NTSE Sample Paper Solutions

7/30/2019 NTSE Sample Paper Solutions.

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HINTS AND SOLUTIONS

1. Sol:(e)5, 7, 20, 21, 80, 63, ? 189

Rule : Multiply odd placed numbers by four to get next odd placed number

Multiply even placed numbers by three to get next even placed number.

2. Sol: (e)Rule : add 2, 4, 8, 1 6, ... to t he respective numbers given.

3. Sol:(b)Rule : numbers are written in the pattern of 13 + 1, 23 + 1, 33 + 1, 43 + 1 ...etc.

4. Sol:(b)Rule: Multiply by 6, divide by 6, multiply by 7, d ivide by 7, multiply by 8, divide by 8.

5. Sol:(b)Rule: Numbers are in pattern of 122, 222, 322, 422, 522, 622.

6. Sol:(c)Rule: Subtract (7) from number to get next number.

7. Sol:(a)Rule : Numbers are increased in the order of + 1, + 2, + 3, + 4 ... etc. to get nextnumbers

8. Sol:(b)Rule : Alphabats of code word are replaced by exactly next alphabet of word

given in their respective places.

9. Sol:(d)Rule : Alphabets placed at even positions in code and in word are same whereas

alphabets placed at odd positions in code are one alphabet before of the alphabets

used in word to write code.

10. Sol: (c)Reason : Red is called yellow as per given statements.

11. Sol:(e)www .pion

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Reason : 37 means which class, 583 means caste and class so we can conclude that

common number is

(c) and common word is class so 3 means class. Therefore caste means 5 or 8.

12. Sol:(c)Reason : As per given codes D is for 7, E is for 3, L is for 5, H is for 4, I is for 1, C is for 8,

A is for 2, L is for 5, U is for 9, T is for 6

Therefore CALICUT is for 8251896

13. Sol:(d)Rule : Code is obtained by writing the alphabets in reverse order as they are written in

given words.

14. Sol:(a)Rule : first alphabets of pairs are continuous alphabets from left to right while second

alphabets of pairs are also continuous alphabets from right to left.

15. Sol:(c)Reason: Gaps in alphabets from left to write are in the order 1, 2, 3, 4. Therefore

fourth letter from G is K.

16. Sol:(e)Reason: Alphabets are at a gap of one alphabet between them in reverse order.

17. Sol:(a)Reason: Add + 4, + 4, + 6, + 6 to get next alphabets to their previous positions.

18. Sol:(e)Rule : Central numbers are the sums of square roots of numbers outside of circles

separately.

19.Sol:(b)Rule : Central numbers are the cube roots of sums of all four numbers outside the

circles.

20. Sol:(a)Rule: Central numbers are square root of sums of all four numbers outside the circles.

21. Sol:(b)www .pion

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Rule: Central numbers are obtained by dividing the product of remaining four numbers

by ten.

22. Sol:(e)Rule: Central numbers are half of sum of squares of all other four remaining numbers

present.

23. Sol:(d)

Rule: d = a b

e = b cf = a c

24. Sol:(c)

Rule: d = a 8

e = b 8

f = c 8

25. Sol:(d)Rule : Numbers in the second row are obtained by subtracting one from twice the

numbers of first row.

26. Sol: (a)Rule : Column one :4 4 (4)2 = 416

Column two : 5 5 (5)2 = 525

Column Three : 3 3 (3)2 = 309

Therefore column four : 2 2 (2)2 = 204.

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27. Sol:(b)Rule :

e + x + b = f + x + c = a + x + d

28. Sol: (c)Reason: Price of SOAP = Rs [4 + 1 + 6 + 5] = Rs. 16/ as cost of each alphet is given.

29. Sol:(d)Reason: OIL = Rs. (1 + 3 + 2) = Rs. 6/

30. Sol:(d)Formula : To find number of parallelograms made by set of m parallel lines onone side and set of n parallel lines other side is given by

(m)(m 1) (n)(n )

4

using m = 4 and n = 4

number of parallelograms

=

4 3 3 24

= 18 parallelogram

31. Sol:(c) count32. Sol:(e)

Rule: as there are three squares in each row and each column as shown in figure

so total number of squares = 32 + 22 + l2 = 14

33. Sol:(d)Reason: Numbers 2, 3, 5, 6 are adjacent faces of number 4 as shown in third and

fourth cube hence only possibility is that 1 will be opposite to number 4.

34. Sol:(a)Reason : Sides of the given hexagon are removed in anticlockwise direction one by one

and place on opposite side from where sides removed.

35. Sol:(d)www .p

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Reason : Use the pattern as shown in four figures.

36. Sol:(a)Reason: Number of dots are increasing one by one inside the pentagon near to vertices.

37. Sol:(c)Reason: Figure (c) does not have any tail and all other figures have tails.

38. Sol:(b)Reason: Position of circle in the figure (2) is not changing hence figure b is odd.

39. Sol: (a)Reason : First row contains some alphabets and second row contains the positions of

alphabets.

40. Sol:(b)Reason: Sequence of numbers is 13, 11, 9, 7, 5, 3 and sequence of alphabets is C, E, G, I,

K and M.

41. Sol:(b)Reason: Numbers allotted to alphabets are squares of their positions in English alphabet

orders from left to right.

42. Sol:(b)43. Sol:(c)44. Sol:(d)45. Sol:(b)

Relative positions of five boys as per given instructions in the question are as under:

So Gaurav won the race.

46. Sol: (b)47. Sol:(d)48. Sol: (c)49. Sol:(e)

Reason: Ravi is son of Kamals sister.

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50. Sol:(c)Reason: Boy is son of Deepaks sister.

51. Sol: (a)Reason: Father wife Mother

Mothers only son Himself.

52. Sol:(a)Reason : Father of man is his father only because man does not have any brother or

sister.

53. Sol:(a)Reason: Rita is wife of her husband so man on stage is brother of her daughter.

Therefore man on stage is here son only.

54. Sol:(e)Reason : All odd placed numbers are increasing by 4 from beginning whereas

all even placed numbers are decreasing by 4 from end hence first 19 will be

replaced by 23.

55. Sol:(a)Reason: Here pattern formed is + 2, + 4, + 2, + 4 ... etc. So 19 will be replaced by 17.

56. Sol:(d)Reason : Pattern formed is 1 = 03 + 1

2 = 13 + 1

9 = 23 + 1

28 = 33 + 1

hence 60 will be replaced by 65 because 65 = 43 + 1.

57. Sol:(e)

Reason : Pattern is + 2, + 4, + 8, + 16, therefore 36 will be replaced by 33.

58. Sol: (a)Reason :0 = 12 1, 3 = 221, 8 = 321

15 = 421, 24 = 52 1, 35 = 621

therefore 1 will be replaced by 0.

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59. Sol:(b)Reason: Any number in series is sum of two numbers before it. Hence number

22 will be replaced by 21.

60. Sol:(c)Reason : Second number 2 = 12

third number 6 = 2 3

fourth number 24 = 64

so fifth number will be 24 5 = 120

hence 119 will be replaced by 120.

The water images of alphabets are inverted alphabets at their own positions.

61. Sol:(b)62. Sol:(e)63. Sol: (d)64. Sol:(d)65. Sol:(c)66. Sol:(b)67. Sol:(d)68. Sol:(b)69. Sol:(a)70. Sol:(c)71. Sol:(c)72. Sol:(a)73. Sol:(c)74. Sol:

(c)75. Sol:(a)76. Sol:(c)

Reason: number of persons in line 11 + 111 = 21

77. Sol:(c)78. Sol:(c) Pattern is bbcc, bbcc, bbcc, bbcc

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79. Sol:(d) Pattern is abc abc abc abc80. Sol:(c) Pattern is aa bb cc dd aa 81. Sol:(a) Pattern is abcd abcd abcd..82. Sol: (a) Pattern is cba abc cba abc 83. Sol:(b)

Reason : HOTEL = 8 + 15 + 20 + 5 +12 = 60

so code for BORE = 2+ 15+ 18+ 5 = 40

84. Sol: (c)Reason : Hotel is coded 300 which is five times of sum of positions occupied of

alphabets used in HOTEL so code for BORE will be 40 5 = 200.

85. Sol:(d)Reason : HOTEL =

60

2= 30

so BORE =40

2= 20

86. Sol:(a)Reason: length of each edge of square = 3d

so area of square = 3d3d = 9d2.

87. Sol:(a) Area of one square is A units hence area of 13 squares = 13ANOTE: In the mirror image alphabets interchanges their positions and also right

and left parts changes their positions.

88. Sol:(a)89. Sol:(b)90. Sol:(a)91. Sol:(c)92. Sol:(d)93. Sol:(a)94. Sol:(c)95. Sol:(a)96. Sol:(c)

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97. Sol:(b)98. Sol:(c)99. Sol:(b)100.

Sol:(c)

181.Sol:(d)

Prime factorization of 2880 is 222222325 therefore 5 is required number.

182.Sol:(a)

Let C.P. of machine be Rs. X

so S.P. when profit is 10%

= Rs.110

x100

now S.P. when less is 10% = Rs.90x

100

hence110 90

x x 400100 100

x = Rs. 2000

183.Sol: (c)

Let number of persons be x.

So according to given condition of inverse variation we can say

x 110

x 10 100

x = 110 person

184.Sol:(b)

Let principal is Rs. x/ and compound interest on principal Rs. x at 4% p.a. for 2 years

be Rs. k/

so by using formulawww .

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nr

CI P 1 1100

25

weget 102.5 x 1 1100

(1)

K = x24

1 1100

(2)

On dividingequation number (1) and (2)

we get k = Rs. 81.60/

185.Sol:(d) Every natural, whole, integer, rational, irrational number is always real.

Therefore ( x y ) is real number.

186.Sol:(b)

Solution: LCM of 6, 12, 15 and 18 is 180 seconds.

Hence number of times in 4 hrs they will ring together =4 60 60

180

= 80 times

Hence they will ring together = 80 + 1 = 81 timesOne is added for first ring when watches first time rings.

187.Sol:(b)

5 610122

7 7

4

70 567

Hence difference =610 218

567 7

So percentage change

=218/7

100 30%610/7

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188.Sol:(d)

Reason: Midpoint of hypotenuse of a right angled triangle is always equidistant

from all three vertices.

189.Sol:(b)

Hint: Start solving from last and step by step

1 10 1 91 so

19 9 1019

Now1 9 19

1 11 10 10

1 9

1 10

1 191

11

9

Hence1 10 29

1 11 19 19

11

1 9

190.Sol: (d)

Solution: Let initial radius is r and when radius is doubled is 2r.

so V1 =34 r

3

and V2 =34 (2r)

3

3

1

32

4r

V 34V

r 83

Volume become 8 times.

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191.Sol:(c)

Sum of all interior angles of a polygon of n sides = (n 2) 180

Sum of all exterior angles of any polygon is always 360.

So as per given condition

(n2) 1 80 =1

2360

(n2) = 1 n = 3

192.Sol:(c)

Solution:

We also have a = 2b 135 5a 2b = 13 (1)

or b =2b 13

5

We also have b =30

a ab = 30. ... (2)

On cubing equation (1) both sides we get

125a3 8b3 30ab (5a 2b) = 2197

125a3 8b3 303013 = 2197

125a3 8b3 = 13897

193.Sol:(a)

Reason: We know that between any two numbers there are finite number of

integers and infinite number of rational and irrational numbers.

194.Sol:(d)

Reason: Kite is a quadrilateral whose diagonals & are perpendicular but kite is other

than square and rhombus.

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Sol:(b)

Method: Here it is required to mention that is always equal to 180 hence 2 is equal to

360 so sum of the given angles is sum of angles of two triangles which is 360 or 2 .

196.Sol:(a)

Hint: Join O to B and then make use of property that angle subtended by an arc at

centre is double the angle subtended by same arc in any part of circle leaving the

given arc.

197.Sol:(a)

Solution: (x4 + x2 + 1)

Adding and subtracting x2

we get x4 + 2x2 + 1 x2

(x2 + 1)2 (x)2

(x2 + x + 1)(x2x+1)

198.Sol:(c)

Solution: Let acid in first jar is px litres y and water is 1x litres.

and acid in second jar is qy litres and water is 1y litres.

Volume of solution in first jar=Volume of solution in second jar.

so px + 1x = qy + 1y

x(p + 1) = y (q + 1)

x q 1

y p 1

Now when we mix the solutions of two jars

we get Acid = (px + qy) litres

Water = (x + y) litres

so ratio of acid to water =px qy

x y

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On dividing numerator and denominator both by y we get

xp q

y

xy

y

But we have x q 1y q 1

On putting this value we get ratio as

p q 2pq

p q 2

199.Sol:(c)

Solution: Let side of square be x unit

so1

2 (x) (x - 5) = 48

x = 12 cm.

200.Sol:(c)

Let seven consecutive numbers are (x), (x +1), (x + 2), (x + 6)

(x) + (x + 1) + (x + 2) + ... + (x +6) = 1617

x = 228

So numbers are 228, 229, 230, 231, 232, 233, 234.

5 of these numbers are not prime.

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NTSE Sample Paper Solutions