2.35 x 2.35 + 0.35 x 0.35 - 0.7 x 2.35

= 9

20 27 a ) 27~ b)W C ) 7 d)

5. Find the value of (2 .8) 2 - ( l -2) 2

2.8x2.8 + 1.2x1.2 + 5.6x1.2 a) 0.5 b)0.25 c)1.6 d)0.4

4 4 3 3 X X

6 7 7 7 7 = 7 4 3 7 7

a) l b)0 c)2 d) Can't be determined 0.704 x 0.704 - 0.296 x 0.296

7 S i m p U f y 0.704-0.296 a) l b)2 c)0 d) Can't be determined

0.25x0.25-0.24x0.24 ' 8. Simplify = ?

(RRB Exam, 1991) a) 0.0006 b)0.49 c)0.01 d)0.1

Answers l .b 2.c 3. a 4.b 5.d 6. a 7. a 8.c

Rule 7 Application of the formula,

{a + bf = a3 +3a2b+3ab2 + b3 = a3 +b3 + 3ab(a + b)

Illustrative Example Ex: Simplify

0.6 x 0.6 x 0.6 + 0.4 x 0.4 x 0.4 + 0.72.

Soln: The above expression can be written as

(0.6)3 + (0.4)3 + 3 x 0.6 x 0.4(0.6 + 0.4)

Now, we suppose 0.6 = a and 0.4 = b and applying the above formula, we have

(0.6 + 0.4)3 = (lf =\

Exercise 1. Find the value of0.7x 0.7 x 0.7 +0.3 x 0.3 x 0.3+ 6.3

a ) l b)3 c)4 d)2

0.6x0.6x0.6 + 0.4x0.4x0.4 + 0.4x0.6x3 _ 0 2 0.6x0.6 + 0.4x0.4 + 0.48

a)2 b ) l c)3 d)4

0.73 x 0.73 x 0.73 + 0.27 x 0.27 x 0.27 + 0.81x0.73 1 0.73 + 0.27

a)2 b ) l c)4 d)0.5

12x12x12 + 27 + 108x15

= 9


a) 15 b)9 5. Find the value of

= 9 225

c)7 d) 12

4. a; Hint:

1.4x 1.4x1.4+ 3 x l . 4 x l . 4 x 1.6 + 3 x l . 4 x l . 6 x l . 6 + (l .6) 3

a ) l b)8 c)27 d)64 6. Find the value of

(1.25)3 +2.25x(l.25) 2 +3.75x(0.75) 2 +(0.75)3

a)4 b)6 c)2 d)8

Answers l .a 2.b 3.b

12x12x12 + 3x3x3 + 3x3x12(12 + 3) 12x12 + 3x3 + 72

5. c 6. d; Hint: Let 1.25 = a, and 0.75 = b, then the given expres-

sion

= a3 +3bxa2 +3axb2 +b3 = a 3 +3a2b + 3ab2 + b3

= (a + 6) 3 = (l .25 + 0.75)3 = (2) 3 = 8

Rule 8 Application of the formula,

(a-bf = a3 -3a2b + 3ab2 -b3

= a3-b3-3ab{a-b)

Illustrative Example Ex: Simplify

(7.6S)3 - 3 x 7.65 x 7.65 x 0.65 + 22.95 x (0.65)2 - (0.65)3 7 x 7 x 7

Simplification 19

Soln:

(l.65f - 3 x 7.65 x 7.65 x 0.65 + 22.95 x (0.65)2 - (0.65)3 7 x 7 x 7

_ (7.65)3 - 3 x (7.65)2 x 0.65 + 3 x 7.65 x (p.65)2 - (0.65)3 7 x 7 x 7

Now, applying the above formula,

(7.65-0.65) 3 _ f 7 V _

W " W =

Exercise 1. Find the value of

(4J5) 2 x4.35 -3 x(4.35) 2 (0.35)+3x4.35x(0.35) 2 -(0.35) 3 16

a) l c)3

b)2 d)4

(1.25)3 -(0.25) 3 -0.75xl.25 _

2 (l.25) 2 +(0.25)2 -0.5x1.25

a) l b) 1.5 c)2 d)4

1.33xl.33xl.33-0.33x0.33x0.33-3x0.33xl.33 _ 0 2.4x2.4x2.4 + 1.6x1.6x1.6 + 4.8x2.4x4

1 1 1 a) l b ) - c)-r d)y

( l .7 ) 3 - (0 .7) 3 -3x0 .7x1 .7 4 Find the value of ( 3 6 ) 2 + ( 0 6 ) 2 _ 2 x 3 6 x { ) 6

1 1 1 a ) y b ) l c)y d ) ^

Answers l . d 2.a 3.d 4.c

Rule 9 Application of the formula, a 3 +b3 = {a + bia1 -ab + b2)

Illustrative Example 0.125 + 0.064

BE Find the value of 0.25 + 0.16-0.2

(0.5) 3+(0.4) 3 Soln: The above expression = ( 0 . 5 ) 2 + ( 0 . 4 ) * _ ( 0 . 5 x 0 . 4 )

Applying the above formula,

a 3 +b3 {a + b)= , , , Here a = 0.5 andb = 0.4

a2+b2-ab .-. Required answer = 0.5 + 0.4 = 0.9

Exercise

1. Simplify

a ) l

3.

4.

6.

7.

8.

Find the value of

a) 160

Find the value of

0.835x0.835x0.835+ 0.165x0.165x0.165 0.835x0.835-0.835x0.165 + 0.165x0.165

b)2 c)3 d)0.5

147x147x147 + 123x123x123 147x147-147x123 + 123x123

b)270 c)140 d) 130

0.08x0.08x0.08 + 0.01x0.01x0.01 0.08 x 0.08 - 0.08 x 0.01 + 0.01 x 0.01

a) 0.08 b)0.01 c)0.02 d)0.09 Find the value of

6.431 x 6.431 x 6.431 + 0.569 x 0.569 x 0.569 6.431 x 6.431 - 6.431 x 0.569 + 0.569 x 0.569 a)7 b)5 c)4 d)6

Find the value of (o.6)3 +(0.4) 3 +3x0.6x0.4

a) l b)2 c)1.5 d)0.5

885x885x885 + 115x115x115 Find the value of

a) 2000

885x885 + 115x115-885x115 b)100 c)1000 d)800

[Clerk's Grade Exam, 19911

1.75x1.75x1.75+ 1.25x1.25x1.25 1.75x1.75+1.25x1.25-1.75x1.25

a ) l b)2 c)3 d)4

0.125 + 0.027 . = 9 (0.5) 2-0.15 + (0.3)2

a)0.4 b) 0.7 c)0.8

(0.623)3 +(0.377)3

d)0.5

Simplify:

a ) l

(0.623)2 -(0.623 x 0.377)+(0.377)2

b)0 c)2 d)0.5 [LIC Exam, 1991]

10.

11.

0.5xQ.5x0.5 + 0.6x0.6x0.6 _ 9 0.5x0.5-0.3 + 0.6x0.6

a ) l b) 1.1 c)2 d) 1.5 [Hotel Management Exam, 1991]

5.7x5.7x5.7 + 2.3x2.3x2.3^1 _ 9 5.7x5.7 + 2.3x2.3-5.7x.23_

a)2.3 b)3.4 c)5.7

12. The simplification of

d)8.0 [CBI Exam, 1990]

(0.87)3 +(0.13)3

(0.87)2 +(0.13)2 -(0.87X0.13) yields the result: a) 0.13 b)0.75 c) 1 d)0.87

[I. Tax & Central Excise Exam. 1988!


13. 1.04 x 1.04 +1.04 x 0.04 + 0.04 x 0.04 1.04 x 1.04 x 1.04 - 0.04 x 0.04 x 0.04

a) 0.001

Answers l .a 2.b 8.c 9. a

b)0.1

3.d lO.b

c ) l d)0.01 [Assistant Grade Exam, 1987]

4. a 11.d

5.a 12.c

6.c 13.c

7.c

Rule 10 Application of the formula, a3 -b3 ={a- bj[a2 + ab + b2)

Illustrative Example Ex: Find the value of

3.254 x 3.254 x 3.254 - 0.746 x 0.746 x 0.746 3.254 x 3.254 + 0.746 x 0.746 + 3.254 + 0.746

Soln: The above expression can be written as

(3.254)3 -(0.746) 3

(3.254)2 +(0.746)2 +(3.254x0.746)

Now, we suppos a = 3.254 and b=0.746 Applying the above formula, we have

(a-b)=-a3-b3

= 3.254-0.746=2.508 a2+b2+ab

.-. Required answer = 2.508

Exercise 0.89 x 0.89 x 0.89 - 0.64 x 0.64 x 0.64

1. Simplify:

a) 0.25

2. Simplify:

3. a) 3 b)2 Find the value of

0.89 x 0.89 + 0.89 x 0.64 + 0.64 x 0.64 b)0.35 c)0.64 d)0.32

(2.3) 3-0.27

(2.3) 2+0.69 + 0.09

c ) l d) Can't be determined

0.7541 x 0.7541 x 0.7541 - 0.2459 x 0.2459 x 0.2459 0.7541x 0.7541 + 0.7541 x 0.2459 + 0.2459x 0.2459

a) 0.2409 b) 0.5082 c) 0.5802 d) 0.5820

1.75x1.75x1.75-1.953125 4. Find the value of 1.75xl.75 + 2.1875 + (l.25) 2

a) l b)0.5 c)1.5 d)0.3 5. Find the value o f

4.645 x 4.645 x 4.645 - 2.345 x 2.345 x 2.345 x 2.345

(4.645)2 +(2.345)2 +(4.645x2.345)

a) 3.2 b)2.5 c)5.2 d)2.3


a ) l

0.86x0.86x0.86-0.14x0.14x0.14 0.86x0.86 + 0.86x0.14 + 0.14x0.14

b)0.72 c)1.7 d) 1.2

0.89 x 0.89 x 0.89 - 0.64 x 0.64 x 0.64' s 0.89 x 0.89 + 0.89 x 0.64 + 0.64 x 0.64,

a) 2.5 b)0.25 c)0.93 d) 1.53

Answers l .a 2.b 3.b 4.b 5.d 6.b 7.b

Rule 11 Application of the formulae

(0 a"

00 a"

(iii)

(iv)

m b_\n a.

f m f

b) K a)

(v) am+b~" =amxbn

Illustrative Examples

Ex.1: Find the value of 1

216

_4 Pi 3 27

Soln: Applying the above fomula (iii), we have

f216^ 3 ' 2 7 ^ ^ -4'f"4't

6 2 + 3 4 6x6

3x3x3x3

Ex. 2: Find the value of 8 5 / 3 + (l 25)~I

Soln: Applying the above formula (v) we have,

8 5 / 3 - (l 25)"J = (2 3 J3 x (l 25f'3 = 32 x 25 = 800

Ex.3: Simplify (243) 1 2 x (243) 0 8

Soln: Applying the above formula (1), we have

(243) 0 1 2 x (243) 0 8 = ( 2 4 3 f 2 + 0 0 8

= ( 2 4 3 ) P = ( 3 5 ) i = 3

Ex. 4: Find the value 4)

Simplification

Soln: Applying the above formula (v), we have

- - L f =(-216)1 = [ ( - 6 ) 3 ] t = ( - 6 ) 2 =36 V 216;

Ex. 5: Find the value of ( - 2) (" 2 )

Soln: Applying the above formula (v), we have

(+2)

22

Illustrative Example Ex.: Simplify each of the following

(0 V3.V4"

(ii) 3Vl28

Soln: (i) V J . V i " = lj3x~4

= V\2

(ii) Vl28 =V64x2 =Vo4 ^2

= V 4 7 . V2 = 4 V2 [Usinglst Law ^4*" = 41

Exercise 1. Simplify: ->/5x3V25

a) 5 b)4 c)3 d) Can't be simplified

2. Find the value of lfjx \J4~9 a)3 b)4 c)5 d)7

3- 5v^TxV729=? a)7 b)9 c)6 d)5

4. V l 6 x V 4 = ?

a)2V2 b)3 c ) l d)4 5. Find the value of 4/121 x if[21

a) 12 b)21 c)19 d) 11

Answers l .a 2.d 3.b 4. a 5.d

Rule 14 If'n'is a positive integer and 'a', 'b' are rational numbers,

l/a" fa"

Illustrative Example Ex.: Simplify each of the following:

4

(ii)

Soln: ( i ) 3

27

27

V23" 3 / ^ " 3 a I = a

v/3888 J3888 = 4 0 0 " W M 48

PRACTICE BOOK ON QUICKER MATHS

Exercise

1. Find the value of -3V27

a > 9 b) C > 3

V32 ^6567

V243 Vl296 b)2 c)3


a ) l

3. Find the value of \J36 x \J216 x

a)2 b)5 c)4

^55296 4. Find the value of

a ) l b)2 c)3

Vl1616 V41904

5 ' 3Vl452 X 3Vl552

a)2 b)3

Answers l .c 2.a 3.d

55296

d)7

Vl25 Vl6

c)6

3Vl728 V625 d)l

d)4

d)5

5 5 2 9 6 . ,1 4. b; Hint: - p ^ - = 32 5 ^ = ( 2 *)? = 2

m 11616 41904 5. c; Hint: = 8 . = 27

1452 1552

Rule 15 If'm', 'n' are positive integers and 'a'is a positive rational

number, then 'qifo - ""/a" = .

Illustrative Example Ex.: Simplify each of the following

( 0 4 v W OOvW

( i i ) 2 M = 6V5 Soln: 0 ) ^ = ^

Exercise

1. Find the value of tftfi x i / ^ f

2. Find the value of ^^256

3. Find the value of ^ /243

Answers

1- l^3xS 2.2 3. VJ

Simplification 23

Rule 16 If'm ', 'n' are positive integers and 'a' is positive rational

number, then "Ma")"' = rfa~* = m{a^

Illustrative Example

EJL: Simplify: y * / ( 2 3 ) 4

Soln: Using the above property, we have

( 2 T = V 2

Exercise

Simplify: f ^ f xf^J

2 Simplify: iftffj

3. Find the value of ^ ( n 5 ] f


5. Find the value of &^(p\}

Answers

1.3; Hint: ^ ) * = ^and ^ / ( 3 2 )f =

2. V25 3. 4.13 5.3

Rule 17 Comparison of Surds of Distinct Orders Illustrative Examples Ex.1: Which surd is larger V3 or ^5 ?

Soln: The orders of the given surds are 3 and 4 respec-tively. Now, LCM of3 and 4 =12. So, we convert each surd into a surd of order 12.

Now, 3/3 = 1 ^ = 1#81

and 4V5 = ' v ^ = 1 2/l25

Clearly, 125 > 81

.-. , 2/l25 > , 2/8T => t/5>373. Ex.2: Which is greater ^3 or i/\Q .

Soln: The orders of the given surds are 2 and 4 respec-tively. LCM of 2 and 4 is 4. So, we convert each surd into a surd of order 4.

Now, J3=ij32~ = $j9.

[: "Va" a] Clearly, 10 > 9

.-. VTo >V9 ^>VTo>V3

Ex.3: Which is greater 3/6 or 4/s? Soln: The orders of the given surds are 3 and 4 respec-

tively. LCM of 3 and 4 is 12. So, we convert each surd into a surd of order 12.

Now, 3/6 = 1 2 /6 4 = , 2 / l296

and, V8 = '^S5" = ,-2/512

Clearly, 1296> 512

.-. 1^/l296 > 12/572 =>3v/6>4V8

j 1^/2 /\l/3

Ex. 4: Which is greater I or

Soln: The orders of the given surds are 2 and 3 respec-tively. LCM of 2 and 3 is 6. So, we convert each surd into a surd of order 6 as given below.

K2J

= 6 '9 '

4 1 Now, - > - [ v 4 x 8 > 9 x 1]

J 4 J 1 61 >6(_ V9 V8

I 1 2

Exercise 1. Arrange the following surds in ascending order of mag-

nitude:

(i) V3,^7 ,^48

(iioVe, V2,3V4

(ii) V5,3VlT,26V3

(iv) VJ, \l9, 6Vl05 2. Arrange the following surds in descending order of mag-

nitude:

( i i i ) 4 ^ , 3 ^ , ^

( i i ) 3 ^ , 4 ^ , ^

(iv) Vi,6SM


Answers 1. (i) The given surds are 73 , %pj, 1^ 48 . The orders of

these surds are 4, 6 and 12 respectively. LCM of 4, 6 and 12 is 12. So, we convert each surd into a surd of order 12. We have,

V3 = ^ = '727 , 6V7 = = ^49 and '748~ is a surd of order 12.

Since 27 < 48 < 49. Therefore,

'#27 < '#48 < '#49 => V3 < '#48

Simplification

or, x- =20-V20-V20-V20 7....oo = 2 0 -x

or, x2 + x - 2 0 = 0

or. x 2 + 5 x - 4 x - 2 0 = 0 or.x(x + 5 ) -4 (x + 5) = 0 .-. x = 4 and -5, we neglect the -ve value of x .-. Required answer = 4 Quicker Method: Applying the above theorem, we havex = 20 = 4 x 5 .-. required answer = 4

Note: To find x = n(n + 1), factorize x and get the required numbers. As for example, 2 20 2 10

5 = 4 x 5

Exercise

1- ^ 6 - 16 ...oo a)2 b)3

2- 1^2-- J l 2 - - V l 2 - . a)3 b)4

c)0

..oo = ? c)2

d ) l

d)5

3 Find the value of ^ 30-^30-4^0^...00

a) 6 b)4 c)5 d)3

4. Find the value of ^42 - \J42 - 742 - ....00 a)6 b)7 c)8 d)9


^12-^12-jvi^ - ^20-720-720^...00^ ^ / A >

a) 8 b)4 c)2 d) Can't be determined _ c ) 2

6- A / l 3 2 - V l 3 2 - V r 3 2 ...00 = ? a ) l l b) 12 c)14

Answers l .a 2. a

d) 18

6. a 3.c 4. a 5.c

Rule 19

If -jx + ylx + Vx+~...oo and x = n(n +1), then the value of

above expression is given by (n +1).

Illustrative Example Ex: Find the value of

1/2O + V2O + V2O+...00

Soln: Detail Method:

Let the 72O + V2O + 720+... = .

.-. x 2 =20 + 720 + 720 + ... =20+x

or, x 2 - 2 0 - x = 0

or, x2 - 5 x + 4x-20 = 0 or,x(x-5) + 4 (x-5) = 0 .-. x = 5or,-4 We neglect the -ve value .-. required answer = 5 Quicker Method: Applying the above theorem, wehavex = 20 = 4 x 5 .-. Required answer = 5

Exercise

1. Find the value of ^ 30 + 730 + 730

c)3 a) 5 b)6 Find the value of

+ ....00

d)4

7no+Vno+Viio+ -.co

^ 5 0 6 - ^ 5 0 6 - 7 5 0 6 - ....00

a) l b)2 c)0 1 6 ) 2

3. Find the value of ^210+-^210 + 7210+...00

c) 12 d) 14 a) 15 b)16 4. Find the value of

^42 + ^42 + 742 + ...00 x -^42-742-^/42 ...00 J V

c) 42 d) Can't be determined a) 48 b)40

5. Find the value of ^380 + ^380 + 7380 + ...00

a)20 b)19 c)18 d) 17

Answers l .b 2.d 3.a 4.c 5.a

Rule 20 To Simplify a Continued Fraction:

Fractions of the form 1 2 + -

1 are called continued

3 + 1

4 + 1 3

fractions. A continued fraction is also written as

( 1 1 i n

J. It is necessary that the sign '+' should be

written in the denominator. To simplify a continued fraction, begin at the bottom and work upwards. Following example will illustrate our poo*.



Ex: Simplify 3 +-7

4 + -

Soln: 3 + -

6 +

= 3 + - = 3+-7

4 + - 4 + 6 + -

10 13

13 31

2 + - ^ 2 + 1 3 + - 5 " 3 + %

2 2 + I R

Exercise Simplify the following fractions:

] 1.

7 + 1

6 + 1

2 . - V 3 . 1 3 + -

1-- 7 + 11

= 3 + ^ = 3 ^ i -204 204

Process: Begin at the bottom. First take up the lowest com-

( \ 5

plex fraction namely 6 + I

V 2 ,

Multiply the numera-

10

tor and denominator by 2 and we get . Next multi-

ply the numerator and denominator of the fraction

13

4 + 10 13;

by 13, and we get . Then multiply the

numerator and denominator of

31

( 1

7 - H 31

by 31,and

we get . Hence the fraction is reduced to 3 + 31 204

or 3 31

204 Note: We may convert a fraction to a continued fraction,

with unity as numerators and all the signs positive. The following example will illustrate the process.

17 Ex: Convert to a continued fraction.

17 1 1 1 1 1 2 + 2 + - V 2 + -

17 17 17 - + 2

4. 5 + -

1 5. 5 + -

6 + 1 6 + -

7. 4 +

9

J_ _ L I 4+ 1+ 2

13

6 + -10

8. Convert to a continued fraction.

Answers 31

1

3 - - J L 2 - -

l .

4.

7.4

222

55 284

3

2. 12

5. 5 81

496

3.1

6.

43 227

14 3 + -

1 + -

F r a c t i o n s

2 + -4

Rule 21

Theorem: If a man spends ~ part of the total salary on

food, ~ part of the total salary on entertainment, yi ' y3

part of the total salary on clothing, and so on. After these expenditures, he is left with a balance amount ofRs B, then Balance Amount

X\ X-t + + ~ L + ... y\2 yi

x Total Amount

Simplification 27

Proof: Here, the amount spent on each item is expressed as a fraction of the total amount (or salary). So, spend-ing on each item is not dependent on the expenditure incurred on the other item. This type of activity is known as independent activity. Independent activities are always added. .-. Total spent part

y\ y* x Total Amount

.-. Balance Amount = Total Amount - Total Spent Part => Total Amount -

X\ X-i + + + . . . y\ yi

x Total Amount

- + + -y\ v 3

+ . . . x Total Amount

In general, for independent activities, :. Balance Amount

1_ f L + i l x Total Amount U i yi y-i


1 1 EJL: A man spends of his salary on food, of his

salary on house rent and of his salary on clothes.

He still has Rs 18000 left with him. Find his salary. Soln: The expenditure incurred on each item is expressed

as part of the total amount (salary), so it is an inde-pendent activity. Using the above theorem, we have

l - l l + - l + l 5 10 5

x Total salary = Rs 18000

1 10

x Total salary = Rs 18000

.-. Total salary = Rs 18000 x 10 = Rs 180000.

Exercise

I i i i i & A man spends of his salary on food, j of his salary

*i . I I

on house rent, of his salary on health and of his

salary on clothes. He still has Rs Rs 15500 left with him.

Find his salary. a)Rs21500 b)Rs21000 c)Rs31000 d)Rs24000

2. A man spent of his savings and still has Rs 1000 left

with him. What were his savings? a)Rsl400 b)Rs2000 c)Rs 21000 d)Rsl800

2 3 3. A man spends of his salary on food, of his salary

1 on house rent and 7 of the salary on clothes. He still

o has Rs 1,400 left with him. Find his salary. a)Rs7000 b)Rs8400 c)Rs8000 d)Rs9800

4. A man spent 7 of his savings and was still left with Rs

2,000. What were his initial saving? a)Rs,5000 b)Rs5500 c)Rs8500 d)Rs8000

^ M V B a i * ^ , ' - ic -w in fca- 1 . 5. A man spends of his income on food and on rent

6 12 and rest he saves. I f he saves Rs 50, find his income. a)Rs500 b)Rs600 c)Rs400 d)Rs800

6. A person went to the market and purchased a pen for Rs

15. I f he is still left with of his total money, find the

total amount of money he had initially. a)Rs30 b)Rs25 c)Rs35 d)Rs40

1 7. The fuel indicator in a car shows th of the fuel tank as

full. When 22 more litres of fuel are poured into the tank, the indicator rests as the three-fourth of the full mark. Find the capacity of the fuel tank, a) 50 litres b) 42 litres c) 40 litres d) 36 litres

1 8. A persons spends of his salary on entertainment,

of his salary on purchasing books and of his sal-8 4

ary on foods and clothing. I f his salary is Rs 16824, find the balance amount with which he left. a)Rs700 b)Rs7001 c)Rs7010 d)Rs710

3 13 9. and part of a pole are respectively in mud and

water. I f the pole is 20 metres high, how much portion of it wil l be above water and mud? a)3m b)5m c)4m d)6m

10. A man distributes 0.375 of his money to his wife and 0.4 to his son. He has still Rs 3,375 left with him. How much


initial money the man have? How much did his wife get? a) Rs 16000, Rs 6525 b) Rs 25000, Rs 7525 c)Rs 15000, Rs 5625 d)Rs21000,Rs8575

1 11. A lamp post has half of its length in mud, - of its length

in water and 3 m above the water. Find the total length

of the post. a) 20 metres b) 25 metres c) 27 metres d) 21 metres

3 12. When Jack travelled 25 km, he found that - of his jour-

ney was still left. What is the total journey to be covered by Jack.

125 45 a) ~y km b) km c)62km

1

135 d) k m

13. A persons expends of his income for board and lodg-

ing, - in clothing and in charity, and saves Rs 3180. o 10

What is his income? a)Rs6200 b)Rs7200 c)Rs7280 d)Rs7270

2 17 14. A man travelled of his journey by coach, by rail

and walked the remaining 1 kilometre, how far did he go? a)22km b)20km c)33km d)27km

15. Of a certain dynasty of the kings were of the same

Mates 1 1 1 name, of another, of another, of a fourth, and

4 8 12 there were 5 besides. How many kings were there? a) 24 b)28 c)16 d)48 2 3

16. of a post are imbedded in mud, are in the water,

and 3 metres are above the surface, what is the length of the post? a) 15 metres b) 20 metres c) 90 metres d) 16 metres

_ 1 17. A man carrying a cask full of milk to the market lost

of the milk due to leakage, he sold 7 litres and found

that half of the cask was still full of milk. Find how much milk did the cask contain? a) 18 litres b) 32 litres c) 16 litres d) 24 litres

18. A man divided a piece of land among his three sons

. , 1 5 thus, he gave 35 sq km to the first, of the whole

4 12 to the second and to the third as much as to the first two together. Find the whole land.

a) 423 sq km c)419sqkm

Answers 1. b 2. a

3 1

3.c

b)421 sqkm d)425 sqkm

4. a 5.b 6.b

7. c; Hint: | J x tank capacity = 22

.-. tank capacity = 40 litres. 8. c 9. c 10. c; Hint: 1 -(0.375 + 0.04) x total money = balance money

or, (1 -0.775) x total money = 3375

3375 or, Total money = l _ 0 1 1 5 = Rs 15000

.-. wife's share = 0.375 x total money = 0.375 x 15000 = Rs5625

11. a; Hint: In this problem, the portion above water is given. Hence, equate amount (length) above water to the part (fraction) above water.

3 I = 3

I 1 1

+ -2 3

x length of post

10 1 10 => length of post = "V T = x ^ =20 metres.

3 6 3 12. a 13. b 14. a 15.a 16.c 17.c 18. a; Hint: Since the sum of the shares of the first two sons

is equal to the share of the third, the share of the third

1 = of the whole

2

i (5 0 - 1 .:. share of the first son = 1 ! 12 + 2 ] ~ 12 ^ ^

whole.

or, of the whole = 35 sqkm

.-. the whole land = 35^-x 12 = 423 sq km 4

Rule 22

Theorem: If a man spends . part of the total salary on

x2 food, part of the remaining (rest) amount on entertain-

yi

ment, ^ part of the remaining (rest) amount on clothing

Amplification 29

mdno on. After these expenditures he has balance amount if ta B. then, M B U Amount (B)

x Total Amount

Here spending on the second item (ie entertainment) depends on the amount left after spending on the first item (ie food). Similarly, spending on the third an (ie clothing) depends on the amount left (remain-ing l after spending on the first item and the second

f \i i i - * * 1 - ^ - X 1 -LI 1

Here, spending on each item (except the first item) depends on the amount remaining, after spending on the previous item. This type of activity is known as dependent activity. Dependent activities are always multiplied. Hence, in general, for dependent activity. Balance Amount

= 1--Vl

1 -yi) v 3

x Total Amount

t ive Examples

A man spends of his income on food, of the rest 3 4

on house rent and of the rest on clothes. He still

has Rs 1760 left with him. Find his income. It is a problem on dependent activity. Hence using the above method, we have

=H>H>H =Rsl760 - Total Income = Rs 4400.

1

x Total Amount

A man spends of his income on food, of the rest

1

on house rent and on clothes. He still has Rs

1760 left with him. Find his income. Here, of the rest amount (after spending on food),

1 B spent on house rent and is spent on clothes. So

y n r i i n g on these items are independent on each r, but dependent on the expenditure incurred on

; first item. It is a problem both on dependent and h >er.;er.t activities.

f

1-.5C X i - - x x Total Income = Rs 1760

.-. Total Income = Rs 4800

Exerc ise

3 ,4 1. A man reads of a book on a day and of the remain-

o 5 der, on the second day. I f the number of pages still un-read are 40, how many pages did the book contain? a) 520 b)320 " c)230 d)250

An fin if i- 1 2. A man spends ~ of his income on food, of the rest on

1 house rent and of the rest on clothes. He still has Rs

1,760 left with him. Find his income. a)Rs4500 b)Rs4600 c)Rs48^0 d)Rs4400

1 1 3. A man spends of his salary o i food and of the

1

remaining on clothing and - of the remaining on enter-

tainment. He is still left with Rs 600. Find his salary.

a)Rs2100 b)Rs2400 c)Rsl800 d)Rsl600 2

4. A man while returning from his factory, travels of the

distance by bus and of the rest, partly by car, and

partly by foot. I f he travels 2 km on foot, find the dis-tance covered by him. a)23km b)26km c)24km d)30km

1 A boy after giving away - of his pocket-money to one

companion and of the remainder to another, has Rs

200 left. How much had he at first? a)Rsl550 b)Rs750 c)Rsl500 d)Rsl750

At his first game a person loses of his money, at the

second - of the remainder, at the third of the rest;


what fraction of his original money has he left? Illustrative Examples

a) 25 b) 25 c) 25 d >2l

7. One-fifth of an estate is left to the eldest son, to the 6

\.5 second and of the remainder to the third, how much

6 was left over?

19 a)

17 180 b>T80 c)

91 180 d)

29 180

3 4 8, I read of a book on one day and of the remainder

8 5 on another day. I f there now were 30 pages unread, how many pages did the book contain? a) 160 pages b) 240 pages c) 640 pages d) 100 pages

9. The highest score in an inning was of the total and

the next highest was of the remainder. The score

differed by 8 runs. What was the total? a) 172 b)142 c)152 d) 162

Answers I . b; Hint: It is a dependent activity, because on the second

4 / day he reads of the remaining pages.

or f l 3 > 1-11 1 X 1-11 I 8j v 5j

x Total pages = P a g e s l e f t u n r e a d

or x x Total pages = 40 8 5

.-. Total pages = 320 2.d 3. a 4. c; Hint: Here distance travelled on foot is given. Hence

equate the part (fraction) travelled on foot to the dis-tance covered on foot, ie Rest part on foot = distance on foot.

5. c 6.c 7. a 8.b 9. d

Rule 23 In case of single activity only ie Part done + Part remained

= 1

Total Amount -Amount Spent _ Amount Remained

Balance Pa. t Part Remained

4 1 Ex. 1: of a pole is in the mud. When of it is pulled out,

250 cm of the pole is still in the mud. Find the full length of the pole.

Soln: Using the above formula, we have

Total Amount : Amount Remained

Part Remained .-. Total length of pole

Length in mud 250 Part in mud 1_I

7 3

= 1050

.-. Length of pole = 1050 cm. Ex. 2: After covering five-eighth of my j ourney, I find that I

have travelled 60 km. How much journey is left? Soln: Using the above formula, we have,

Amount Remained _ Amount done Part Remained Part done

Let the journey left = x km

1 8

60 5 8

x = 36

:. Journey still left = 36 km

Exercise 4

1. In a school, of the children are boys. I f the number of

girls is 200, find the number of boys. a) 900 b)700 c)850 d)800

^ ...

2. A drum of water is full; when 3 8 litres are drawn from

1

it, it is just - full. Find the total capacity of the drum in

litres. a) 80 litres b) 85 litres c) 75 litres d) 90 litres

3. In a flag post, - is red, - is green and the rest is white.

I f the white part is 7 metres long, find the length of the flag post.

a) 16 metres b) 14 metres c) 20 metres d) 15 metres

2 4. A man pays off of his debt and has to pay Rs 240 to

pay off the debt completely. Find the total amount of debt. a)Rs400 b)Rs450 c)Rs500 d)Rs550

5 1 5. A man spends of his income on food and on rent

HS 31

out,

full

wmd rest he saves. I f he saves Rs 50, find his income. a)Rs6O0 b)Rs800 c)Rs575 d)625

covering th of my journey, I find that I have

175 km. How much journey is left? b)20km c)30km d)35km

3.d 4. a 5. a

Rule 24

6. a

Xj X2 tfwu+gnrn fractions be and and we want to insert

(Inserting one fraction between - and )

1 1+5 5 5+4 4 3 ' 3 + 8 ' 8 ' 8 + 5 ' 5

1 5 (Inserting one fraction between and 77. and one

fraction between and ) o 5

J _6_ 5 _9_ 4 ~ 3 ' l l ' 8' 13' 5

that I

ber of

nfrom

rum in

es

white,

of the

stres

240 to

>unt of

50

on rent

Xj X2

mpmcnon tying between and , the following steps

awirf t-e Taken. Sfc^ L The numerators of two given fractions are added to get -ie numerator of the resulting fraction ie numerator of nn tsaiting fraction = xx+x2. Saej Th The denominators of two given fractions are added tfget the denominator of the resulting fraction. That is Ammunaior of the resulting fraction = yx + y2

Xj + X2 Stty HI: Resulting Fraction = ~

y\

- resulting fraction so obtained has its magnitude fmtime) lying between the two given fractions. By this

any number of fractions can be inserted between gpeen fractions.

itive Examples 1 4

Insert one fraction between and y .

Using the above method,

I 111 1 - 1 1 1 ' V 3 + 5 ' 5 ~ 3 ' 8' 5

Thus, the resulting fraction is more than and o 3

(Three fractions inserted between - and )

less than in magnitude (value).

1 4

Insert three fractions between and .

Using the above method _1 111 1 - 1 1 1 ~ 3 ' 3 + 5 ' 5 ~ 3 ' 8' 5

Exercise 1 5

1. Insert one fraction between and ~ . 4 6

2 4 2. Insert two fractions between and .

5 9 3. Insert three fractions between ~ and

1 5 4. Insert four fractions between ~ and .

3 0 - A -S f-J****'tK

5. Insert five fractions between and y .

I l l 3 " 4 ' 9 ' 5

4. 2 ]_ 2 7 5 ' 2 ' 3 ' 9

Answers 3 3_ 7

! 5 2 - 4 ' 9

_7_ 6_ H_ _5_ _9_ 5 ' 22 '17 '29 '12 '19 Note: Answers may be different from what it is written here.

Rule 25

Ex.: What must be subtracted from the sum of 13 and 00

4: to have a remainder equal to their difference? 66

Soln: Detail Method:

Di f fe rence=13^-4A = ( 1 3 _ 4 ) + [ ^ - A

= 9+A = 9 l 66 33


7 5 12 2 Sum= 13 + 4 = 1 7 = 17

66 66 66 11 .-. the required answer

= i 7 A _ 9 _ L = 1 7 A _ 9 _ L = 8 A 11 33 33 33 33

Direct Formula: The required answer = 2 * smaller value

= 2 x 4 = 8 66 33

Exercise

11 1 o 4

1. What must be subtracted from the sum of 12 and

to have a remainder equal to their difference? What must be subtracted from the sum of 14 and

5 to have a remainder equal to their difference?

3. What must be subtracted from the sum of 17 and

15 to have a remainder equal to their difference?

What must be subtracted from the sum of 5 and 3 2 5

to have a remainder equal to their difference?

What must be subtracted from tl of29- and

19-j to have a remainder equal to their difference?

Answers

2. 10 14 33

3. 30-2 3

4.64 5 .394

Rule 26 In the group of fractions /

x x + a x + 2a x + 3a x + na y' y + b' y + 2b' y + 3b'"" y + nb

x + na y + n o has the highest value.

Wherei) a = b or ii) a>b

Illustrative Examples Ex. 1: Which one of the following fractions is the greatest?

3 4 5 , T and 7 4 5 6

Soln: We see that the numerators as well as denominators of the above fractions increase by 1, so the last frac-

5 tion, ie 7 , is the greatest fraction,

o [Here, a = b]

Ex. 2: Which one of the following fractions is the greatest?

1 1 2 . 8' 9 ' 10

Soln: We see that the, numerator increases by 3 (a constant value) and the denominator also increases by a con-

7

stant value (1), so the last fraction ie. isthe great-

est fraction.

[Here,a>b]

Exercise 1. Which one of the following fractions is the greatest?

I ! 3 2 ' 3 ' 4

2. Which one of the following fractions is the greatest?

1 1 A 11 8'9*10'11


1 A _?_ 11 9'11'13'15


J_ 2 4_ _3_ 7 ' 9 ' H ' l O

5. Which one of the following fractions is the greatest'

1_ 1 0 4 7 5 ' 1 1 ' 7 ' i

Answers

10 3

4

4 T T

11

10 11

13 3 -15

Rule 27 The fraction whose numerator after cross-multiplication gives the greater value is greater.


5 9 Ex.: Which is greater or ?

8 14 Soln: Students generally solve this question by changing

Simplification

the fractions into decimal values or by equating the denominators. But we suggest you a better method far getting the answer more quickly. Step I: Cross-multiply the two given fractions.

- | -^X^^- ,wehave5 x 14 = 70and8 x 9 = 72

Mep I I : As 72 is greater than 70 and the numerator

9 involved with the greater value is 9, the fraction

is the greater of the two.

Exercise

6 5 w~hich is greater or .

Which of the following is lesser or, .

15 5 Which of the following is greater, or,

17 8 Which of the following is greater or

14

25 51 Which of the following is lesser or, y^y .

41 53 Which of the following is lesser or .

Answers

5 4 L 2. 2.

13

17 25 i.

35 5. 51

5 3 9

41 53

Rule 28 tm the group of fractions

x x + a x + 2a x + 3a x + na y' y + b' y + 2b' y + 3b'"" y + nb

a


7.

JL i i 11 7 ' l f 15'19 Which of the following fractions is the least?

2 4_ 6_ _8_ 5 '11'17'23

Which of the following fractions is the least?

1 A A A 8 '17 '26 '35 Which of the following fractions is the greatest?

4 _8_ 12 16 20 7 '15 '23 '3T ' 39

Which of the following fractions is the greatest?

3 6 9 12 7 15 23 31

Answers

' 7 ^ 1 4

3 3.

23 35 20 39

7.

Rule 29 To determine the missing figures which are indi-

cated by asterisks. There is no any general rule that will apply to all type of questions. We can better understand the rule to determine the missing figures by the illustrative example.

Illustrative Example Ex.: Supply the two missing figures which are indicated

3 1 by asterisks in the equality 5 x*= 19, the frac-

2 tion being in their lowest terms.

Soln: Since (5 + a fraction) is contained (3 + a fraction) times in 19, the integral portion of the second mixed number must be 3.

3 1 Hence, 5 - x 3 - = 19

* 2 3 2 3

o r , 5 l = 19x4 = 5 l ' * 7 7

.-. the required figures are 7, 3.

Exercise 1. Determine the missing figures (denoted by stars) in the

following equations, the fractions being given in their lowest terms:

( i ) 6 - x * - = 30 (ii) 3 3 *

* - x 2 - = 14 7 * 14'

" f f l ' H r 5 *3 '

o 9 t 2 , 16 (v) 8-=-* = 7, v ; ** 27 17

3 1 * fiv) - x * - = U ; 8 9 12'

1 1 ** ( v f ) 4 - * = 1 K } 6* 17 6*

2. In a book on Arithmetic a question was printed thus:

'Add together 1 1 1 1

2' r 1 4 - 1 9 -

3 4

: the denominator of 13-

one fraction being accidentally omitted. The answer

11 given at the end of the book was . Find the missing

Z o

denominator.

Answers

1. (07,4;

2 5 7 ( i i i ) 7 - - 4 = 3; v ' 3 11 33

o 9 , 2 , 16 (v) 8+ 1 = 7: y ) 17 27 17

c 3 o 3 , J 3 (ii) 5 - x 2 - = 14 ; W 7 4 1 4 '

3 , 1 5 (iv) - x l - = ; y ' 8 9 12

A 1 n 1 1 6 5

(vi) 4 2 = 1 K ' 68 17 68

2.5

S o m e m o r e q u e s t i o n s o n B a s i c C a l c u

l a t i o n s a n d S i m p l i f i c a t i o n

S E T - I 1. 171-19x9 = ?

a ) l b) 18

2. 5005-5000-10.00 = ? a) 0.5 b)50

3. 8 - 4 ( 3 - 2 ) x4 + 3 - 7 = ? a)-3 b)5

10 12 ? J 4 x x= 16

3 5 4 a) 6 b)2

a) 1

7 ( 3 2]_2 9 I 9 + 9 J 9

b)

c)81 d)0 [SBIPO Exam, 1987]

c)5000 d)4505 [BankPO Exam,1988|

c)4 d)^l [Railways, 19911

c)8 d)4 [Railways, 1991]

[Clerical Grade Exam, 1991 ]

9 7

12x8-19x2 6 x 7 - 6 . 5 x 2

c) d) 1 9 ' 1

[BSRB BankPO Exam, 1990]

Simplification

a) 2

1

b)

4 + 20 2 = , 22x44

1 x4 + 20

81 3 a ) 88 b ) I T

3+ ( 8 - 5 ) , ( 4 - 2 ) ,

c) 29

d) None of these

(BankPO Exam, 19901

c) 161 176

d ) l

2 + V 13

[Hotel Management Exam, 1991]

a) 11 17 b) 17 13

68 13 d > o I

[Hotel Management. 1991 ]

29 10 a ) 79 b ) 17.28,? _

15. - = 2

c) 29 10

10 d ) y

[SBIPO Exam, 1988) 3.6x0.2

a) 120 b)1.20 c)12 d)0.12

5 6 . 8 . 3 3 1 7 16 + x? , 1 + x3 = 2

a 6 7 9 5 4 3 9 '

a) b)

1 7 4 l + 3 l + ? + 2 l = 1 3 -2 6 3 5

[GIC,AAO Exam, 1988]

c ) l d) None of these

[SBIPO Exam, 1887]

a ) 3 f b) i f c ) 4 l d ) 4 -

1 1 , 1 of -

5 5 5 _o 1 , 1 1 ' - of - + 5 5 5

[Railways, 1991] 18- The value of is: [SSCExam, 1987]

a) l b)5 *5 d)25

ML (20 + 5)+2 + (16 +8)X2 + (10 + 5)X(3-!-2) = ? [Clerical Grade Exam, 1991]

c) 15 d) 18

[BankPO Exam, 1988]

107

a) 9 b) 12

3 i x A + l , 2 0 = ? 10 10 5

a)0 b ) l

16 -6x2 + 3

c)100 d)

12 2 3 - 3 x 2

a ) ! 7 b) 23 40 c)

200

[BankPO Exam, 1988]

14 23

2

u. V

a)

2 - , l l - I 4 2

b ) l

l l - l - l ] . 2 3 6 j

[Central Excise Exam, 1989]

1 0 4 - d ) l

77 228

.4 The value of 1 + - is: 1 + -

1 + -

3 + -2 + -

2

a) b) c) 19 d) 19

1 1 19. How many 's are there in 37 ?

a) 300 b)400 c)500 1

[SBIPO Exam, 1988] d) Can't be determined

1 20. In a college, 7 of the girls and 7 of the boys took part

5 o in a social camp. What of the total number of students in the college took part in the camp?

[SBIPO Exam, 1988] 13 13 2

a ) 4 0 b ) 8 0 C ) 1 3 d) Data inadequate

21. In a certain office (1/3) of the workers are women, (1/2) of the women are married and (1/3) of the married women have children. I f (3/4) of the men are married and (2/3) of the married men have children, what part of workers are without children? [SBIPO Exam, 1987]

5 4 11 17 a ) l i b ) 9 C ) I d)3~6

22. I f we multiply a fraction by itself and divide the product , . 2 6

by its reciprocal, the fraction thus obtained is 1 8 ^ y -


The original fraction is:

8 a) 4

[LIC AAO Exam, 1988]

1 27 3 3

23. What fraction must be subtracted from the sum of

d)None of these

1

1 1 and to have an average of of all the three frac-

tion?

4 b ) T c) 1

[SBIPO Exam, 1988]

1 d > 6

24. 2 + V 2 + U + -2 + VT v ^ - T r P B A E n m , l 3 ]

a) 2 b)4 c)0 d) Can't be determined

25. Iffx-jy = 1 and 4x + -Jy = \1, then ^/xy = ?

b)64

26.

a) 72

7+12 0.2x3.6

a) 17.82

[CET Exam, 1997] c)82 d)96

= 2

b) 17.22

27. V?x7xl8 = 84 a)3.11 b)3.12

28. 2 - X v x J

[CET Exam 1997]

c) 17.28 d) 17.12

[MBA Exam, 1982]

c)3.13 d)3.14

y ? J = 7 T , find the value of x and y.

[MBA Exam, 1987]

a) (3,15) b)(3,14) c)(14,3) d)(24,6)

29. I f x * y = (x + 2f{y-2) then7*5 = ?

[MBA Exam, 1983] a) 234 b)243 c)343 d)423

30. I f m and n are whole numbers such that m " = 121, then

( / w - l ) " + 1 =?

a) 10 b) 102

V1296 ? [CET Exam, 1996]

[MBA Exam, 1988] c) 103 d) 104

31 ? 2.25

a) 6 b)3

a

c)9

_17 a + b 32. If r - , what is equal to?

a + b 23 a-b

a) 23 17

c) 23 11

d) 12

23 d > 1 7

[MAT 1995]

33. h g ? 1

' J a I o 2 l = J

4 ^

[BSRBPO Exam, 1992]

a) 64 b)34 c)54 d)94

^ (o.ss^+Co.o?) 2,^) 2 _ ? (0.055)2 + (0.007)2 + (0.0027)2

a) 10 b)100 c)50 d)1000 [MBA Exam, 1992]

35. J ^ = 0 . 2 6

a) 10 b ) 1 0 2

18-3x4 + 2 . 36.

6 x 5 - 3 x 8

[SBIPOExam, 1988]

c) 103 d) 502

[BSRBPO Exam, 1986]

b)

37. W W . ! ? ?

a) 6 4 b) 4 6 C) 2 6

d > J

d) 6 2 [ITI Exam, 1988]

25 x 38. Find the value of x in the equation J l + = 1 + 144

a) l b)0.5

64 9 121 64

39. 8 3 + -11 8

c)2 12 ' d)4

[SBIPO Exam, 1986]

a) 88 31 b)

31 88

41 C > 9 9 d)

99 41

4- I f V l 8 x l 4 x x = 84, then x is equal to? a) 82 b)28 c)32 d)42

(Auditors 1986)

41- V 9 8 - V 5 0 = ? x V 2 a)2 b)4 c ) l d)3

[BankPO 1980| 42. Express the number 51 as the difference of squares of

two numbers.

a) 37 2 - 1 4 2 b) 36 2 - 1 5 2

c) 26 2 - 2 5 2 d) Can't be determined [BankPO 1982]

43. Thehighest score in an innings was of the total score

mad the next highest was of the remainder. These

scores differed by 8 runs. What was the total score in me innings? 162 b)152 c)142

| J _ j + ( 6 4 ) " ^ + ( - 3 2 ) ^

b ) ^

d)132 [NDA1988]

I f _ 4 , then find the value of n

; S b)10 c)12

h l x 4 + 7 = 2 2 -3 10 3

a) 2.4 b)4.2 c)2.6

d) 16 [MAT 1992]

d)2.8 [NDA 1983]

(l .06 + 0.04)2 - ? = 4 x 1.06 x 0.04 a) 1.04 b)1.4 c)1.5 d) Can't be determined

[CDS 1980]

I - I l . 1 1 I f a2+b2 =45 and ab= 18, find - + - .

d) Can't be determined

[MBA 1987]

a) b ) ? c ) -

-i9 If a2+b2 ab

c2 + d2 cd of c and d only.

a + b then find the value of r in terms a-b

a) : + d ~c~d~ b)

cd c + d c)

c-d c + d

50. Simplify

flV2+fl-V2 i_ a-V2 - + -

1-a

a)

1 + Va

a-l b) c)

c + d d) -

c - a [MBA 1987]

[MBA 1987]

2

a - l y 2

51. Solve 5V* + 1 2 ^ = 13^* a) 4 b)2 c ) l

52. 0.05 x 0.09 x 5 = ?

a - l d) 1-a

d)6 [MBA 1987]

a) 0.025 b) 0.225

2 , ( 2 x 2 ) _ 0 5 1 ( 2 , 2 ) x 2

a) 2 b ) l

54. 0.9-0.3x0.3 = ? a) 0.1 b)0.9

c) 0.005

c)4

c)0.09

[SSC 1994] 55. 2 2 N 5 + 3 y

2 2 hi - + -

5 3

a ) l b) 2 15 c ) 2

15

56. 7386 + 3333-7=10010 a)619 b)609 c)719

57. 4.16x0.75 = ? a)3.12 b)0.0312 c)31.2 d) 0.312 e) None of these

58 2- + 3- + 4- = l 2 3 4

a) 11

d) 10

12

1

b)5^ 0 9-

e) None of these

8.4x4.2-2.1 59 = ? 2.1x4.2,8.4

a) 16 b)8 c)1.6 d) 1 e) None of these

60. 400

0 4 [UTI 1990]

d)709 ]UTI 1990[

[UTI 1990]

[UTI 1990]

[UTI 1990]

289 425 a) 6800 b)256 c)272 d)225 e) None of these

[Assistant Grade Exam, 1980) 61. 1012x988 = ?

a)988866 b) 989996 c) 999856 d) 992786 e) None of tese

[SBI POExam.19-9]

62. ^0.361/0.00169=?


65.

68.

69.

70.

71.

19

190

b) L9 13 c)

19 13

d ) IT e) None of these [GIC1987]

63. 196 ?

a) 28 d) 16.5

? 36

= 9

b)84 c)56 e) None of these

[IAS, 1982]

64. I f Vl 5625 =125, then

Vl 5625 + Vl 56.25 + Vl .5625 = ? a) 1.3875 d) 156.25

b) 13.875 c) 138.75 e) None of these

[GIC1988] 2592

a) 18 d) 16

= 324

b)144 e)64

c)8

66. VT21 45 13

[RRB 1980]

11 a) 10.31 d)3

169 V225 b ) l c) 35.96

67. I f J1 + 25 144

e) None of these

1 + thenx=? 12

[UDC1983]

[BankPO 1981] a) l b)2 c)5 d) 7 e) None of these The cost of telephone calls in an industrial town is 30 paise per call for the first 100 calls, 25 paise per call for the next 100 calls, and 20 paise per call for calls exceed-ing 200. How many calls can one make for Rs 50?

[BSRBPO Exam, 1988] a) 175 b)180 c)200 d)225 e)250 0.538x0.5380-462x0.462

1-0.924 a) 2 b)1.08 d) 0.987 e ) l 6 - [ 5 - { ? - ( 2 - 1 . 5 ) } ] = 3 a)2 b ) l d)1.5 e)3.5

16.6 166

a) 0.1 b ) l

c) 0.076

c)2.5

c)0.01

[BankPO, 1983]

[BankPO, 1986]

[ITI, 1988}

d)100 c)10

72. I f (4 ) 3 x( > ^" | J =2" , tnenn = ? [NIC, 1982]

a) 10 b) 11 c)24 d) 12 e) 14

73. A number of men went to a hotel and each spent as many rupees as there were men. I f the money spent was Rs 15625, find the number of men.

[BankPO, 1980] a) 115 b) 125 c)130 d) 135 e) 145

74. The smallest possible decimal fraction up to three deci-mal places is, a)0.001 b)0.101 c)0.011 d) 0.111 e) None of these

[IITCE,1990] 75. A glass full of water weighs 100 gm. An empty glass

weighs 45 gm 320 mg. How much water can the glass hold? (Fashion Tech, 1993] a) 54.68 gm b) 145.32 gm c) 55.68 gm d) 60 gm e) None of these

76. A, B and C have to distribute Rs 1,000 between them, A and C together have Rs 400 and B and C Rs 700. How much does C have? [MBA Exam, 1987] a)Rsl00 b)Rs50 c)Rs200 d) Rs 300 e) None of these

8 3 , n a 3 77. I f - = 1 0 , t h e n - = ?

a a 8 a [BankPO, 1983]

a)-10

24 d) ~2

a 2 - 8

b)

e)

95 16

a 2 - 2 4 8a

c) _16 95

78. I f8700 ,x = 300and4,590-y=170,then(;c-y) *(x+y)

[BSRB Exam, 1988] c) 112

= ?

a) 29 b)56 d)27 e)81

3 7 ,.2 I f 9 - x y -

x 9 = 1 6 -

3 '

a) (7,1) d ) (5 , l )

b ) (8 , l ) e)(2,l)

]IRS, 1990] c)(13,l)

3 1 y &x x 9 ~ I 2 ' t b e n t b e v a m e f ' s '

[Assistant Grade Exam, 1992] a ) l , 13 b) 1,5 c ) l , 7 d ) l , l l e) 1,1

81. ^ - which of the following can replace all the ques-

Simplification

tion marks? [BankPO Exam, 1989]

a) 6 b)7 c)2 d) 42 e) None of these

xy* 3x 82. Divide by

yz * 4z

a) 3z 4

2 d > 3

b)

e)

4z 2z

83. V0.0064 is equal to, a) 0.08 b)+0.08 d) + 0.8 e) + 8

[BankPO 1975]

c)0.8

1 1 + -2 3

[SSC, 1994]

a)-5

d)

b)5 c) 25

:e) None of these

1 2 x ( 3 7 ) 2 - -

86. W. is equal to which of the following num-2 x 3 7 - 1

ber? a) 36.5 b)38 c)37.5 d) 37 e) None of these

87. Find the number one-seventh of which exceeds its elev-enth part by 100. a) 1925 b)1295 c)1952 d)1592 e) None of these

16 88. A boy on being asked to multiply of a certain frac-

tion made the mistake of dividing the fraction by 16 17

33 swer by . Find the correct answer.

a)

d)

64 85

64 58

46 b >5l

46 C>85"

e) None of these

Simplify 2 i

7

[BankPO 1975] a) 630 d)105

7 a ) L2

7 7 C ) 4 l

Answers l .c 2.d

7 7 8. a 9.d

d > l i e ) 27 15.c

if IO 2 ^ - 25 then what is the value of io^ ? 16. b; Hint: Let

89. A woman sells to the first customer half her stock and half an apple, to the second half her remaining stock and half an apple, and so also to a third, and to a fourth customer. She finds that she has now 15 apples left. How many had she at first? a) 255 b)552 c)525 d)265 e) None of these

5 4 2 4 90. of of a number is 8 more than of of the

same number. What is half of that number? b)315 c)210 e) None of these

3.c 4.c 5.d 6. a 7.c lO.a 11.b 12.a 13.c 14.a

5 6 8 8 3 10 25 y . T h e n , xx + x

6 7 9 5 4 3

5 7 8 5 3 10 25 X X X X 1 X = 6 6 9 8 4 3 9

35 5 5 25 or xx + =

' 36 9 2 9

35 25 5 5 50 + 10-45 15 or ~x '

' 36 9~ + 9 2 18 18

.". x 15 36 6 18 ^ 35 ~ 7

67 5

9 19 7 + + X + 2 6 3

(9 19 + + - = 6 3 j

18. c

1 75 1 19. a; Hint: 3 7 2 = y = g x

17 2 -10 = = 3 -

5 5

2 ^ - =1x300 2 j 8

and so got an answer which exceeded the correct an-Thus, the number of - ' s in 37^ is 300.


20. c; Hint: Out of 5 girls, 1 took part in the camp and out of 8 boys, 1 took part in the camp. Thus out of 13 stu-

2 dents, 2 took part in the camp. i.e. of total students

joined the camp. 21. c; Hint: Let the total number of workers be x.

1 Then, number of women = x ;

And, number of men = x

Number of women having children

1 A , 1 1 = of of X X

3 2 3 18 2 3 2 1

Number of men having children = ofofx = -x

1 1 7 Number of workers having children = x + x = x

1 \1 x = x

18 No. of workers having no children = ! x~ fgj

22. b; Hint: Let the fraction be I T Then,

^ x U * = 1 8 - S 6 5 1 2 b b 21 21

w 3 -U J , 3 ,

a a a 512 or x x = o r

' b b b 21

-=*=2-"b 3 3

1 1 , 1 23.d;Hint:Let T + ^ - x : = 3 x J 2 " - T h e n ,

1 1 1 1 + x = or x = 4 6 4 6'

24. a; Hint . 2 + V2-V 2 - 2 + 2 + V2

(2 + V2)(V2-2)

[Since a = ^2 a n d b = 2

.-. (a + b ) ( a - b ) = a 2 _ / , 2 ]

= 2 + V 2 " + ^ = > 2 + V ? + ^ = : 2 - 4 -2

25. a; Hint: V I + V>' = 1 7

and, 4x-Jy=\)

Adding equations (i) and (ii), we get Vx = 9

Subtracting equation (ii) from (i), we get Jy = 8

Substituting these values,

Jxy = yfxxjy = 9x8 = 72

26. c; Hint: Putting* in place of?

jc-5-12 _ 2 0.2x3.6 ~

or, 12 = 2x0.2x3.6 => = 2x0.2x3.6 12

or, x= 12x2x0.2x3.6 = 17.28 27. a; Hint: Substituting x for ?, we get,

V * x 7 x l 8 = 84

Ir 84 or, V*x7 =

18

or, x x 7 = 84

.". x-84x84

= 3.11 18x18x7

28. c; Hint: Taking the quotients 2, y and 7, we get 2y = 7, which gives the quotient as 3 .-. y - 3. Substituting the value of y, we get,

o 3 7 1 T 3 2 x3 = 7

x 2 4

Now, 4- = 2 - = > 2 A = 2 I 3 I x 14 x

2

.-.x = 14 v = 3 29. b; Hint: Substituting x = 7 and y = 2, we get,

7*5 = (7 + 2 ) 2 ( 5 - 2 ) = (9) 2 x3=243

30. c;Hint:Giventhatm n= 121 m n = ( l l ) 2 Hence, m = 11 and n = 2, substituting these values,

( m - i y ^ H - l f + U l o ^ l O O O

3 l .c; Hint: Putting x for (?),

Vl296x2.25 = x 2

or, 36x2.25 = x 2

Simplification

or, x = ^36x2.25

or, x = 6 x 1.5 [since ^1296 =36]

.-. x = 9

a _ 17 I : Hint: Given that r _ 77

a + o 23 i.e., i fa= 17, then a + b = 23 or, b = 6 a - b = 1 7 - 6 = 1 1 a + _ 2 3 " a - 6 11

33. c; Hint: Putting x t o T ^ > 3 ^ f e f a % % ^ *,w.e,ige.t.

x x' * - 1 18X162 ~

or, x 2 =18x162

or, x2 =18x18x9 or ,x=18x3

. \ = 54 34. b; Hint: Let 0.55 = a, 0.07 = b and 0.027 = c

Then, the given expression becomes

a 2 + b 2 + c 2 _ l a ^ + c 2 !

(0 .1xa) 2 +(0.1x6) 2 +(0.1xc) 2 0 . 0 l [ a 2 + f c 2 + c 2 ]

0,01 = 100

35. c; Hint: Putting x for (?) and solving it for x,

67.6 x

67.6 or,

= 0.26

= (0.26)2

X = AZA_ o r x = 1000 0.0676

36. d; Hint: Putting x for (?) and applying VBODMAS rule,

18-32 + 2 18 + 2-12 20-12 or,x = -^zr - or,x = or.x =

30-24

or,x = .'. x = 6 3

30-24 30-24

37. b; Hint: Putting x for (?), and since all base are equal to 4, hence, put a = .

o r , x = W 4 2 N t 5 ) r

or, x = an -=-a 6xl or,x = a 1 2 - 6 or, x = a6

[since (a5J =\]

[since a m + a " = a " ~ " ]

..x = 4

144 + 25 . x 38. a; Hint: J - ^ - = l + -

169 , x 13 x or, J = 1 + or, = 1 +

' V144 12 12 12 or, x = l

( 6 4 2 - 9 x l 2 l ) 8x11 39.b;Hint:x= x ( 8 x 8 + 3 x l l )

'{6A1 - 3x3x1 l x l l ) 3x11 OT'*~ (11x11x8x8) X ( 6 4 + 33)

or, x = (64 + 33X64-33)

11x8 (64 + 33)

or, x = 31

1 8 x T 4 x x = 84x84 or, x -

40. b;Hint: V l 8 x l 4 x x = 84 Since x is under square root, so, squaring both sides we get

84x84 18x14

.'. x = 28 41. a; Hint: Putting x for (?) and solving it for x,

^ 7 x 7 x 2 - ^ 5 x 5 x 2 = x x ^ 2

or, 7V2-5V2 = x x V 2

.". x = 2 42. c; Hint: Using the formula,

, where N = original number 'N+l

2 ~N-\ 2 2 N =

PutN = 51

or,51 = "51+r 2 " 5 1 - f

2 2

or, 51 = (26)2 - (25) 2 43. a; Hint: Let the total score be x runs, such that

2 2 ( 2 J C X X X | = ' i

9 9 I 9 Or,

2 2 7 . x xx = 8 9 9 9

2 2 or xx= 8 or, x = 162 "> ft n


.-. The total score in the innings was 162.

44. a; W + ( 6 4 ) * + ( - 32)*

= l + ( 8 2 ^ + ( - l x 3 2 ^

= 1+ 8 ' + l ( - l ) ^ x ( 3 2 ) ^ J

45. c; Hint: =64

49. d; Hint: a2+b2 ab a2+b2 lab c2+d2 cd ' c 2 + r f 2 2co"

or, -=-

= 1 8 +

( c + d ) {c-d)

or, , ( 2 " f = 2

= i + l + b 4 l = n l

,6 _ n 1

:.n = X2 46. a; Hint: Putting x for (?) and solving for it gives

l l I x 4 A + A ; = 2 2 -3 10 3

1 8 2 or, H - x 4 = 2 2 y x x [sincea-b = cthena = b * c ]

1 8 or x = x4

2 10

1

a2+b2+lab a2+b2-lab or, 5 5 = r =

c2+d2+lcd c2+d2-lcd

or,

a+b_+c+d a-b c-d

aA+a'Yi \-a-Yi 50. d; Hint: ; + - = -

1-a X + Ja

- + \ a^Jx-a

l - a = ( l ) 2 - ( ^ ) 2 = ( l + ^ ) ( l - ^ ) since

11 since 3 _ 1

2 2 2 2 3

or, x = 2.4 47. a; Hint: Putting x for (?) and solving for it

(l .06 + 0.04)2 - x = 4 x 1.06 x 0.04 Here, 1.06 = a and 0.04 = b

:.(a + bf -x = 4ab

;. x = (a - bf = (l .06 - 0.04)2 = 1.0404

\a + bf-{a-bf=4ab\

t . 1 1 1 - a + fc _ yla2+b2+lab

tfi+a ^ + ( l - a ~ ^ Y l - a ^

\ a^Jx-a^

a^+a^+X-a^-a^+X 2 X-a (1-a)

51. a; Hint: 5V* + \4x - j3V*

The given equation is of the form

5 2 +12 2 =13 2 [By the Pythogoras theorem of numbers]

Comparing the two equations, we find

4x = 2 ;. x = 4

52. d 53. d 54. b 55. b 56. d 57. a 58. d 59. a 60. c 61. c; Hint: 1012 x 988=(1000 + 12) x (1000 -12) [Use (a+b)

( a - b ) = a 2 - b 2 ] 48. c; Hint:

since

a b ab a o

a + b = yj(a+bf

62. d;Hint:

63. b

0.361 0.00169

' ^ x l O 2 U 3

V45 + 2 x l 8 _ 9 _ + 1 18 ~J&~~2

64. c; Hint: ^/with two decimal places = one decimal place

V156.25 =12.5

^/with four decimal places = Two decimal place

plification *3

.-. Vl .5625 = 1.25 66. d 67. a 68. b 69. e 70. c

f l6 .6f C ; H i n t : ? = l l ^ J

a ; H m t : ( 2 2 y x ^ ] 8 = 2 ' ' ^ 2 6 + 4 = 2 ' ,

~3 b; Hint: x 2 =1562 _ 4 .a 75. a 76. a;Hint: [(A + C) + (B + C ) ] - ( A + B + C) = C] 77. b 78. c 79. b 80. b 81. d; Hint: Put x for? 82. b 83. b 84. d

85b;Hint: io y == VlO2^7 == ^25 == 5 86. c

1 1 4 S a; Hint:

7 11 77

77 of the number = 100

77-the number = 100 x = 1925

4

88. a; Hint: Since 17 16 289-256 33 16 17 16x17 16x17

33 33 .-.the fraction x - x i ? ^

33 16x17 _ 4 t h e f r a c t i o n =

16 4 64 .-. the correct answer = of = r r

17 5 85 89. a; Hint: Begin with the fourth customer.

Her stock before the 4th customer came was 2

or31

Her stock before the 3rd customer came was 2 31 +

or63 Her stock before the 2nd customer came was

2^63 + - J o r l 2 7

Her stock before^ the 1st customer came was

2 127 + - or 255.

90. d; Hint: Let the number be x.

5 4 2 4 . x xx x - x x = 8 7 15 5 9

8x315 . . . or, x = = 210

12 .-. Half of the number =105

S E T - I I Directions (Q. 1-10): Four of the five choices are ex-

actly equal. Which one of the parts is not equal to the other four? The number of that choice is the answer. 1. a)5280-3129 + 933

b) 80% of5000 + 4% of 150 - 461 x 2

c) 8^ of558-1680

d) 1950 + 300 + 50% of 1700 - 8 x 2 e) 22 x 30 + 30 x 15 + 75 x 3 5 - 6 5 1

2. a) 75 x 75 - 50% of2200 - 5% of 500 b) 80 x 30 +15 x 40 + 60% of 1800 + 420 c) 25 x 85 + 90 x 20 + 50% of 1150 d) 35 x 3 5 + 2 1 x 9 0 - 5 % o f 100+1392 e) 1 1 0 0 x 5 + 2 x 3 0 - 5 0 % o f 2 0 8 8 - 2 x 2 x 2 x 2

3. a) 1.3x5 + 2 .3x5 b) 4 ! -3 ! c) 23 + l 3 + 2 2 + 1 8 - ( 3 2 + P ) d) 40% of40 + 20% of 20 -10% of 20 e) 10%of20 + 20%of80

4. a)0.5 + 0.55 + 0.05 b) 0.6+ 0.04+ 0.05+ 0.3+ 0.01 c) 0.1x 1.0x0.01x1000 d) 0.3+0.27+ 0.03+ 0.4 e) 0.5x2.0

1 1 1 1 5. a) r * T + - + - 7 * a b e d

e+b+c+d

b)

d)

1 -xe a+b+c+d

(a + b + c + d)

bde

ac

6. a) 87-i'% of 528 b) 6 6 y % of 522 + 6x19

c) 23 2 - 8 2 -3 d) 1 6 - % of 2772

e) 6 2 - % of 496 + 3 7 - % of 288-4 + 8x6 5 5


a)21 x5 + 12x6 + 2x 12 b)19x9 + 2 x l 5 c)13x l l + 1 5 x 4 - 2 x 1 d) 11 x 19-2 3 + 12 e) 1 5 x 1 2 + 1 1 x 2 - 1

63 a) 40% of 36 + 5% of 175 -

b) V625-25% o /20

11 c) 20% of 125+ of207-104

d) - o f424- 5 a /256

1 11 e) - of320- of 76

2 10 a)35%of48 +15%of76 b) 3 2 + 4 - % of 25

3 14

c) 46.68-19.57+1.09 d) 78^-% of3.6

b)5680-3510+1930 d) 1080+2320 + 710

e) V795.24 10. 3)3130+2060-1090

c) 11450-5090 -+2260 e)8645-3155-1390 Directions (Q. 11-30): In the following questions one of

the choices among (a), (b), (c) and (d) is different from other three. Mark the choice which is different If the four choices are equal, the answer is i.e. No error. 11. 3)7x0.5+1.5x0.5 + 2.5x0.3

b) (1 .4x5)-2 + 15-10

c) ( i 2 + 2 2 + 3 2 ) + 3 - ( 3 x 4 ) - 6

d) 2(l.5+ 1.3)-3x0.2 e) No error

12. 3)15%ofl50 + 25x0.3 b ) ( l x 2 x 3 x 4 x 5 ) - 4

c) 4 3 _ 6 2 + 2 d) l2 + 2 2 + 3 2 + 4 2

e) No error 13. 3)(1.6x6 + 6.2 x5)-0.5

b) 13x3 + 1 4 x 2 + 1 . 4 x 5 +1.2x5 C) 4 3 + 2 4

d) ( l 2 + 3 2 + 5 2 + 7 2 ) - 2 2 e) No error

14. 3 ) 10% of45 + 55% of30 b) 70% of30 + 40% of 20 - 8 c) 15%of40+ 13% of 50+ 17% of 50 d) l x2 + 2 x 3 + 3 x 4 e) No error

15. 3)1.2x0.003 x20x0.01 b) 0.0032 x0.02 3 x l O 4

c) 0.062 x 10 2 x 0.002 d) 0.3 x 2400 x 0.0012 e) No error

16. a) (a + b + cf -(a-b-cf

b) (a + b-cf -{a-b-cf +4c{a + b)

c) (b + c-af -(a + b-cf + 4c(a-b)

d) 4a(b + c)

e) No error

17. 3 ) a2(b-c)-b2(a-c)+c2(a-b)

b) -(a-b\b- c\c - a)

c) ab(a-b)+ bc(b-c)+ ca(c-a)

d) c2(a-b)+a2(b-c)-b2{a-c) e) No error

18. a)6.5x 1.5 + 3.5x2.5-5x0.5

b) 3 2 + 80% o/12 + 7% of 20 - 2 2

c) ^ - 1 2 . 5 % o/32

d) 3 V 7 8 4 - 8 2 - 2 2 e) No error

19- 3 ) 4 8 2 b H 2 ^ 3 ) 2 c ) 6 ( 2 3 ) + 9 3

d) 9 x 2 8 e) No error

20. 3 ) 1 2 + 3 2 + 5 2 + 7 2 b ) l 3 + 3 3 + 4 3 _ 2 3

c) 2 2 + 4 2 + 6 2 + 5 2 +3 1 d) 2 3 + 4 3 + l 3 + 3 2 + 2 ' e) No error

21. a)1.4x4 + 2.3x4-1.6x0.5 b) 2 . 7 x 4 - 1 . 8 x 3 + 2 x 4 . 3 c) 1 .2x 8 -1 .952 + 2 x 8 . 2 d) 2 .8x2+1.2x7 e) No error

22. a) - 2 n ( 2 + 3 m 2 )

b) - ( n - m f -{n + mf

c) - 2m[m2 + 3 2 )

d) 4r? -6mn(m-n)-6n2(m + n) / e) No eJtfor

l23. *){x-yf -{x + yf -(y + 2xl2x-y)

b) 3y2+(y-2x\x-2y)-x(2x + 9y)

c) y2 -4x(x + y)

. : : : r .

30. a)20%of45 + 17%of9 b)25-5 x 2.894

b) 6 4 - 2 4

Ifcf32 x 62.5% of25.6 d) 2 7 x 5

-*X*-*Xc -4 - - ( z -* ) -Xft-xX*-c ) . . ( r -*)}

65+15% of 9 b ) 5 2 _ 5 x 0 . 8 3

xl.175 d)45%of 4 6 -

b) 1000% of 15

50% of -3 3 d) 0.09375 of 7 -

- r y z - ( y + 3 x ) 2

x) 2 - (3y - x ) 2 + 5(y - xX* + v)-12xv

+2y)

^ + ( v - 2 x ) 2 + v

wf -(n + mf

-In2)-3n(n + 3m) 2 - ( m - n f

a 4 . 2 + 3 m 2 )

".f -m2(m + 9n)-3n2(n + m)

d)50%of 2 1 -c)9 + 3x0.51

e) No error

Answers 1. c; A l l others are equal to 3084. 2. d; Al l ohers are equal to 4500. 3. c; Al l others are eual to 18. 4. a; A l l others are equal to 1.

bde 5. c; A l l others are equal to .

ac 6. e; Al l others are equal to 462. 7. d; Al l others are equal 201. 8. d; Al l others are equal to 20. 9. d; A l l others are equal to 28.2. 10. d; A l l others are equal to 4100. 11. c; Al l others are equal to 5. 12. e; A l l are equal to 30. 13. e; A l l are equal to 80. 14. d; A l l others are equal to 21. 15. b; Al l others are equal to 0.00072. 16. c; A l l others are equal to 4ab + 4ac. 17. b; A l l others are equal to

a2b - a2c - ab2 +b2c + ac2 - be2 18. e; A l l are equal to 16. 19. c; Al l are equal to 2304. 20. e; Al l are equal to 8. 21. c; A l l others are equal to 14.

22. c; A l l others are equal to - 2 3 -6m2n

23. e; Al l are equal to - 4 x 2 + y 2 - 4xy

24. d; A l l others are equal to 1280. 25. e; A l l others are equal to 0. 26. c; Al l these are equal to 20.85.

27. a; A l l are equal to 0.6

28. e; A l l are equal to - 5x 2 - lOxy

29. b; Al l others are equal to _ 2 w 3 - 6m2n 30. d; Al l others are eual to 10.53.

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Chapter 2