General Mental Abilities Test (GMAT) Detailed solutions

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General Mental Abilities Test (GMAT)

Detailed solutions

NTSE Stage – II

Exam conducted on 14.02.2021

1. (3)

M E N T A L

13 5 14 20 1 12

Sum of position numbers = 13 + 5 + 14 + 20 + 1 + 12

= 65

Multiply with number of letters 65 × 6 = 390

C O M P E T E N C Y

3 15 13 16 5 20 5 14 3 25

3 + 15 + 13 + 16 + 5 + 20 + 5 + 14 + 3 + 25 = 119

119 × 10 = 1190

2. (3)

22 + 12 + 42 = 4 + 1 + 16 = 21

82 + 32 + 52 = 64 + 9 + 25 = 98

62 + 72 + 32 = 36 + 49 + 9 = 94

Missing number is 7

Solution for questions 3 to 4

3. (1)

8 small cubes do not have coloured faces

4. (3)

There are 3 × 4 = 12 cubes with red and yellow colour faces (out of which 8 cubes have black colour

face also)

5. (1)

Code of “project work” is “34”

From Statement I.

Code of “completed project on time” is “173”

Then implies that “project” code is “3”

Hence work’s code is “4”

Statement I alone is sufficient.

Red

Yellow

Black

NTSE Stage II 2020 (GMAT) solution

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6. (2)

All tortoises like to jump

So, tortoises who like to fly also like to jump

7. (4)

The angle at 3 o’clock is 90°

When the angle changes by 99°, it becomes 9° and the minute hand will be ahead of the hour hand.

Angle changes by 5.5° in 1 minute

∴ Time required to change the angle by 99° 1

99 185.5

= × = minutes.

So, at 3 : 18, the angle is 9°

8. (4)

Letters in the rhombus and numbers in square are paired as under.

2H, 1F, 3I and 5J.

Number are replaced by letters having same position number. Letters are replaced by their position

numbers.

2H → B 8 ; 1 F → A 6

3 I → C 9; 5 J → E 10

9. (2)

Numbers on the dial are replaced as under

1 = C; 2 = D; 3 = G; 4 = H; 5 = K; 6 = L

7 = M; 8 = N; 9 = Q; 10 = R; 11 = S; 12 = T

Class start at N : T means at 8 : 00 AM

Class closes at S : K means at 11 : 25 AM

Duration of class = 3 hr 25 min

Break = 7 + 9 + 11 + 13 = 40 min

Duration of 5 periods = 3 : 25 – 0 : 40

= 2 hrs 45 min.

Duration of each period = 165

335

= minutes

10. (1)

We have 5 letters in BOARD; so, either sum or difference should be 5.

B + C = 5

O – J = 5

A + D = 5

R – M = 5

D + A = 5

B O A R D

A M D J C

2 15 1 18 4

1 13 4 10 3


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11. (2)

Dancing Acting Singing Tabla Guitar

M � � � �

P � � � �

B � �

K � �

L �

V � �

Only P is good in dancing, singing and playing guitar

12. (2)

There are 72 squares

13. (4)

25 + 10 × 4 = 19 ⇒ 25 – 10 + 4 = 19

10 ÷ 3 – 3 = 10 ⇒ 10 × 3 ÷ 3 = 10

+ means –; × means +

÷ means ×; – means ÷

16 × 5 + 40 – 10 + 2

= 16 + 5 – 40 ÷ 10 × 2

= 16 + 5 – 4 × 2

= 25 – 8 = 13

14. (2)

‘>’ occurs four times nine symbols after second ‘>’ is $

Five symbols before $ is S.

15. (1)

Sum of numbers is 45 (Horizontally / Vertically / Diagonally)

16. (1)

Statements

T ≤ R; R ≥ M; M = D; D < H

D ≤ R is correct

17. (3)

Statements

M = B, B > N, N ≥ R, R < K

Conclusion II R < B (R © B) and

Conclusion IV N < M (N © M) are correct

Conclusion III M $ R implies that M ≥ R only.

So, only conclusion II and IV hold good since there no such option, the nearest answer (option 3) may be

taken.

18. (1)

19. (3)

20. (4)


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21. (2)

23rd April Monday

Number of days from 23rd April to 13th June is

Apr 7

May 31

June 13

51 = 2 odd days

Hence 13th June falls on Wednesday


Sun Mon Tue Wed Thu Fri Sat

Dr. Ashutosh 12 to 4 PM � � � � � � �

Dr. Dhanwantri 10 AM to 2 PM � � � � � �

Dr. Shehnaz 9 AM to 12 : 30 � � � �

Dr. Shehnaz 2 PM to 4 PM � � �

22. (2)

23. (3)

24. (3)

25. (3)

Percentage of candidates not qualified in all the subjects = 100 – 62 = 38%

From the Venn diagram, number of candidates not qualified

= 30 + 75 + 50 + 10 + 8 + 12 + 5

= 190

38% is equivalent to 190

∴ 100% 100

38= × 100 = 500

Number of candidates appeared = 500

Number of candidates not qualified at least in two subjects = 10 + 8 + 12 + 5 = 35

∴ Percentage 35

100= × 100 = 7

26. (2)

By analysis

27. (3)

50 + 37 = 41 + x

x = 87 − 41 = 46

28. (3)

(# − 13) × 6 + 8 = 44

⇒⇒⇒⇒ # = 19

29. (2)

By observation


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30. (1)

Number of students having A grade = 72

Number of students having D grade = 180

Number of students having C grade = 252

Number of students having B grade = 336

31. (3)

By analysis

32. (2)

2,3,5,7,11,13,17,19

Difference between above prime nos =

1,2,2,4,4,4,2 – which is the given series

33. (2)

By observation

34. (1) or (3)

By observation, since both Option (1) and (3) are the same.

35. (2)

52 − 4 × 5 + 15 ÷ 3 = 37

52 − 20 + 5 = 37

37 = 37

36. (3)

Number of students representing atleast two clubs = (Exactly two clubs) + (Exactly three clubs)

+ (Exactly four clubs)

= (14 + 36 + 16 + 24) + (8 + 11 + 10 + 5) + (12)

= 90 + 44 + 12

= 146

37. (3)

Number of boys in 2016 17 10

Number of girls in 2016 17 10

−=

−

Number of boys in 2017 18 28

Number of girls in 2017 18 20

−=

−

Number of boys in 2017 18 28 14

Number of boys in 2016 17 10 5

−= =

−

38. (2)

Distance travelled = 27 + 2(2.7) + 2(0.27) + 2(0.027) + ....

= 27 + 2(2.7 + 0.27 + 0.027 + ......)

= 2.7

27 21

110

+ −

= 27 + 2 × 0.3 × 10

= 33


6

39. (1)

X = (10 + 7)2 = (17)2 = 289

Y = 10 − 7 = 3

40. (2)

13 + 1 = 002, Sum of digits = 0+0+2 = 2 = B

23 + 1 = 009, Sum of digits = 0+0+9 = 9 = I

33 + 1 = 028, Sum of digits = 0+2+8 = 10 = J

43 + 1 = 065, Sum of digits = 0+6+5 = 11 = K

53 + 1 = 126, Sum of digits = 1+2+6 = 9 = I

41. (2)

By observation

42. (2)

16 (7B + 9G)

Karthik − 17th

43(28B + 15G)

G 15

B 28=

43. (4)

Sum of digits of LHS = RHS => 9876 : 12234567 = 30

Sum of digits of LHS = RHS => 9993 : 8886 =30

44. (4)

By observation

45. (4)

By deduction, we get L = 2, = 8 and = 16

Option (1) and (2) are ruled out as they have got square + pentagon + other figures (which is 24)

Option (3) total = 8 + 7 = 15 so ruled out

Option (4) first 3 figures + pentagon = 7 + 16 = 23

46. (1)

By observation

47. (4)

By observation

48. (4)

By observation

49. (4)

By observation

50. (1)

12 × 13 + 15 ÷ 5 = 180 − 21

156 + 3 = 159

159 = 159

51. (4)

52. (4)

53. (4)


7

54. (2)

55. (3)

56. (3)

3 3nn

2+

3 44 3

2= × ×

64 + 6 = 70

57. (4)

x + 63 = 53 + 50

⇒ x = 103 − 63 = 40

102 + y = 140 + 118

y = 258 − 102 = 156

58. (2)

(7 + 3 + 14 + 10 + 5 + 9) = 24 × 2

(1 + 6 + 8 + 18 + 16 + 9 + 7 + 3) = 34 × 2

(5 + 16 + 12 + 7) = 20 × 2

59. BONUS – None of the Options hold good

60. (3)

Let Ashwinder’s present age be x years

Ayaz’s age = Ashwinder’s age + 8

= x + 8

Aman’s present age = Ayaz’s age + 6

= x + 8 + 6

= x + 14

Given that,

Sum of Aman’s and Ayaz’s present age = 5(Ashwinder’s age 4 years ago)

= x + 14 + x + 8 = 5(x − 4)

2x + 22 = 5x − 20

2x − 5x = − 20 − 22

− 3x = − 42

42

x 143

+= =

+

x = 14

∴ Ashwinder’s present age = 14 years

Aman’s present age = x + 14

= 14 + 14 = 28 years


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61. (3)

+3 +2 −4 +7 +3 +2 −4 +7 +8 −4 +5 −7

B F I D L : N R U P X : : A J E K C : MVQWO

62. (3)

Number of female between below 60 years and below 18 years

1150 − 220 = 930

Total population = 3000

Number of male and female below 60 years = 1250 + 1150 = 2400

Number of male and female above 60 years = 3000 − 2400 = 600

Equal number of men and equal number of women are there in above 60 years group

∴ Number of men = number of women = 300

Number of women between 18 years to 60 years = 1230 − 220 = 1010

Difference between the female population of above 60 years and below 60 years = 930 − 300 = 630

63. (3)

Sum Of Digits is same

784 = 7 + 8 + 4 = 19

676 = 6 + 7 + 6 = 19

289 = 2 + 8 + 9 = 19

64. (2)

4 × 2 + 1 = 9

9 × 3 − 2 = 25

25 × 4 + 1 = 101

101 × 5 − 2 = 503

65. (4)

By observation

66. (4)

67. (1)

Monday Tuesday Wednesday Thursday Friday

V Y I R G

68. (1)

69. (1)

BC ⇒ 23 written as 32 on the top; B = 2 written as is, at the bottom

KM ⇒ 1113 written as 1311, O = 15

70. (1)

24 × 3 − 10 = 120 ÷ 2

72 − 10 ≠ = 60

503 9

101 25

4


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71. (2)

72. (4)

(2 × 8 = 16) (2 × 9 = 18)

(7 × 8 = 56) (7 × 9 = 63)

(12 × 8 = 96) (12 × 9 = 108)

(43 × 2 = 86) (11 × 9 = 99)

Except 4th Options all others have (n × 8) and (n × 9)

73. (1)

By observation

74. (2) or (4)

Logic for Option (2) (Count of both types of figures = 8)

Logic for Option (4) (Column count, 1st column top cell = 2 = middle cell + bottom cell)

75. (2)

x⊗ −

� = 2x

∆ = 8x

= 16x

� = 4x

76. (1)

77. (1)

First set of addition => incremental numbers, 2nd set of addition repeats after 3rd one

3 + 7 = 10 ; 15 + 20 = 35

4 + 7 = 11 ; 0 + 55 = 55

4 + 8 = 12 ; 45 + 30 = 75

5 + 8 = 13 ; 30 + 5 = 35

5 + 9 = 14 ; 15 + 40 = 55

78. (3)

79. (4)

80. Either (1) or (3)

The order from upper staircase to lower is given below:

Srinivas

Yaima

Jeet

Ranjan

Aloka

Danial

Barisha

Option (1) if 5th position is taken from the top.


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Option (3) if 5th position is taken from the bottom.

81. (1)

3 male, 2 female different = 1

82. (2*)

This is the closest option.

83. (1)


Fine arts Social since Chemistry Biology Physics

A � � � �

B � � �

C � � �

D � � �

E � � �

84. (3)

85. (4)

86. (2)

87. (4)

21 + 1 = 3; 42 + 2 = 18, 63 + 3 = 219

84 + 4 = 4100, 105 + 5 = 100005

88. (2)

Exactly 2 faces painted = 24

No faces painted = 8

Ratio = 8 : 24

= 1 : 3

89. (3)

DF = 30 m, CB = 22 m, EF = 8 m

CD = 26 m, BA = 20 m, AE = 6 m 2 2 2 2AF AE EF 6 8 100 10 m= + + = =

90. (1)

24, 22, 25, 23, 26, ___, ___, 25

Dual series

22, 23, 24, 25

24, 25, 26, 27

XXIV, XXVII

91. (3)

92. (1)

93. (4)

A E I O U

01 05 09 15 21

↓ ↓ ↓ ↓ ↓

10 50 90 51 12

F D

A

B C

26 20

E

30

22


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Central angle = ( )

12

10 50 90 51 12+ + + + × 360 = approximately 20 degrees

94. (2)

0 4 2

95. (2)

With 1 side fixed we will get

4 × 8 = 32 triangles

96. (2)

97. (4)

98. (3)

1 2 3 4 5 6

Time keeper Moderator SS SS

E B C F D

A

A

D

99. (1)

100. (1)

* * *

Seerat

Ruhani

21

9

2

+ 3

Shaurya

6

General Mental Abilities Test (GMAT) Detailed solutions

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