Answers:

1. Option A

2. Option D

3. Option C

4. Option D

5. Option B

6. Option A

7. Option B

8. Option D

9. Option C

10. Option A

11. Option C

12. Option B

13. Option E

14. Option B

15. Option B

16. Option C

17. Option B

18. Option C

19. Option D

20. Option C

21. Option E

22. Option A

23. Option B

24. Option D

25. Option C

26. Option E

27. Option B

28. Option D

29. Option C

30. Option A

31. Option D

E X A M I N A T I O N

1st 3rd 6th 8th

Meaningful words = NEAT, ANTE

32. Option C

33. Option D

Rakesh>Mukesh> Suresh

Amar >Rakesh> Harish

The relation between Harish and Suresh cannot be established from the given information.

So, it is not possible to find out the shortest person.

Rules

(i) Odd number + Composite odd number

(ii) Even number + Odd number

(iii) Even Number → Perfect square number, then Perfect square number – Even number

(iv) Odd number ÷ Prime odd number

(v) Odd number – Even number

34. Option A

1st row 15 8 21 → 15 – 8 = 7 (rule V)

7 + 21 = 28 (rule i)

= p

2nd row p 3 27 → 28 3 27

28 + 3 = 31 (rule ii)

31 + 27 = 58 (rule i)

Resultant of 2nd row = 58

35. Option E

1st row 12 64 17 → 64 – 12 = 52 (rule iii)

52 + 17 = 69 (rule ii)

= m

2nd row 20 m 16 → 20 69 16

20 + 69 = 89 (rule ii)

89 – 16 = 73 (rule v)

Resultant of 2nd row = 73

36. Option B

1st row 85 17 35 → 85 ÷ 17 = 5 (rule iv)

5 + 35 = 40 (rule i)

= r

2nd row 16 19 r → 16 19 40

16 + 19 = 35 (rule ii)

35 – 40 = − 5 (rule v)

Resultant of second row = − 5

37. Option C

1st row 24 15 3 → 24 + 15 = 39 (rule ii)

39 ÷ 3 = 13 (rule iv)

= d

2nd row d 6 15 → 13 6 15

13 – 6 = 7 (rule v)

7 + 15 = 22 (rule i)

Resultant of 2nd row = 22

38. Option D

1st row 28 49 15 → 49 – 28 = 21 (rule iii)

21 + 15 = 36 (rule i)

= h

2nd row h 3 12 → 36 3 12

36 + 3 = 39 (rule ii)

39 – 12 = 27 (rule v)

Resultant of 2nd row = 27

39. Option A

1st row 36 15 3 → 36 + 15 = 51 (rule ii)

51 ÷ 3 = 17 (rule iv)

= n

2nd row 12 3 n → 12 3 17

12 + 3 = 15 (rule ii)

15 ÷ 17 = 15/17 (rule iv)

Resultant of 2nd row = 15/17

40. Option C

41. Option D

42. Option E

43. Option A

44. Option C

45. Option C

Time of ringing bell = (7:45 – 0:45) = 7:00 am

But it happened 5 min. before the priest gave the information to the devotees.

Time of giving information = 7:00 + 0:05 = 7:05 am

46. Option C

Day before yesterday = Saturday

Yesterday = Saturday + 1 = Sunday

Today = Sunday + 1 = Monday

Tomorrow = Monday + 1 = Tuesday

Day after tomorrow = Tuesday + 1 = Wednesday

47. Option A

48. Option B

49. Option A

50. Option E

51. Option A

52. Option C

Candidates (i) (ii) (iii)(a) (iv)(b) (v)

Saurav True True False True True

Jagat True True True True True

Sudesh True True True True True

Mohd. Ghous True False — — —

Nimesh True True True True True

Sushil True True True True

Mohan True True True True True

Francis

D’Costa

True True True True True

Sukhdev True True True ? True

Neeraj True True True — —

53. Option B

54. Option E

55. Option D

56. Option B

57. Option E

58. Option C

59. Option A

60. Option E

61. Option C

62. Option D

63. Option D

64. Option D

65. Option E

66. Option C

L = 45 H and L + H = 1150

46 H = 1150

H = 25

Let x be other number, L × H = Product of two numbers

(45 × 25) × 25 = x × 125

x = 225

67. Option D

68. Option B

√1.1 + √1100 + √0.011

+ √11 × 100 +

. + 3.313 × 10 + .

1.0488 + 33.16 + .10488

= 34.31

69. Option D

5 years ago, father’s age = 40 years

Son’s age = 40 – 20 = 25

After 5 years, sum = 40 + 20 + 10 + 10 = 80

70. Option C

Let numerator = x, denominator = x + 3

= 2

x = 5, fraction = , sum = 13

71. Option A

Investment of last person i.e. Raghu = Rs.100

Mohit = 100 – 10 = 90, Pradeep = 1.2 × 90 = 108

Total investment = 100 + 90 + 108 = 298, then investment of Raghu = Rs.100

When total investment = 17880, investment of Raghu = × 17880 = Rs.6000

72. Option B

CP =

/

= 160

% profit = × 100 = 10%

73. Option C

Let the even numbers are 2x, 2x + 2 …. 2x + 8

10x + 20 = 66 × 5

x = 31

(2x + 2) (2x + 8) = 64 × 70 = 4480

74. Option D

W : S = 2 : 1, S : D = 2 : 1

W : S : D = 4 : 2 : 1

2x = 4500

There are 2 sons, so ratio distribution is 4 : 2 × 2 : 1 = 4 : 4 : 1

9x = × 9 = Rs.20250

75. Option B

C – (A + B) = 1500

7x – (2x + 3x) = 1500

x = 750

A’s share = 2x = Rs.1500

76. Option D

% = 40%

77. Option B

75% × 40 + x% × 35 = 80% × 75

x = 86

78. Option B

x = 26 × 17 = 13 × x

x = 34, 34 – 26 = 8

79. Option A

T = ( ) = × = 90 hours

80. Option B

× × = × ×

=

= Rs.8000

81. Option A

Interest difference = 1597.20 – 1452 = 145.20 for 1 year

R = × .

×

= 10%

82. Option A

3 × (8 – 5) % 1100 + 3 × 8% of x = Rs.173

x = 300

83. Option D

(2 × 40 + (n – 1) 5 = 385

n = 7

84. Option B

t =

× =

+ 5 = 50

= 45 km/h

85. Option A

Using simple logic of relative velocity, current opposes the man, s

Time =

t = = 3 hours

86. Option B

Total number of students from all the colleges together = 2040 + 2300 + 2400 + 2200 + 2090

+ 2120 = 13150

87. Option D

Required percentage growth = ×

= × = 3.21

88. Option C

Required percentage = ×

= × = 18.43% = 18%

89. Option A

Required ratio = 2250 + 2480 : 2260 + 2440

= 4730 : 4700

= 473 : 470

90. Option E

Required average number =

= = 2278.33 = 2278

91. Option D- in 1999, average employment per factory is greater than 1998 and less than 2000.

92. Option B- in 1996, for 20100 factories , employees = 13,245,900; average = 609, total employees = 18,361,350. So average for remaining factories = 509.

93. Option A- total employment in 1999= 36240 x 736.

Total employment in 1996 = 609 x 30150,

Ratio = (36240×736)/(30150×609)=1.45

94. Option D- From the figure it can be seen that the maximum increase in the number of factories is in 1995-2000.

95. Option D- we do not have year wise data in between 1975- 1980 & similarly between 1980-1985,etc.

96. Option C

In each ordinary years no of odd day = 1

So, 10th June, 2001 ____ 10th June, 2002 ____ 10th June, 2003

2004 is a leap year, in a leap year, no. of odd days = 2

10th June 2004 = Thursday

97. Option C

Le the lengths are 3x, 4x then A’s speed =

B’s speed = ratio = × =

98. Option C

= l × b = (a + 5) (a – 3)

= + 2a – 15

a =

Perimeter = 2 (a + 5 + a – 3) = 2 (2a + 2) = 34

99. Option A

100. Option D

No. of ways = 3 × 2 × 5 = 3240