National Talent Search Examination Stage I - 2013 (For Class X Students)
Q1. Probable dates of sister’s birthday according to Anshu =
16, 17, 18, 19
Probable dates of sister’s birthday according to her father
= 16
Thus, Anshu’s sister’s birthday falls on 16 Oct which is
common as per given information.
Option 2
Q2. Option 4
Q3. Option 3
Q4. P(16) S(19) : T(20) W(23)
(1) J(10)
M(13) : R(18) Q(17)
(2) A(1)
D(4) : D(4) I(9)
(3) A(1)
D(4) : E(5) H(8)
(4) F(6)
C(3) : Z(26) E(5)
It is clearly evident that the relation between PS & TW
is same as between AD & EH.
Option 3.
Q5. BYA, CXB, ?, EVD
The pattern is highlighted in the following figure
Option 4
Q6. (1) 729 = (9)3
(2) 343 = (7)3
(3) 576 = (24)3
(4) 512 = (8)3
Option 3
Q7. Except 1321, rest is divisible by 3.
Option 3
Q8. The pattern in highlighted in the following figure
Option 3
Q9. The entire family relationship is shown below:
Option (1)
Option 4
Q11. The pattern is square of the natural numbers in
decreasing order starting from 12. Thus, the number completing the series would
be 64 or (8)2.
Option 3
Option 4
Option 4
Option 4
Q15. Option 3
Q16. Option 1
Q17. Option 4
Q18. Option 2
Q19. Option 4
Q20. Option 2
Q21. Except ‘Copper’, rest is alloy.
Option 2
Q22. Except ‘Seismograph’, rest is natural calamity.
Option 1
Q23. Option 1 represents no symbol, rests are symbols.
Option 1
Q24. Except ‘Mumbai’, rests are non-metro cities.
Option 2
Q25. 145 = 5 × 29
120 = 5 × 24
90 = 5 × 18
105 = 5 × 21
Option 1
Q26. Except 69, rests are prime numbers.
Option 3
Q27. A = 6 × S
After four years,
(A + 4) + (S + 4) = 43
A + S + 35
6S + S = 35
S = 5 & A = 30
Option 1
Q28. The pictorial diagram is shown below:
Distance to reach the starting point = 12 – 7 = 5 km
Option 4
Q29. If A is 3 times more efficient than B, then he must
take 3 times less number of days than B to complete the same work.
Let us assume B takes 3x days to complete the work.
Thus, A takes x days to complete the work
3x – x = 60
2x = 60
x = 30
Time taken if both A & B work together = 1/{(1/30) +
(1/90)} = 45/2 days
Option 3
Q30.
Q31. The pattern follows as the given numbers are cubes of
natural numbers starting from 13 onwards. Thus, the number completing the
series will be 4913 or (17)3.
Option 3
Q32. Cells ®Tissues®Organ®Organ system®Body
Option 4
Q33. After re-arrangement GOVERNMENT will be written as
OGEVNREMTN.
Thus, 3rd letter from right end is M.
Option 1
Option 1
Q35. Let the number of boys and girls in the college be 5x
and 2x respectively.
5x – 2x = 450
3x = 450
x = 150
Total number of students in the college = 5x + 2x = 7x or 7
× 150 = 1050
Option 2
Q36. IN 60 minutes, train travels = 92.4 km
In 10 minutes, train travels = 15.4 km or 15400 m
Option 1
Q37. x = 2y & y =
2z
x = 4z, y = 2z
Putting these values in given equation in terms of ‘z’, we
get:
4z + 2z – z + 100 = 350
5z = 250
z = 50
Option 2
Q38. Suppose his monthly salary is Rs. 100.
He spends Rs. 20 on food.
Remaining amount = 100 – 20 = Rs. 80
He spends 40% of 80 on food = Rs. 32
Remaining amount = 80 – 32 = Rs. 48
He spends Rs. 48 on routine expense.
48 = 5760
100 = 12000
His monthly salary = Rs. 12000
His annual salary = 12 × 12000 = Rs. 144000
Option 3
Q39. 800 × 100 = P × 4 × 5
P = Rs. 4000
Option 1
Q40. Percentage error = {(40.8 – 40)/40} × 100 = 2%
Option 4
Q41. Let the numbers be x & y.
xy = 120 & x2 + y2 = 289
(x – y) 2 = x2 + y2 – 2xy =
289 – 240 = 49
(x – y) = 7
Option 1
Q42. C/A = C/B × B/A = 9/8 × 7/6 = 21/16
Option 3
Q43. Applying identity A2 – B2 = (A +
B) (A – B), we get
{(A2 – B2)/(A – B)} = (A + B) = 0.525
+ 0.275 = 0.8
Option 4
Q44. LCM of (12, 16, 18) = 144
Series of numbers divided by 144 = 144 × n where ‘n’ is a
natural number.
Second number of series = 2 × 144 = 288
288 + 7 = 295
Option 3
Q45. Given information can be pictorially represented using
venn diagram as:
100 = (75.8 – x) + x + (49.4 – x)
x = 25.2%
Number of students who took both French & Latin = 25.2%
of 250 = 63
Option 3
Q46. Option 1
Q47. Option 3
Q48. Option 2
Q49. Option 4
Q50. 27 ÷ 3 × 9 – 17 + 18 = 81 – 17 + 18 = 82
Option 3
Q51. Let the total number of fruits on the basket = 12
LCM of denominators (4, 3) = 12
Apples = ¾ of 12 = 9
Peach = 2/3 of (12 – 9 = 3) = 2
Oranges
= (3 – 2) = 1
1 = 300
12 = 3600
Option 4
Q52. 12 × 16 = 4 × n
n = 48 days
Option 4
Q53. Option 3
Q54. The coding follows as
Option 4
Q55. Option 2
Q56.
Q57. Option 2
Q58. Let the Weight of body be 100 kg. Now, distribution is
classified below:
Distribution of elements in human body
Water = 70 kg
Proteins = 16 kg
Other Dry Elements = 14 kg
Distribution of protein in human body
Muscles = 16/3 kg
Skin = 16/10 kg
Bones = 16/6 kg
Hormones Enzymes & other proteins = 32/5 kg
Ratio = 16/3 : 16/6 = 2 : 1
Option 2
Q59. Required percentage = 0.016% of 100 kg = 1.6
This is equivalent to the weight of skin in human body.
Option 3
Q60. Total percentage value of two items = 16 + 14 = 30%
100% =360°
30% = 108°
Option 3
Q61. Quantity of water = 70% of 50 kg = 35 kg
Option 1
Q62.
Option 3
Q63.
Option 2
Q64. After re-arrangement:
PAL TUB RAC
NOS DIH
Option 3
Q65. After re-arrangement:
BUT CAR HID
LAP SON
Option 1
Q66. Option 3
Q67. After re-arrangement:
LBP BVT CBR
SPN HJD
Option 4
Q68. Vehicle moves on ‘road’ which is coded as ‘water’.
Option 4
Q69. Option 4
Q70. Milk in the mixture = 7/9 × 729 = 567 ml
Water in the mixture = 729 – 567 = 162 ml
Let the quantity of water added be x ml.
{567/(162 + x)} = 7/3
x = 81 ml
Option 2
Q71. Surface area of cube = 6a2
6a2 = 384
a2 = 64
a = 8
Volume of cube = a3 = 83 = 512 m3
Option 3
Q72. Sales = Price of TV × No. of TV sold
Initially
S1 = P × N
After change,
S2 = 1.3P × 0.8N = 1.04 P × N
Percentage increase = {(1.04 – 1)/1} × 100 = 4%
Option 4
Q73. Option 4
Q74. Option 3
Q75. Option 3
Q76. Option 4
Q77. Option 1
Q78. Option 1
Q79. Option 2
Q80. Option 4
Q81. Option 3
Q82. Option 2
Q83. Option 4
Q84. Option 2
Q85. Option 4
Q86. Option 2
Q87. Option 3
Q88. Option 4
Q89. Option 2
Q90. Option 2
