Vedic Maths: Addition, Subtraction, Multiplication and

101 Pages·2016·5.18 MB·English

QA / Exercise - 1
CEX-5301/P1BS/17 / Page 1

Quantitative Aptitude – 1

Vedic Maths: Addition, Subtraction, Multiplication and Squares

Answers and Explanations

1. d
2. d
3. b
4. b

5. d
6. c

7. a
1
1
1
1
1
1
1
1
9
6
12
20
30
42
50
72
+
+
+
+
+
+
+

=
+
+
+
+
+
×
×
×
×
×
1
1
1
1
1
1
...
9
2
3
3
4
4
5
5
6
8
9

=
+
−
+
−
+
+
−
1
1
1
1
1
1
1
...
9
2
3
3
4
8
9 = 1
2

8. d
+
+
+
+
+
×
×
×
×
×
×
1
1
1
1
1
1
5
6
6
7
7
8
8
9
9
10
10
11

=
+
−
+
−
+
−
+
−
+
−
1
1
1
1
1
1
1
1
1
1
1
5
6
7
7
8
8
9
9
10
10
11

= 1
1
11
5
6 .
5
11
55
55
−
−
=
=

9. b
Expression

=
+
+
0.63
0.37
0.80

=
+
+
63
37
80
99
99
99

=
+
+
63
37
80
99
= 180
99

= 81
199 = 1.81

10. a
11. d
12. a
13. b
14. b

15. a
16. d
17. b
18. d

19. a
? = 5 – [4 – {3 – (3 – 3 – 6})]
= 5 – [4 – {3 – (– 6})]
= 5 – [4 – {3 – 6}]
= 5 – [4 – 9]
= 5 + 5 = 10

20. a
? = 1 – [5 – {2 + (– 1)2}]
= 1 – [5 – {2 – 2}]
= 1 – [5 – 0]
= 1 – 5 = – 4

21. b
243 × 658
Step 1 : 8 × 3 = 24

Step 2:

4
3

5
8

= 4 × 8 + 5 × 3

= 32 + 15 = 47

Step 3:

4
3

5
8

2

6

= 2 × 8 + 6 × 3 + 4 × 5

= 16 + 18 + 20 = 54

Step 4:

2
4

6
5

= 2× 5 + 4 × 6 = 10 + 24 = 34

Step 5: 2 × 6 = 12
Now the answer is 12|34|54|47|24 = 159894
Though here we have first done each step and then
at the end write the answer, each step has to be done
to keep getting each digit of the answer (starting from
the units place simultaneously).

1
d
2
d
3
b
4
b
5
d
6
c
7
a
8
d
9
b
10
a
11
d
12
a
13
b
14
b
15
a
16
d
17
b
18
d
19
a
20
a
21
b
22
c
23
d
24
c
25
b
26
c
27
c
28
a
29
c
30
c
31
d
32
b
33
d
34
a
35
a
36
a
37
d
38
b
39
b
40
a
41
c
42
a
43
b
44
a
45
b
46
b
47
b
48
b
49
c
50
d
51
d
52
c
53
b
54
a
55
c
56
b
57
d
58
c
59
c
60
b

P-1 (BS)

QA / Exercise - 1
CEX-5301/P1BS/17 / Page 2

22. c
96 × 108
Base = 100, Deficit = 96 – 100 = –4,
Surplus = 108 – 100 = 8

(
)

96
| –4
108
| 8

108
4 | 32

×
+

−
−

or
(96 + 8)
⇒ 104| – 32 ⇒ 103|100 – 32 ⇒ 103|68 = 10368
Right part will now be (–4) × 8 i.e. – 32. To take care of
the negative we will borrow 1 from the left part, which
is equivalent to borrowing 100 (because we are
borrowing from the hundredth place digit of the
answer). Thus, this part will be 100 – 32 = 68.

23. d
126 = 2 × 32 × 7
Since 2 and 7 are not in pairs.
∴ Required number to be multiplied = 2 × 7 = 14.

24. c
112 × 113
Base = 100, Surplus = 12 and 13

(
)
1

112
| 12
113
| 13

112
13
|
56

+
×
+

+

⇒ 125 + 1 | 56 = 12656.

25. b
92 × 97
Base = 100, Deficit = 92 – 100
= – 8 and 97 – 100 = – 3

92
| 8
97
| 3

(92
3)
| 24

−
×
−

−
+

⇒ 89 | 24
= 8924.

26. c

= 8 × 7 | 8 × 3 + 7 × 1 | 3 × 1

56
+3
59

3 1

= 5913.

3

81×73

27. c
63 × 72 = 6 × 7 | 6 × 2 + 3 × 7 | 3 × 2
= 42 | 33 | 6
= 42 + 3 | 36
= 4536.

3
6

7
2

28. a
524 × 368 = 5 × 3 | 5 × 6 + 3 × 2 | 5 × 8 + 2 × 6 + 4 × 3
| 2 × 8 + 6 × 4 | 8 × 4
= 15 | 36 | 64 | 40 | 32

= 15 | 36 | 64 | 40 + 3(= 43) | 2

4
5

3
8

2

6

= 15 | 36 | 64 + 4(=68) | 3 | 2
= 15 | 36 + 6(=42) | 8 | 3 | 2
= 15 + 4 | 2 | 8 | 3 | 2 = 192832.

29. c
88 × 93 = 8 × 9 | 8 × 3 + 8 × 9 | 8 × 3
= 72 | 96 | 24

= 72 | 96 + 2 (= 98)| 4

8

9
3

8

= 72 + 9 (= 81) | 8 | 4
= 8184.

30. c
Using the base of 100, the answer is
(106 + 45) × 100 + 6 × 45 = 15100 + 270 = 15370.

31. d
426 × 543 = 4 × 5 | 4 × 4 + 5 × 2 | 4 × 3 + 2 × 4 + 6 × 5
| 2 × 3 + 6 × 4 | 6 × 3
= 20 | 26 | 50 | 30 | 18
= 20 | 26 | 50 | 30 + 1(= 31) | 8
= 20 | 26 | 50 + 3( = 53) | 1 | 8
= 20 | 26 + 5(=31) | 3 | 1 | 8
= 20 + 3 | 1 | 3 | 1 | 8 = 231318.

32. b
1000
52500
52.5 125 52.5
6562.5.
8
8
×
=
×
=
=

33. d
197 × 203 = (200 – 3)(200 + 3) = 2002 – 32

202 × 198 = (200 – 2)(200 + 2) = 2002 – 22

⇒ (197 × 203) + (202 × 198)
= 2002 – 32 + 2002 – 22

= 2 × 40000 – 13 = 79987.

34. a
It is obvious that one needs to do multiplication for
solving the given equation. But observing the equation
we can see that 121 is a multiple of 11 and 111 is a
multiple of 3 and, among the given options only option
(a) satisfies this condition.

35. a
375 × 24 × 11 = 125 × 3 × 3 × 8 × 11
= 9 × 11 × 1000 = 99000.
Note: If you get multiples of 5 and even numbers, try
to maximize the number of trailing zeroes and reduce
the calculation.

36. a
Largest number of 4 digits will be 9999

(
)
(
)
2
2
99
9999
100
<
<

Hence, the required number will be (
)
2
99
9801.
=

QA / Exercise - 1
CEX-5301/P1BS/17 / Page 3

37. d
Here, base = 100
Therefore, (108)2 = (100 + 2 × 8) | 82

= 116 | 64 = 11664.

38. b
Here, base = 100

Therefore, (
)
(
)
2
2
92
100
2
( 8) |
8
=
+
× −
−

= 84 | 64
= 8464.

39. b
Here, base = 1000
Therefore,

(
)
( )
2
2
1008
1000
2
8 | 8
1016 | 064
1016064.
=
+
×
=
=

40. a
Here, base = 1000

Therefore, (
)
(
)
2
2
993
1000
2
7 |
7
986 | 049
=
−
×
−
=

= 986049.

41. c
Here, base = 1000
Therefore, (1012)2 = 1000 + 2 × 12 | (12)2

= 1024 | 144 = 1024144.

42. a
1162 = (100 + 16)2 = 1002 + 2 × 16 × 100 + 162

= 100(100 + 32) + 256
= 13200 + 256 = 13456.

43. b
942 = (100 – 6)2 = 1002 – 2 × 6 × 100 + 62

= 100(100 – 12) + 36
= 8800 + 36 = 8836.

44. a
542 = (50 + 4)2 = 502 + 2 × 50 × 4 + 42

= 2500 + 400 + 16 = 100(25 + 4) + 16
= 2916.
Alternate method of using (base 50):
542 = 25 + 4|42 = 2916.

45. b
The square of the number of the form ‘a5’
= a × (a + 1)| 52

Hence, 752 = 7 × (7 + 1)| 52 = 5625.
Note: Number ‘a’ can have more than one digit as
well.

46. b
3034 – (1002 ÷ 20.04)

=
1002
3034 – 20.04

=
×
1002
3034 –
100
2004

= 3034 – 50 = 2984

47. b
1
1
1
1
1
1
3
1
1
4
4
2
2
3
6






÷
−
−
−













=







÷
−
−
−













17 13
5
1 3
1
1
2
4
4
2 2
3
6

=





−
−


−
÷
−













17
13
5
1 9
2
1
2
4
4
2
6

=





−
÷
−
×









17
13
5
1
6
2
4
4
2
6

=





−
÷
−









17
13
5
1
2
4
4
2

=



−


−
÷ 








17
13
5
2
2
4
4

=



−
÷





17
13
3
2
4
4

17
13
4
2
4
3


−
×



 =
−
17
13
2
3

=
−
51
26
6
= 25
6 =
1
4 6

48. b
0.008 × 0.01 × 0.072 ÷ (0.12 × 0.0004)
= 0.008 × 0.01 × 0.072 ÷ (0.000048)

= 0.008 × 0.01 ×
0.072
0.000048

= 0.00000576
0.000048
= 0.12

49. c
Expression
= 25 – 5[2 + 3 {2 – 2 (5 – 3) + 5} – 10] ÷ 4
= 25 – 5 [2 + 3 {2 – 2 × 2 + 5} – 10] ÷ 4
= 25 – 5[2 + 9 – 10] ÷ 4

= 25 – 5 ÷ 4 =
− 5
25
4

=
−
100
5
4
= 95
4 = 23.75

50. d
We have

÷
×
5
2
*
3
7
7 = 5
2
6
4
3
×
×

⇒
×
×
5
7
*
3
2
7 =
×
×
×
5
2
6
4
3

∴ * =
×
×
×
×
×
×
×
×
5
2
6
3
2
7
5
7
4
3
= 6

QA / Exercise - 1
CEX-5301/P1BS/17 / Page 4

51. d
−
+
3.36
2.05
1.33

=
−
+
36
05
33
3
2
1
99
99
99

=
+
−
−
+ +
36
5
33
3
2
1
99
99
99

=



−
+
+
−
+





36
5
33
(3
2
1)
99
99
99

=
−
+


+ 




36
5
33
2
99

=
+ 64
2
99 =
64
2 99 = 2.64

52. c
[0.9 – {2.3 – 3.2 – (7.1 – 8.9)}]
= [0.9 – {2.3 – 3.2 + 1.8}]
= [0.9 – 0.9] = 0

53. b



+
−
+







5
3
25
15
9
2
2
4
4
4

=
× 19
4
4 = 19

54. a
Expression =

+
+
(0.04
0.01)
(0.01
0.02)

55. c
Expression

=





+
−
−









1
19
19
7
2
4
6
3

=



−


+
−









1
19
19
14
2
4
6

=



+
−





1
19
5
2
4
6

=
+
−
1
19
5
2
4
6

=
+
−
6
57
10
12
=
=
53
5
4
12
12

56. b
Expression

=
×
+
×
−
×
×
17
17
2
2
17
2
2
15
15
15
15
15
15

= 

−





2
17
2
15
15

=
−




=









2
2
17
2
15
15
15
= 1

57. d
Let
11
415 = a and 15
71 = b.

∴ Expression
= (a + b)2 – (a – b)2

= (a2 + b2 + 2ab) – (a2 + b2 – 2ab) = 4ab

=
×
×
11
15
4
415
71 = 4 ×
×
71
15
15
71 = 4

58.c

2
1
2
2


+





=
+
+
×
×
=
1
1
1
2
2
2
4
2
2
2

59. c
Expression
= 71 × 29 + 27 × 15 + 8 × 4
= 2059 + 405 + 32 = 2496

60. b
Expression
= 0.05 × 5 – 0.005 × 5
= 0.25 – 0.025 = 0.225

QA / Exercise - 2
CEX-5302/P1BS/17 / Page 1

Quantitative Aptitude – 2

Vedic Maths : Square Roots, Cubes,
Cube Roots and Reciprocals

Answers and Explanations

1. b
2304
2
2
2
2
2
2
2
2
3
3
=
×
×
×
×
×
×
×
×
×

8
2
2
3
=
×
= 24 × 3

= 48.

2. a
7
4 3
+

3
4
2.2 3
=
+
+

(
)
( )

2
2
3
2
2.2. 3
=
+
+

(
)
2
2
3
=
+
{� a2 + b2 + 2ab = (a + b)2}

∴ Square root of 7
4 3
2
3.
+
=
+

3. c
30
6 289

=

2
2
2

2
1764
2
2
3
3
7
7
2
3
7
289
17
17
17

×
×
×
×
×
×
×
=
=
×

= 2
3
7
42
8
2
.
17
17
17
×
×
=
=

4. c
156.25
0.81
3.61
+
−

=

6
2
2

2
2
2
15625
81
361
5
9
19
100
100
100
10
10
10
+
−
=
+
−

= 125
9
19
10
10
10
+
−
= 12.5 + 0.9 – 1.9 = 11.5.

5. d
The square root of 0.0036 × 10–4 =
8
36
10−
×
= 6 × 10–4.

6. d
Cube root of 250
686 will be equal to 3 250
686

3
3
3
3
3
125
5
5
5
5
5.
343
7
7
7
7
7

×
×
=
=
=
=
×
×

7. a
The number of boys in a row

=
20736
2
2
2
2
2
2
2
2
3
3
3
3
=
×
×
×
×
×
×
×
×
×
×
×

=
8
4
4
2
2
3
2
3
144.
×
=
×
=

8. a
The last term of the given sequence of infinite terms

=
2
3 2 2
(1
2)
1
2
+
=
+
= +

On being added with the previous term i.e. 2
2
+

it again becomes 3 2 2
+
This cycle repeats itself and finally the answer will be

2 1
+
.

9. a
147
432
×
=
3 7 7
2 2 2 2 3 3 3
× × ×
× × × × × ×

= 4 × 7 × 9 = 252.

10. b
614656 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 7 × 7 × 7 × 7

Hence,
614656 = 2 × 2 × 2 × 2 × 7 × 7 = 784.

11. b
x = 144 + 25 = 169
⇒ x = 13 × 13

x
∴
= 13.

12. a
239
961
31 31
2 361
361
19
19
×
=
=
×

31
12
1
.
19
19
=
=

1
b
2
a
3
c
4
c
5
d
6
d
7
a
8
a
9
a
10
b
11
b
12
a
13
a
14
b
15
c
16
b
17
a
18
b
19
b
20
a
21
b
22
d
23
b
24
d
25
d
26
b
27
a
28
b
29
b
30
a
31
a
32
b
33
c
34
a
35
c
36
d
37
b
38
b
39
a
40
b
41
c
42
c
43
d
44
b
45
b
46
c
47
a
48
c
49
d
50
b
51
b
52
a
53
b
54
b
55
d
56
d
57
d
58
c
59
a
60
c

P-1 (BS)

QA / Exercise - 2
CEX-5302/P1BS/17 / Page 2

13. a
Let 0.2 = a and 0.08 = 2b
∴ Expression

=
×
×
+
×
×
×
×
+
×
×
3
a
a
a
b
b
b
2a
2a
2a
2b
2b
2b

=

+

+

3
3
3
3
3
a
b

8(a
b ) =
=
3 1
1
8
2 = 0.5

14. b
×
2 0.014
0.14x

= 0.014 × 0.14 2 y

On squaring both sides.
0.014 × 0.014x
= (0.014)2 × 0.142 × y

∴

x
y = 0.014 × 0.14 0.00196

15. c
13
1300
0.013
+
+

=
130
130
10 13
10
10000
+
+

=
+
+
1
1
130
10 13
130
10
100

=
+
×
+
11.40
11.40
3.605 10
10
100
= 1.140 + 36.05 – 0.1140
= 37.304

16. b
Let 0.03 = x ⇒ 0.003 = x
10

0.21 = y ⇒ 0.021 = y
10

and 0.065 = z ⇒ 0.0065 = z
10
∴ Expression

=

+
+







+
+













2
2
2

2
2
2
x
y
z

x
y
z
10
10
10

=
+
+

+
+

2
2
2

2
2
2
(x
y
z )
100
(x
y
z )

=
100 = 10

17. a
+
+
+
+
5
11
19
29
7

=
+
+
+
5
11
19
6

=
+
+
5
11
25

=
+
+
5
11
5 =
+
5
4

=
9 = 3

18. b
=
18225
135

∴
=
182.25
13.5;

=
1.8225
1.35,

=
0.018225
0.135
∴ Expression
= 135 + 13.5 + 1.35 + 0.135
= 149.985

19. b
Let the number of days of tour be x.
∴ Total expenditure = x2

∴ x2 = 361 ⇒ x =
=
361
19

20. a
Number of students in the last row =
=
1369
37

21. b
Base 100. Answer is (100 + 36) | 3 × 144 | 1728
= 1404928.

22. d
(
)
( )
3
3
3
(30
1)
30
1
3.30.1(30
1)
+
=
+
+
+

= 27000 + 1 + 2790 = 29791.

23. b
Base 1000. Answer is (1000 + 15) | 3 × 25 | 125
= 1015075125.

24. d
Base 1000. Answer is (1000-9) | 3 × (
)
(
)
2
3
3
|
3
−
−

= 991 | 27 | – 27
= 991 | 026 | 1000 – 27 = 991026973.

25. d
Base 100. Answer is (100 – 27) | 3 × (
)
(
)
2
3
9
|
9
−
−

= 73 | 243 | – 729
= 73 | 235 | 800 – 729 { Taking carry 8 from the other
side}
= 73 | 235 | 71 = 73 + 2( = 75) | 35 | 71 = 753571.

26. b
1440 = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 = 23 × 22 × 32 × 5
To make 1440 a perfect cube, it must be multiplied by 2
× 3 × 5 × 5 = 150
The required sum = 1 + 5 + 0 = 6

27. a
3000 = 3 × 1000 = 3 × 103

Clearly, when we divide 3000 by natural number 3,
the quotient is 1000 which is a perfect cube.

28. b
12 × 12 × 12 = 1728
∴ Required number
= 1728 – 1720 = 8

QA / Exercise - 2
CEX-5302/P1BS/17 / Page 3

29. b
4320 = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5
= 25 × 33 × 5
∴ Required number = 2 × 5 × 5 = 50

30. a
Look at the pattern:
1001 × 1001 = 1002001
1001 × 1001 × 1001 = 1003003001

31. a
3 1728
3 2
2
2
2
2
2
3
3
3
=
×
×
×
×
×
×
×
×

3
6
3
2
3
=
×
= 22 × 3

= 12.

32. b
3
3
144
1500
×

= 3
3
2
2
2
2
3
3
2
2
3
5
5
5
×
×
×
×
×
×
×
×
×
×
×

= 3
3
4
2
2
3
2
3
2
3
5
×
×
×
×

= 3
6
3
3
2
2
3
5
2
3
5
60.
×
×
=
×
×
=

33. c
3
3
486 36 64

= 3
3
3
3
2
2
3
486 36 4
486 36
4
486 6
2
=
×
=
×

=
3
3
6
2
3 486
6
2
2
3
2
3
18.
×
×
=
×
=
×
=

34. a
3 0.000729 = 3
729
1000000

=
( )
(
)

3

3
3
9

100

9
0.09.
100
=
=

35. c
3
3
32
250
×
3
3
5
3
2
2
5
=
×
×

3
6
3
2
5
=
×
= 22 × 5 = 20.

36. d
4
3 12 5625 81×10

= 3 4×3 5×5×5×5×3×3 900×900

= 3 4×3 5×5×5×5×3×3×900

3 4×3×5×5×3×30
=

= 3 2 2 2×3 3×3 5×5×5
2 3 5 30.
× ×
×
×
= × × =

37. b
3
3
2

729

9

=
3
3
2 2

9 9 9

(3 )

× ×
= 9
3 = 3.

38. b
3
3
3
24
9
2 2 2 3 3 3
×
=
× × × × ×
= 6.

39. a

3
3
3
3
512
8
8
0.08.
1000000
100
100
=
=
=

40. b
3 36
9
16
×
×
=
3
3 36
9
4
36
6
6.
×
×
=
×
=

41. c
We know that
a3 + b3 + c3 – 3abc
= (a + b + c) (a2 + b2 + c2 – ab – bc – ca)

= 1
2 (a + b + c) [(a – b)2 + (b – c)2 + (c – a)2]

∴
3
3
3
3 (333)
(333)
(334)
3
333
333
334
+
+
−
×
×
×

=

+
+
−
+
−

+
−

2
2

3
2

1(333
333
334) [(333
333)
(333
334)
2

(334
333) ]

=
3
3 1
1000
2
1000
2 ×
×
=
=
×
×
=
3 10
10
10
10

42. c
Here,
=
3 175616
56

∴
=
3 175.616
5.6

and
=
3 0.175616
0.56
∴ Required sum
= 5.6 + 0.56 + 0.056 = 6.216

43. d
=
3
3
0.000064
0.008

=
×
×
3 0.2
0.2
0.2 = 0.2

44. b
Expression

=
+
+
3 15612
154
225

=
+
+
3 15612
154
15

=
+
3 15612
13

=
=
3 15625
25

45. b
Here, 3 571787
83
=

∴ 3 571.787
8.3
=

and 3 0.571787
0.83
=
∴ Required sum
= 8.3 + 0.83 + 0.083 = 9.213

46. c
Since 338
2
473
3
≈

2
338
27
338
356
3
473
473
27
500

+
×
∴
=
=
+
= 0.712 i.e. 71.2%.

QA / Exercise - 2
CEX-5302/P1BS/17 / Page 4

47. a
Since 1
14.28%
7 =

∴

3
14.28
3
42.84%
7 =
×
=
.

48. c
Since 443
1
898
2
≈

Therefore,

1
443
102
443
494
2
898
898
102
1000

+
×
=
=
+

= 0.494 i.e, 49.4%.

49. d
Since 1
5.26%
19 ≈

8
8
5.26
42.08%.
19
∴
=
×
=

50. b
Since 547
3
1973
11
≈

Therefore,

3
547
27
547
547
7.36
11
1973
1973
27
2000

+
×
+
=
=
+

554.36
0.27718
27.72%
2000
=
=
≈
.

51. b
3
625
1331
+
=
3
5 5 5 5
11 11 11
× × ×
+
×
×

= 5 × 5 + 11 = 36.

52. a
243 = 3 × 3 × 3 × 3 × 3
Hence, the number is 3 and the resultant number will
be 243 × 3 = 729 = (27)2 = (9)3.

53. b
Let, 32 = a
79 = b, – 111 = c
When (a + b + c) = 0
then a3 + b3 + c3 – 3abc = 0
Here, a + b + c = 32 + 79 – 111 = 0
∴ (32)3 + (79)3 – (111)3 + 3 × 32 × 79 × 111 = 0

54. b
If a + b + c = 0, then
a3 + b3 + c3 = 3abc
Here, 0.111 + 0.222 + (– 0.333) = 0
∴ (0.111)3 + (0.222)3 + (– 0.333)3

= – 3 × 0.111 × 0.222 × 0.333
= – (0.333)2 × 0.222
∴ Expression
= [–(0.333)2 × 0.222 + (0.333)2 × 0.222]3 = 0

55. d
Expression
= (0.98)3 + (0.02)3 + 3 × 0.98 × 0.02 – 1
= (0.98)3 + (0.02)3 + 3 × 0.98 × 0.02 (0.98 + 0.02) – 1
= (0.98 + 0.02)3 – 1 = 1 – 1 = 0

56. d
Let 0.0347 = a
and, 0.9653 = b

∴ Expression =

3
3

2
2
a
b

a
ab
b

+

−
+

=
+
−
+

−
+

2
2

2
2
(a
b)(a
ab
b )

a
ab
b
= a + b

= 0.0347 + 0.9653 = 1

57. d
Expression

=

−

+
×
+

3
3

2
2
(3.2)
(0.2)

(3.2)
3.2
0.2
(0.2)

Let 3.2 = a and 0.2 = b

∴ Expression =
−

+
+

3
3

2
2
a
b

a
ab
b

=
−
+
+

+
+

2
2

2
2
(a
b)(a
ab
b )

a
ab
b
= a – b

= 3.2 – 0.2 = 3

58. c
Expression

=

3
2
2
(0.75)
(1
0.75)[0.75
0.75
1 1 ]
1
0.75
+
−
+
× +
−

=
+
−
3
3
3
(0.75)
1
(0.75)
0.25

=
=
=
1
100
4
0.25
25

∴ Required square root

=
=
4
2

59. a
Multiples of 11 whose square root are whole number
First = 11 × 11 = 121
Second = 11 × 11 × 4 = 484

60. c
Let the number be x. Then,
x2 = (75.15)2 – (60.12)2

= (75.15 + 60.12)(75.15 – 60.12)
= 135.27 × 15.03
= 2033.1081

⇒ x =
2033.1081 = 45.09

QA / Exercise - 3
CEX-5303/P1BS/17 / Page 1

Quantitative Aptitude – 3
Approximation and VBODMAS

Answers and Explanations

1. b
0.2 + 2 × (0.02)2 – 3 × (0.2)3

= 0.2 + 2 × 0.0004 – 3 × 0.008
= 0.2 + 0.0008 – 0.024 = 0.2008 – 0.024 = 0.1768.

2. c

1
1
5
0.075
1
0.3
5
4
6
0.001
5
3




×
÷
+
×
−
×
×









13
5
0.075
(1.5
5)
6
0.001
3
=
×
÷
×
−
×
×

13
5
0.075
7.5
6
0.001
3
=
×
÷
−
×
×

13
5
0.01
6
0.001
3
=
×
−
×
×
= 0.05 – 0.026 = 0.024.

3. a

1
6
3
1
of
2
5
4
2
3
3
24
8
3
8

×
÷

÷
×
+
−

1
6
3
2
5
8
3
3
32
3
8

×
÷
=
÷
×
−

1
16
8
8
2
5
5
.
1
4
3
5
32
3
8

×
=
=
=
−
×
−

4. d
2
3
2
11
11
11
11 11of 11
+
÷
−
+

=
2
3
2
2
11
11
11
11
11
+
÷
−
+
= 112 + 11 – 11 + 112

= 2 × 112 = 242.

5. b
17
4
3
7
2
5
20
5
20
−
+
×
−
÷

17
4
3
5
5
20
5
20
=
−
+
×
÷

17
4
3
17
16
3
20
5
20
20
−
+
=
−
+
=
4
1.
20
5
=
=

6. a
36
4
8 of 2
12
2 of 3
÷
×
+
−

= 36
4 16
12
6
÷
×
+
−
= 9 × 16 + 12 – 6
= 144 + 12 – 6 = 150.

7. c
{
}
8
7
4
2
5
(6
2)
4 of 2


×
−
+
÷
×
+
−



=
[
]
8
5
5
8
4 of 2
×
÷
×
−

=
[
]
8
5
5
8
8
×
÷
×
−

=
[
]
8
1 8
8
8
(8
8)
×
×
−
=
×
−

= 8 × 0 = 0.

8. b
4
2
3
2
6
2
17
11
8
6
3
3

−
−
−
−
−

4
2
4
3
2
8
6
18
2
1
24
17
18
11
7
7
3
3
3

−
−
−

=
−
=
−
−
−

18
18
18
18
36.
7
7
7
7
7
−
=
−
=
+
=

9. c
[
]
5.4
4.8
3.2
1.6
5.7
3.2 3
2
1.6
5.7
1.8
2.4


−
−
×
=
−
−
×




= 3.2 – 1.6 × 5.7
= 1.6(2 – 5.7)
= 1.6 × (–3.7) = –5.92.

10. d

18
30
15
27
18
30
15
27

8
8
27
54
2 2
2 2
3 3
3 6





÷
÷
×
=
÷
÷
×









9
15
5
9

2
2
3
6



=
÷
÷
×





9
15 3
9
9
27
2
2
5 2
6
2



=
÷
×
=
÷







9
2
2 .
27
3
2
=
×
=

1
b
2
c
3
a
4
d
5
b
6
a
7
c
8
b
9
c
10
d
11
b
12
a
13
c, d
14
c
15
b
16
c
17
b
18
d
19
d
20
b
21
c
22
c
23
a
24
d
25
c
26
a
27
a
28
c
29
a
30
d
31
d
32
d
33
b
34
a
35
b
36
b
37
d
38
a
39
a
40
b
41
c
42
b
43
b
44
d
45
a
46
b
47
d
48
b
49
a
50
d
51
d
52
a
53
d
54
c
55
a
56
c
57
b
58
d
59
b
60
c

P-1 (BS)

QA / Exercise - 3
CEX-5303/P1BS/17 / Page 2

11. b

1
5
1
1
5
1
of
of
1
1
3
7
3
7
4
4
3
5
3
1
5
5

−
=
−
−
−
−
−

1
5
1
1
5
1
of
of
5
8
5
3
7
3
7
4
2
2

=
−
=
−
−
−

1
5
2
1
10
of
3
7
3
3
21
=
−
=
−

7
10
3
1.
21
21
7
−
−
=
=
= −

12. a

5
9
2
1
3
2
1
2
1
2
4
3
4
5
5






+
−
÷
÷
−
−













30
27
8
5
3
12
5
12
4
5


+
−
−
−




=
÷
÷













=

49
4
14
49
5
12
5
5
15
14




×
÷ −
= −
×









7.
6
= −

13. (i) c 8888 + 888 + 88 + 8 = 9872.
(ii) d 11111 – 1111 + 111 – 11 + 1 =10101.

14. c
5
6
3
4
1
4
9
×
−
÷
+
÷
+

5
3
4
1
4
9
=
×
÷
+ ÷
+

3
1
15
1
36
5
9
4
4
4
+ +
=
×
+
+
=

52
13.
4
=
=

15. b
2
3125
25 of 25
5
÷
−

2
3125
625
5
=
÷
−

5
25
20.
=
−
= −

16. c
9.5
6.8
9.5
3.2
1.9
5
1.9
4
×
+
×
×
−
×

9.5(6.8
3.2)
5
10
50.
1.9(5
4)
+
=
=
×
=
−

17. b

3
3

2
2
a
b
a
b
a
b
ab

+
=
+
+
−
= 0.73 + 0.27 = 1.

Here, a = 0.73 and b = 0.27.

18. d
5.76
5.9
0.576
41
2.4
3
24
0.2
×
+
×
×
−
×

5.76(5.9
4.1)
2.4
10
24.
2.4(3
2)
1
+
×
=
=
=
−

19. d
(a) 502 – 442 = (50 – 44) (50 + 44)
= 6 × 94 (True)
(b) (a + b + c)2

= a2 + b2 + c2 + 2ab + 2bc + 2ac (True)
(c) (a + b)2 – (a – b)2 = 4ab (True)

(d)

2
3
2
2


+




= 4 + 9
4 + 6 (False)

20. b

(
)
(
)

2
2
1
1
1
1
1
1
6
of
6
of
3
12
18
3
12
18
1
1
of 216
12
8
8
of 18
8
8
13
13

×
÷
×
÷
=
÷
+
−
+
−

2
2
1
1
1
1
1
6
of
6
3
12
18
36
18
1
2
8
of 26
8
13

×
÷
×
÷
=
=
−
−

1
36
18
18
2
3.
2
8
2
8
6

×
=
=
=
= −
−
−
−

21. c

1
1 of 3
(?)
3 – 5
3
18
–4
1
18
(2
7) of
8
2

÷
×
+
=
÷
+
−

1
1
(?)
3 – 5
3
6
–4
9
18
8
2

÷
×
+
⇒
=
÷
−

2
(?)
3 – 5
–4
4 – 8
×
+
⇒
=
2(?) – 2
–4
–4
⇒
=

2(?)
18
⇒
=

(?)
9.
⇒
=

22. c

1
1
7
1
7 of
9
6
18
9
6
18
24
8
24
8
1
1
1
1
1
7
14
2
7
14
2
of
8
24
7
8
24
7

÷
÷
+
÷
+
÷
−
=
−
÷
×
÷
×
÷
÷
×

1
7
9
3
3
1
24
7
2
2
8

+
=
−
×
×

1
7
1
9
3
3
21
=
+
−
×

1
1
81 3
1
83
9
.
3
9
9
9
+
−
=
+
−
=
=

23. a
0.66 × 0.66 + 0.28 × 0.28 + 0.06 × 0.06 + 2(0.66 × 0.28
+ 0.66 × 0.06 + 0.28 × 0.06)
= (0.66 + 0.28 + 0.06)2 = (1.00)2 = 1.

QA / Exercise - 3
CEX-5303/P1BS/17 / Page 3

24. d
(
)

(
)

343
17
7
3
24 of 2
24
1
45
6
9 of
4

÷
+
×
−
−

÷
+

(
)
343
38
24 of 2
24
1
45
15 of
4

÷
−
−
=
÷

49
24
343
28
24
4
15
4
45
45
4
15

−
÷
−
=
=
÷
×

49
96
47
47
4
4
.
3
4
12
48
1
1

−
−
=
=
= −
×

25. c

3
3

2
3
a
b
a
b
a
ab
b

−
=
−
+
+
�

Here, a = 0.03 and b = 0.01
∴ a – b = 0.03 – 0.01 = 0.02
∴ (?) = ab = (0.03)(0.01) = 0.0003.

26. a

1
1
4
4
2
2
2
2
2
2
1
6
1
2
3
3

=
−
−
−
−
−
−

1
1
4
4
2
2
6
10
6
2
5
5

=
=
−
−
−
−

1
1.
2
5
3
=
= −
−

For questions 27 to 30: After replacing the signs with
respective meanings the expression becomes:

27. a
13 × 26 ÷ 78 – 143 + 156

= 13
26
13
78
×
+

13
3
3
=
+
= 52.
3

28. c
1001 ÷ 78 × 36 ÷ 77 × 3

= 1001
36
3
78
77
×
×

= 7
11 13
6
6
3
13
6
11 7
×
×
×
×
×
×
×
×
= 18.

29. a
39
13
16
3
8
17
34
5
÷
×
+

= 39
17
16
3
8
13
34
5
×
×
+
=
3
3
5
+
= 18.
5

30. d
(a + b)2 – (a – b)2 = 4ab.

31. d
5
9
2
1
3
2
1
2
1
2
4
3
4
5
5







+
−
÷
÷ −
−
−













30
27
8
5
3
12
–
1
12
4
5
5



+
−




=
÷
÷
−
−













49
5
–3
12
5
12
4
5
−
−




=
÷
÷








49
4
20
12
5
5
−




=
×
÷ 








(
)
49
49
1
49
4
.
15
15
4
60
=
÷ −
= −
×
= −

32. d

1
1
3
1
3
4
of
of
of 13
of (?)
2
2
2
2
4
3




+
=
+









1
1
3
13
9
16
of
of
of (?)
2
2
2
2
12
+




⇒
+
=









1
1
25
of
of 8
of (?)
2
2
12
⇒
=

1
1
25(?)
8
2
2
12
⇒
×
×
=

⇒ 2 × 12 = 25(?)

24
(?)
.
25
⇒
=

33. b
152 – 132 + 172 = (?)2 – 112 – 110
⇒ 225 – 169 + 289 = (?)2 – 121 – 110

⇒ (?)2 = 225 + 289 + 121 + 110 – 169

⇒ (?)2 = 745 – 169

⇒ (?)2 = 576 = 242 ⇒ (?) = 24.

34. a
(a) 173 – 153 – 1535
= (17 – 15) (172 + 17 × 15 + 152) – 1535
= 2[289 + 255 + 225] – 1535
= 1538 – 1535 = 3.

(b)

2
6
6
16
3
6
×

8
6
6
6
2
3
4.
2
3
=
×
=
×

(c) 1889 – 5 × 13 × 29 = 1889 – 1885 = 4.

(d)
1
1
3
3
1
2
3
2
2
1
3

−
=
−
−
−
−

= 3 + 1 = 4.

QA / Exercise - 3
CEX-5303/P1BS/17 / Page 4

35. b
(
)

1
1 of 10
7
8
5
10
5
1
1
1
1
3
of 10
of
3
10
20
10
1
1 of 10
15
5
10
5
1
1
1
1
3
of 10
of
3
10
20
10

÷
+
÷
×
×
÷

÷
÷
=
×
×
÷

1
1
1
1
2
15
15
5
10
5
10
2
10
1
1
1
10
1
1
3
3
10
3
10
20
10
3
10
20

÷
×
×
÷
=
×
=
×
×
×
÷
×
×
×

1
3
3
1
20
20
0.3.
1
10
10
20
1
20

=
×
=
×
×
=

36. b
3
25
2
7
0.35
5
12
3
3
1000


×
÷
×
−
×
×





1
7
0.35
0.2
5
=
×
×
−

= 7 × 0.07 – 0.2
= 0.49 – 0.2 = 0.29.

37. d

3
69
1
17
1
1
(?)
1
1
5

=
−
+
+

3
69
1
17
1
5
(?)
6

⇒
=
−
+

3
69
6
17
1
6(?)
5

⇒
=
−
+

17
6
1
23
6(?)
5
⇒
=
−
+

6
6
6(?)
5
23
⇒
=
+

⇒ 6(?) + 5 = 23

⇒ 6(?) = 18

⇒ (?) = 3.

38. a
1 + 0.1 + 0.01 + 0.001 = 2 + 0.2 × x
⇒ 1.111 = 2 + 0.2x

⇒ 0.2x = –0.889

0.889
x
0.2
⇒
= −
⇒ x = – 4.445

x
–4.5.
⇒
≈

39. a

3
3

2
2
a
b
a
b
a
ab
b

+
=
+
−
+
= 0.05 + 0.01 = 0.06.

40. b

0.4
6.4
0.0004
0.0289
11
1
25

×
+
+

2.56
0.0004
0.0289
36
25

+
=
+

1.6
0.02
1.62
5
0.17
0.17
6
6
5

+
×
=
+
=
+

8.10
1.02
9.12
1.52.
6
6
+
=
=
=

41. c
(
)
1
1
3
of 20
11
9
5
of 6
7
5
3


+
−
+
−
+
−





1
1
3
of 20
15
of 6
7
5
3


=
+
−
+
−





[
]
1
3
of 20
15
2
7
5
=
+
−
+
−
1
3
of 7
7
5
=
+
−

7
15
7
35
13
3
7
2.6.
5
5
5
+
−
=
+
−
=
= −
= −

42. b

729
729
999
48
27
999
16
27
27
9
−
×
−
=
−
×
−

= 999 – 81 × 16 – 27 = 999 – 1296 – 27
= 999 – 1323 = –324.

43. b
359
x
1059
360
x
1060
1
x
3
x
0.33.

=

⇒
≈

⇒
≈

⇒
≈

44. d
x2 + 0.1 = 0.776
⇒ x2 = 0.676

x
0.8.
⇒
≈

45. a
35.56 × 41.29 = x × 53.2

35
41
x
53
×
⇒
=
(By approximation)

x
27.6.
⇒
≈

QA / Exercise - 3
CEX-5303/P1BS/17 / Page 5

46. b
33.33 + 444.4 + 5.555 = 66.6 × x

66.6
x
483.285
⇒
×
=

483.285
x
66.6
⇒
=

x
7.25.
⇒
≈

47. d
987654321 123456789
x
9876543210
+
=

987654321
123456789
x
9876543210
9876543210
⇒
+
=

0.1
0.01
x
⇒
+
≈
x
0.11.
⇒
≈

48. b
0.0961
x
0.0529
x
×
=

2
0.0961
961
x
0.0529
529
⇒
=
=

31
x
23
⇒
=
x
1.35.
⇒
≈

49. a
3.5
x
0.23
2.4
6.9
×
=

0.23
6.9
2.4
x
3.5
×
×
⇒
=

3.8088
x
3.5
⇒
=
x
1.08.
⇒
≈

50. d
3569 × 2387 × 5389 = x

3
3
3
x
3.6
10
2.4
10
5.4
10
⇒
≈
×
×
×
×
×

9
x
46.656
10
⇒
≈
×

10
x
5 10 .
⇒
≈
×

51. d
? =
+
+
+
+

1
1
2
1
3
2
4
1
5

=
+
+
×
+
+

1
1
2
1
3
5
2
5
4

=
+
+
+

1
1
2
1
5
2
3

=
+
×
+
+

1
1
2
3
1
6
5

=
×
+
+
1 11
1
11
6

=
+ 11
1
17 =
11
117

52. a
? =
+
=
×
+
+

2
2
1
1
3
5
5
1
1
9
3

=
×
+
=
2
3
7
1
8
4

53. d
+
+
−

1
17
1
22
3
1
2
7
9

=
+
+
−

1
17
1
22
3
9
2
7

=
+
+
−

1
17
1
22
3
14
9
7

=
+
=
+
+
+

1
17
1
17
1
7
22
22
3
3
5
5
7

=
+
+
1
17
15
7
22
5

= 5
17
22
22
22
22
+
=
= 1

54. c
x =
+
+
+
+

1
1
1
1
1
1
1
1
2

1
1
1
1
2
1
3

=
+
+
+

=
1
1
3
1
5

+
+

=
+ 1
1
8
5

=
+ 5
1
8 = 13
8

∴
+ 7
2x
4 =
×
+
13
7
2
8
4

=
+
13
7
4
= 20
4 = 5

QA / Exercise - 3
CEX-5303/P1BS/17 / Page 6

55. a
÷
+
+
+

19
2
1
43
2
1
3
1
1
4

=
÷
+
+

19
2
1
43
2
4
3
5

=
÷
+

19
1
5
43
2
19

=
÷
19
19
43
43

=
×
19
43
43
19 = 1

56. c

+
−

5
3
3
3
2
3

=

+

5
3
3
1
3

=
+
×
5
3
3
3 =
+
5
3
9 = 5
12

57. b
2 =
+
+
+

1
x
1
1
1
3
4

⇒ 2 =
+
+
+

1
x
1
1
12
1
4

⇒ 2 =
+
+

1
x
4
1
13

⇒ 2 =
+
+
1
x
13
4
13

⇒ 2 =
+ 1
x
17
13

⇒ 2 =
+ 13
x
17 ⇒ x =
− 13
2
17

=
−
34
13
17
= 21
17

58. d
× 

+
×
÷





2
3
1
5
3
5
1
1
6
2
4
2

=
×
+
÷

2
3
5
5
1
2
4
4

=
×
×

2
3
5
4
3
4
5

=
×
2
3
3
= 2

59. b
+
−
×
+
−

4
1
1
5
3
2
2
10
1
2

+
−
+

4
5
1
6
2
2
9

=
+
−
+

4
5
1
2
2
2
3

=
+
−
4
5
1
8
2
3

=
×
+
−
4
3
5
1
8
2

=
+
−
3
5
1
2
2 =
+
−
2
3
5
2
= 0

60. c
Suppose that

1
1
1
10
10

+
+
= 111
101 = a

and,
−
+

1
1
1
10
10

= 91
101 = b

=
−
+

2
2
a
b
(a
b) =

+
−
+
(a
b)(a
b)
(a
b)

= (a – b)

=
−
111
91
101
101 = 20
101

QA / Exercise - 4
CEX-5304/P1BS/17 / Page 1

Quantitative Aptitude – 4
Numbers : Properties, Surds and Indices

Answers and Explanations

1. c
Prime numbers between 60 and 75 are 61, 67, 71 and
73.
∴ Sum = 61 + 67 + 71 + 73 = 272.

2. b
6
16
0.86;
0.84;
7
19
=
=

19
11
7
0.90;
0.78 and
0.7.
21
14
10
=
=
=

∴ Descending order is 19
6
16
11
7 .
21
7
19
14
10
>
>
>
>

3. b
78
67
1.13;
1.15;
69
58
=
=

25
41
1.56 and
1.28.
16
32
=
=

∴ Ascending order is 78
67
41
25.
69
58
32
16
<
<
<

4. c
Sum of first n odd number = n2 .
Hence, sum of first 38 odd number = (38)2 = 1444.

5. c
Whole number.

6. d
323 = 17 × 19
Hence, it is a composite number.

7. c
2 is the smallest prime number.

8. c
Since, the difference between the numerator and
denominator of all the five fractions is a constant i.e.
13, then the value of the fraction with the biggest
numerator will be the highest.

121
92
87
76
51.
134
105
100
89
64
∴−
< −
< −
< −
< −

9. b
Dividing Numerator and Denominator by 73

1095
15.
1168
16
=

10. b
180
80
20
=
+
, when we simplify LHS

20
80
20
2 20
3 20
9 20
180.
+
=
+
=
=
×
=

11. b
Number = 35 × 12 = 420

∴ Correct quotient
420
20.
21
=
=

12. d
1 × 2 × 3 × 4 = 24; 2 × 3 × 4 × 5 = 120; 3 × 4 × 5 × 6
= 360 and so on.
Hence, the required number = 24.

13. a
6
5
10
LCM of (6, 5, 10)
30
LCM of
,
,
.
7 14
21
HCF of (7, 14, 21)
7

 =
=





14. a
1000 + 1 = 1001 > 1000
Hence, number is 1.

15. d
Odd numbers =

71 1
36
2
+
=

Even numbers =

71 1
35.
2
−
=

16. c
F
0.84181
=

84181
841
83340
99000
99000
−
=
=
463
550
=

∴ Denominator – Numerator = 550 – 463 = 87.

1
c
2
b
3
b
4
c
5
c
6
d
7
c
8
c
9
b
10
b
11
b
12
d
13
a
14
a
15
d
16
c
17
d
18
c
19
b
20
d
21
b
22
b
23
d
24
a
25
d
26
c
27
d
28
c
29
c
30
b
31
a
32
c
33
a
34
a
35
a
36
b
37
a
38
c
39
d
40
a
41
b
42
a
43
a
44
d
45
c
46
b
47
b
48
a
49
c
50
b
51
c
52
c
53
c
54
b
55
a
56
d
57
c
58
d
59
b
60
a
61
d
62
c
63
b
64
a
65
a
66
d
67
a
68
d
69
d
70
c

P-1 (BS)