ATMA 2006

1. [4] The statement says that life can be found underground, so the surface is not the only place where life can be found. [3] If employees arrive any time, they will not be able to interact with each other and decrease their productivity. [4] The argument says that people live longer, which is weakened if we give another cause of longevity, given in [4] [2] Though home owners may buy more ice cream, it is possible that children may be consuming the ice cream, hence [2]. [3] The question is how to prepare for obsolescence, which is best answered by training the people. [1] [1] [4] [3] [4] [2] [3] I follows as all members were women and employed, II follows as all members were women. [3] I follows as blundering competent, and II follows as blundering competent experienced. [3] As poets contribute to new magazines, I is correct, also II is correct because of D. [1] Only I is true as it satisfies the condition; II cannot be inferred. [4] None of the statements can be inferred from the given statements [4] Both statements are false as faculty members like music, and Faculty members are not Don Juans. [3] If they can't read we can say that they are not well educated or read the magazine. 7. 10. 13. 16. 19. [3] [2] [1] [2] [2] 8. 11. 14. 17. 20. [4] [2] [4] [3] [2]

2.

3.

4.

5.

6. 9. 12. 15. 18. 21. 22.

23.

24. 25. 26. 27.

28.

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29. [2] It is required in small quantities and is harmful in large quantities. [3] [3] P > J, J > K. If M = 6, L = l, N = 7. Hence O can fourth third or fourth. [1] M = 2, L = l, N = 7, we can place O before P. All other choices violate a condition. [3] If L = 2, K. = 5, then M = l, O = 7, and the only place for N is 6, [1] If L = l, N = 7, and P >J > K, then P cannot be lower than third. [4] If P = 5, then J = 6 and K = 7, then M = l [2] [1]

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30. 31.

32.

33.

34.

35. 36. 37. 38.

[4] we have found that the rotational velocity in spiral galaxies either remains constant with increasing distance from the centre or increases slightly. [1] can be inferred from the first paragraph. [2] [3] [3] [4] since the end [4] [1] to an average of... [3] increasingly focused [4] [4] [3] [4] [1] 53. 56. 59. 62. [4] [2] [3] [1] 54. 57. 60. 63. [3] [1] [3] [2] 41. 44. 47. [4] [1] [1] 42. 45. [2] [3]

39. 40. 43. 46. 48. 49. 50. 51. 52. 55. 58. 61. 64.

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65. [3] x + 5 > 2 implies x > 3 and x 3 < 7 implies x < 10. The interval is thus 3 and 10.

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66.

[2] Substituting x = 4 in the equation we get: k = 18. The equation is x2 + 3x 28 = 0. Using the options we get x = 7 as the other option. [2] Sum of prime numbers 2 + 11 = 13, 2 + 7 = 9 and 5 + 11 = 16. [1] We can obtain 4 5 = 20 sums of these, 9 can occur 4 times, hence probability = 4/20 = 1/5. [2] n = 5 7 = 35, so it must be divisible by 35. [1] I < 5n + 5 < 25 implies 4 < 5n < 20, so n can be 0, 1, 2, 3. [2] There are 250 numbers between 101 and 350. Numbers with 2 in hundreds place = 200 299 or 100 numbers. Hence reqd probability = 100/250 = 0.4 [3] If p/q < l then q/p > 1. [2] x 3y 7 = 0; Substituting the points we get: a 3b 7 = 0 and a + 3 3b 3k 7 = 0. On solving the equations we get k = 1/3. [2] From A we get n = 3 and 103/3 is not an integer. From B we get n = 4 and 104/4 is an integer, hence either statement answers the question. [3] A: 560 x < 280 or x < 1/2. B: 700 x > 280 or x > 260/700 = 0.37. Combining the statements we get x must lie between 0.37 and 0.5 so we get the answer. [4] From A: 0.25 < x < 0.5 so x can be 3/8 or 2/5. From B: 0.33 < x < 0.6 so x can again be 3/8 or 2/5 so the question cannot be answered. [3] Both statements give Information about number of students. [4] From A: the number can be 12, 21 or 30. B does not help in getting one number out of the three. [3] From A: The number could be 17, 18 or 19 [by adding 3 we get at least 20]. From B; the number could be 16 or 17 [by reducing 3 we should have less than 15]. By combining we get the answer as 17.

67.

68.

69.

70.

71.

72.

73.

74.

75.

76.

77.

78.

79.

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80. [3] We can make a Venn diagram by combining both statements, hence [3]. [1] From A: we get n = 2 or 3. From B: we get n = 2. [1] We get y = x + 3 from A alone, which is enough to give us the largest side. [2] A: x/y + y/x = 2 which is possible if x = y = l. From B: we get x = y, hence answer is obtained form either statement. [4] For n = 0, we get (a + b)/2 which is the arithmetic mean of a and b. [4] The area of the shaded portion would depend on where the circles intersect. [2] If coefficient of x was wrong the product was correct. Product = 30, which happens in (2). [4] B + C = 140. As they are bisected, 1/2 (B + C) = 70 so O = 180 70 = 110. [3] By combining the, two conditions, we get the fraction as 1/4.

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81.

82.

83.

84.

85.

86.

87.

88.

89.

[1] Surface area of 3 separate cubes = 3 6a2. Surface area of joined cubes = 14a:. Ratio =18:14 = 9:7. [1] [2] Let the speed of the stream be x. Then, 2D/(6 + x) = D/(6 x).On solving, we get x = 2. [4] From the sum, we get the number as 253, since 253 + 99 = 352. The digit in the hundreds place is 2. [4] 212n 64n = 212 64 when n = 1. On solving we get the value as 2800, which is divisible by 100. [4] [2] We get the triplet 5, 12, 13. So the smaller side is 5. [3] dy/dx = 2x + 2 = 0, or x = 1. Substitute in the expression to get the value as 5.5. [4] We can take 2, 3 or 4 women, hence number of ways = 7C4 4C2 + 7C3 4C3 + 7C4 + 4C4 = 371.

90. 91.

92.

93.

94. 95.

96.

97.

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98.

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[1] In 1/10 of an hour, the man walks 0.2 km. He sees up to a distance of 0.6, so total is 0.8 km. In l hour the car travels 0.8 10 = 8 km. [3] 100. 103. 106. 109. 112. [4] [4] [4] [3] [3] 101. 104. 107. 110. 113. [2] [1] [1] [2] [2]

99.

102. [1] 105. [3] 108. [2] 111. [3] Solutions (Qs. 114 to 121): We can make a diagram as follows: D O M F E

G N 114. [1] 117. [4] 115. 118. P [1] [2] 116. 119. [4] [2]

120. [2] Work from the options. All the other options violate one condition or the other. 121. [1] BCF + DH because CE are not together and GD must go together. 122. [3] Either model E must be selected or model H must be selected, but both cannot be selected. 123. [2] 126. [4] 129. [4] 130. [3] Only [3] and [4] start by long operations. [4] is wrong because 7 and 8 are consecutive, which violates the condition. 131. [1] Tendulkar performs the long operation on 1, hence he cannot be scheduled on Monday. Only [1] is correct. 132. [2] There are no conditions for Tuesday.
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124. 127.

[1] [2]

125. 128.

[3] [1]

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133. [2] 136. [3] 139. [3] 142. [3] 145. [4] 148. [2] 151. [3] 154. [2] 157. [2] 160. [4] 163. [3] 164. [4] The intersection will contain squares 9, 25, 36, 49 hence 4 numbers. 134. 137. 140. 143. 146. 149. 152. 155. 158. 161. [4] [2] [1] [1] [2] [4] [1] [4] [3] [4] 135. 138. 141. 144. 147. 150. 153. 156. 159. 162. [2] [1] [1] [4] [1] [4] [2] [3] [1] [2]

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165. [1] Average (or middle term) of first 5 integers = 560/5 = 112. So the next 5 numbers start from 115 and the middle term is 117. Sum =117 5 = 585. 166. [3] 31 cannot be written in the form 2k +1. 167. [3] 168. [2] Number of people in 2006 = 45/1.15 = 39 approx. 169. [1] From A we get the answer: applicants who have at least 4 years' experience and a degree = 18 + 14 30 = 2. 170. [3] We need the values of both x and y, which are given by both statements. 171. [1] We get the answer from A alone: if x = y + 2 then x > y. 172. [3] We need the number of members and those under 35, which are given by both statements. 173. [4] We cannot get the values of x or y even by using both statements.

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174. [2] From B: x y > 0 means x > y so x/y > 1. 175. [3] l/x + l/y = (x + y)/xy. This is given by both statements. 176. [2] Area =
1 (BD AC). This is given by B alone. 2

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177. [1] Midpoint of the points = (2 +l)/2, ((3 + l)/2 = 3/2, 2. Substitute in the given choices to see that only (1) satisfies the point. 178. [1] 179. [3] Area of hexagon = 3/4 (6)2 6 = 54 3 approx. 180. [2] The numbers less than 9 are l, 2, 3, ... 8. Subtract this from the total numbers = 80 8 = 72, 181. [4] Factors of 12 = 3 22. So there will be 220 in 1210. 182. [4] 183. [1] Area = 1/2 base ht = 1/2 6 2 = 6. 184. [2] 185. [1] As it reduces 10% every. time, in 3 times it will become:(12 0.93) = 87.48. 186. [1] If h = 5n, the other sides are 3n and 4n. So b (smaller side) = 3n. 187. [1] Let 6A = 8B = 10C = x. Then the ratio is: 1/6 :1/8: 1/10 = 20: 15: 12. 188. [2] We have wx = k and 2w + 2x = p. On substituting the value of x in the second equation, we get 2w + 2k/w = p, on solving, 2w2 pw + 2k = 0. 189. [4] Mark the points on a number line to get the answer. 190. [1] By elimination, we see that the term is divisible by 3 only when n is odd.

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191. [2] We can make a Venn diagram as follows: So the minimum number in the common set is 110 and maximum is 130. 170 130 150

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192. [4] Probability of all three balls being green = 7/12 6/11 5/10 = 7/44. 193. [1] We will have 2 possibilities: TL or LT = 60% 20% + 40% 80% = 44%. 194. [3] The prizes can be given in 54 ways. 195. [1]

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