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IIFT SAMPLE PAPER (EXPLANATION) ANSWER KEY 1. [2] 6. [2] 11. [4] 16. [1] 21. [4] 26. [3] 31. [4] 36. [4] 41. [2] 46. [2]

2. [1] 7. [3] 12. [4] 17. [1] 22. [1] 27. [2] 32. [3] 37. [1] 42. [2] 47. [1]

SECTION I 3. [1] 8. [4] 13. [2] 18. [1] 23. [1] 28. [2] 33. [1] 38. [2] 43. [3] 48. [2]

4. [2] 9. [2] 14. [1] 19. [1] 24. [1] 29. [4] 34. [2] 39. [2] 44. [3] 49. [4]

5. [3] 10. [3] 15. [4] 20. [3] 25. [3] 30. [3] 35. [3] 40. [1] 45. [2] 50. [4]

51. [4] 56. [4]

52. [1] 57. [3]

SECTION II 53. [1] 58. [4]

54. [1] 59. [1]

55. [3] 60. [4]

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61. [1] 66. [2] 71. [3] 76. [3] 81. [4] 86. [3] 91. [1] 96. [3]

62. [2] 67. [2] 72. [4] 77. [2] 82. [3] 87. [1] 92. [1] 97. [2]

63. [1] 68. [4] 73. [3] 78. [4] 83. [3] 88. [3] 93. [2] 98. [1]

64. [3] 69. [1] 74. [4] 79. [1] 84. [4] 89. [1] 94. [1] 99. [3]

65. [3] 70. [4] 75. [3] 80. [4] 85. [4] 90. [1] 95. [3] 100. [3]

101. [4] 106. [3] 111. [3] 116. [3] 121. [1] 126. [3] 131. [3] 136. [4] 141. [3] 146. [2]

102. [1] 107. [1] 112. [1] 117. [2] 122. [1] 127. [2] 132. [3] 137. [2] 142. [1] 147. [3]

SECTION III 103. [2] 108. [2] 113. [2] 118. [3] 123. [3] 128. [1] 133. [4] 138. [3] 143. [3] 148. [4]

104. [3] 109. [1] 114. [2] 119. [4] 124. [4] 129. [2] 134. [3] 139. [3] 144. [4] 149. [1]

105. [3] 110. [3] 115. [3] 120. [2] 125. [2] 130. [1] 135. [2] 140. [1] 145. [2] 150. [2]

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151. [2] 156. [1] 161. [1] 166. [1] 171. [3] 176. [3] 181. [1] 186. [2] 191. [4] 196. [1]

152. [2] 157. [1] 162. [3] 167. [2] 172. [4] 177. [1] 182. [3] 187. [3] 192. [4] 197. [1]

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SECTION IV 153. [3] 158. [2] 163. [1] 168. [2] 173. [1] 178. [3] 183. [2] 188. [3] 193. [4] 198. [3]

1

154. [3] 159. [2] 164. [3] 169. [4] 174. [4] 179. [3] 184. [3] 189. [1] 194. [3] 199. [1]

155. [1] 160. [2] 165. [4] 170. [4] 175. [2] 180. [4] 185. [2] 190. [1] 195. [4] 200. [2]

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SOULUTIONS SECTION I 1.

We know that the height of the pole = AC + BC. Let BC = x, then in right triangle ABC tan 60° = √3. x 10

⇒ x = 10

Now let AC = y, then cos 60° =

y = 20. The height of the pole = 20 + 10√3. 10 y

⇒

Ans. [2] 2.

Let the side of the square = a units. Then the diagonal of the square = √2a units. Hence, the ratio of the areas of both the squares = =1:2 2 a 2 ( 2a ) Ans [1]

3.

Let the number be x. Then,

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1

3 1 × × ×x =8 3 5 7

⇒ x = 8× 35 = 280 4.

Ans. [2]

5.

Given f(x) =

. Therefore, f[h(x)] = x ⇒

.

x

h( x)

1− x

1 − h( x )

= x ⇒ h( x ) =

x 1+ x

Ans. [3] 6.

x2 + ax + b = 0 For Manju, x2 + ax + b1 = 0 Sum of roots = –a 6 + 2 = –a ⇒ a = –8 For Anju, x2 + a1 x + b = 0 Multiplication of roots = b (– 7) × (– 1) = b ⇒ b = 7 ∴ Equation is x2 – 8x + 7 = 0 ⇒ x2 – 7x – x + 7 = 0 or (x – 7) (x – 1) = 0 ⇒ x = 7 or 1

7.

Let the height of the small cone = h units. Then by the condition given in the question :

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units. 1

1 1 2 30 10 2 πr h = × πr 30 ⇒ h = = 3 27 3 27 9

Ans [3] 8.

Given 5 singers are below the age of 25 years and total 20 members are below the age of 25 years. So, the number of dancers below 25 years = 20 – 5 = 15. Required Percentage = 15 × 100/60 = 25 %. Ans. [4]

9.

Let the cost price = Rs. x and marked price = Rs. y. Then by the condition given in question: 1.33x = 0.95y . Hence, the marked price should be = 40% more than the ⇒

x y

=

95 133

⇒

5

c7 − 5h × 100

7

5

cost price. Ans [2] 10.

Let Y get Rs. y and Z get Rs. z. Then by the condition given in the question y – 20 = z + 20 = x...(1) Again x + y + z = 120...(2) by equation (1) and (2) we get x = 40, y = 60, z = 20. Ans [3]

11.

Since it is not given how many students are between Ravi and Reena we cannot determine Reena’s rank from the front. Ans. [4]

12.

While returning, the speed of the traveler = 50 × 120/100 = 60 kmph. Average speed during the whole journey = 2 × 50 × 60/(50 + 60) = 54.5 kmph. Ans. [4]

13.

In an hour, the bells ring =

= 4.28 = 4 times. Ans [2] 60 × 60 LCM of 6, 7, 8, 10

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14.

Let the usual speed of the person = x kmph and the distance = d km. Then time = d/x hours. Again by the condition given in question 4d/3x = d/x + 1/3 ⇒ d/x = 1 hour. Ans. [1]

15.

B

C

Ungrazed Area

A D Let, the side of the square garden = x units. Then the length of the rope =

units. Then, the area ungrazed

x 2 = x2 -

units.

x x (4 − x) x⇒ 4 4 2

2

Ans. [4]

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16.

Let the Tony’s salary = Rs. x. Then, Tina’s salary = 0.6x and the salary of Tito =

x = 0.5x. Now by the

0.6 1.2 condition given in the question : x + 0.6x + 0.5x = 6300 ⇒ 2.1x = 6300 ⇒ x = 3000. Hence Tony’s salary = Rs. 3000. Ans. [1] 17.

Degrees traveled by the minute hand between 4 : 45 to 5 : 45 = 360° and degrees traveled by the minute hand between 5 : 45 and 6 : 15 = 180°. Hence, total degrees traveled = 540°. Ans. [1]

18.

Total amount of bill = 380 + 5% of 125 = 380 + 5.125 = Rs. 385.125. Ans. [1]

2 6, 7, 8, 9, 12 2 3, 7, 4, 9, 6 3 3, 7, 2, 9, 3 1, 7, 2, 3,1

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LCM of 6, 7, 8, 9 and 12

= 2 × 2 × 3 × 7 × 2 × 3 = 504 ∴ Number of times = 3600 504

20.

=7

Let the radius of the circle be R1 = r. ∴ Area of circle A1 = π r2 Radius is increased by 100% ∴ R2 = 2r A2 = π × (2r)2 Area of the circle increased = A 2 − A1 × 100 A1 = 2

4ππ − πr πr

2

2

× 100

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21.

For P = 1, the value of equation 4 – 7 + 5 = 2, i.e., the positive value. Ans. [4]

22.

Since the average of the three numbers is 410, hence the total of the three numbers = 410 × 3 = 1230. Let the other two number are x and x + 25. Then, by the condition given in the question : 2x + 25 = 1230 – 185 ⇒ x = 510. Hence the smallest number = 185. Ans. [1]

23.

Let the numbers be x and y. Then by the condition given in the question x + y/ 2y = 3/5 ⇒ x/y = 1/5. Ans. [1]

24.

104. Ratio of shares = A : B : C A : B = 100 : 50 B : C = 100 : 40 ⇒ A : B : C = 200 : 100 : 40 = 10 : 5 : 2 ∴ B’s share = 1683 = Rs. 495

5 × 17 25.

Length of the smaller cube = 1 mm Length of the bigger cube = 1 m = 1,000 mm So number of cubes = Volume of the bigger cube Volume of the smaller cube

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=

1,00,00,00,000 1000 × 1000 × 1000 1×1×1

26.

=

Since each smaller cube is having a side of 1 mm, while joining these small cubes end-to-end, total length of this cube-rope will be 1 × 1,00,00,00,000 mm = 1,000 km. Use a + b + ab 100

⇒ 10 − 10 −

27.

10 × 10 100

= −1%

Total number of valid votes = Difference Percentage difference

× 100

= 440 72 − 28

× 100

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= 440 44

× 100

= 1,000 ∴ Total number of voters = 1000 + 350 = 1,350 28.

A rectangular solid has 12 sides in which 4 sides represent length, 4 represent breadth and 4 represent height. Sum of all sides = 4l + 4b + 4h = 4 x (l + b + h) a multiple of 4.

29.

10% of Rs. 500 = Rs. 50 and 5% of Rs. 50 = Rs. 2.5

30.

1 hour work of the leak = 1 6

1 hour work of the tap = 1 10

The tap can fill the tank in

= 15 hr 1 1 6

−

1

MA NY

6 × 10 10 − 6

10

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∴ Capacity of the tank = 15 × 60 × 3

= 3,150 L 1 2

31.

The smallest such number is 1.

32.

Work with choices

33.

Simple interest on Rs. 645 for 1 year = 690 – 645 = Rs. 45 Simple interest for 4 years = 45 × 4 = Rs. 180 ∴ Principal = 645 – 180 = Rs. 465

34.

2x + 3y + 4z = 27 … (i) and 3x + 2y + z = 33 … (ii) Adding (i) and (ii), we get 5(x + y + z) = 60 ⇒ (x + y + z) = 12 ∴ Average = =4

( x + y + z ) = 12 3

35.

Time 1 p.m. 2 p.m. 3 p.m. 4 p.m.

3

Relative speed 0 km/hr –3 km/hr –2 km/hr –1 km/hr

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Distance between A and B 8 km 11 km 13 km 14 km

2

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5 p.m. 0 km/hr 14 km 6 p.m. +1 km/hr 13 km 7 p.m. +2 km/hr 11 km 8 p.m. +3 km/hr 8 km 9 p.m. +4 km/hr 4 km Hence, between 9 p.m. and 10 p.m. the relative speed is 5 km/hr. As distance between them is 4 km, the time required is = 48 min. 4 5

36.

× 60

Area grazed by one horse = πr

2

4

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a 2

| |

a 2

| |

Area grazed by four horses = π r2 Here, r = . 63 2

When a is side of square ABCD. ∴ Area left ungrazed = a2 – π r2 = 632 – π ×

 63    2

2

= 632

 

× 1 −

22 7

×

1



4

= 632 6  28 − 22  2  = 63 × 28  28 

×

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= 632 ×

3 14

=

= 850.5 m2 63 × 9 × 3 2

37.

A |

| B

Speed of A Speed of B

=

Time taken by B to completethe remainingdis. after meeting Time taken by A to completethe remainingdis. after meeting

=

4 1

=2

Speed of A = 2 × Speed of B. So, the first train is twice as fast as the second train.

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38. 1+

⇒ ⇒

x 12

x 12 x 12

169

=

= =

144

13 12

=

13 12

−1

1 12

⇒ x =1 39.

-25 (x2 – 2

x + 2)

2 ⇒ x2 - 2

0

≥

x + 2≤ 0

2 ⇒ (x -

)2 = 0

2 As the square of a real number cannot be negative. ∴ (x ) = 0 ⇒x =

2

2

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40.

g(x) = x3 + 2 ⇒ g(2) = 23 + 2 = 10 f(x, y) = xy ⇒ f(3, 10) = 3 × 10 = 30

41.

To calculate the number of games, it will be (n –1), i.e. 150 – 1 = 149. Alternative method: As there are 150 members, so in order to decide the champion there has to be one less than the total number of members, i.e., 149 games.

42.

328 = (34)7 = (81)7 = (80 + 1)7 When divided by 80, remainder is 1.

43.

(

) (

)

 7+ 5 2+ 7− 5 2  7+ 5 7 − 5  +  = 7−5 7 + 5    7− 5   = 7+5+2 7

5 +7+5−2

7

5

7−5 = 24 2

= 12

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44.

In ∆ ABC, BC is the circumference of a cylindrical pole.

4C B A

For every 5 ft of spiral distance, effective height covered by the insect = 4 ft Hence, if the height of this pole is 16 ft, the insect will travel = 16 = 20 ft ×

45.

Circumference of wheel = π d Distance traveled by wheel in n revolutions = nπ d

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= 100 × 22 7

× 63

= 19,800 cm = 198 m 46.

= 1280 2

x

2

x +2 −2

⇒2

x 5  2  = 1280

⇒2

x

⇒2

x

=

1280 × 2 5

= 512 ⇒ 2

= 512 x

=2

9

⇒x=9 47. [1] No of batches = 12 symbol 180 \f "Symbol" \s 8× 1.5 = 18; no. of sugar cups = 18 symbol 180 \f "Symbol" \s 8× 6 = 108. 48. [2] No. of people who do not prefer chicken or fish = 50 + 60 - 90 = 20. 49. [4] We get x/4 - x/5 = ¼; solving we get x = 5. [x is the distance].

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50. [4] Squaring, (x - 1/x)² = x² + 1/x² - 2 = 4. Squaring once again, (x² + 1/x²) = x4 + 1/x4 + 2 = 36 Hence x4 + 1/x = 34. SECTION II 51. [4] Ram says that only 2 alternatives are possible but Sita gives a third one. 52. [1] The survey could have underestimated dishonest students if they admitted that they were honest. 53. [1] If you believe Anu Malik, then J or L are insane. But from statement I, both must be insane together. 54. [1] It is assumed that a married person loses his identity. Statements II & III are not in the statement. 55. [3] If some other benefit is obtained, then the disadvantage of marriage can be overcome, hence [3]. 56. [4] If a majority of people complained about losing their identity, then the statement would be strengthened. 57. [3] If the painter works on Thursday, then the carpenter and the plumber have come before him and D must also be before E. This leaves E for Friday. 58. [4] If the painter works on Friday, all others are before him. but E can work on Wednesday or Tuesday, hence e) is wrong. 59. [1] We get : I, E, PL, C, P, hence a). 60. [4] T and A do not live in any of the end houses, hence B and D live there. They own grey and mauve cars, hence e) is wrong. 61. [1] If A owns the violet car, only one color (orange) is left for T. P r e p ato r I nwww.manyagroup.com Logon

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62. [2] B is at the extreme, so T can be adjacent to him, hence b) is not necessarily true. 63. [1] Since A is at 5, places 4 and 6 must not have smokers. Hence TVM can occupy 1, 2 or 3 places only and S and K can occupy 4 or 6. As K is not an answer choice, we can say that A must be next to S. 64. [3] From Q. 135. 65. [3] V and K must be together, hence V must be at 3 and K at 4. 66. [2] If T takes over the duties of S, then he must talk on the phone and therefore he cannot be next to M, who requires silence. Thus T or M must ask for a change in seating plan. 67. [2] We get OQPNLM. Hence Q is before P. 68. [4] NLPOWM. Then Q is immediately before M. 69. [1] From Q. 127 70. [4] B must precede C, hence must be at 4. 71. [3] B cannot be at 3 because it would mean then C must come after lunch, which is not allowed. 72. [4] D must be at 1, F at 2; then B and C must come after lunch, which means that C must be at 5. 73. [3] We get DBCAFE from this condition. 74. [4] Zeenat is between Akbar and Anthony while Parveen is unwilling to stay next to any occupied cottage, hence statement V can be derived.

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75. [3] We get - Amar Akbar Zeenat Anthony - Parveen - Hema. There are 4 people next to vacant cottages. 76. [3] From Q. 149. 77. [2] It is given that S and T must be together and K is at 3. Only b) satisfies all the conditions. 78. [4] In this choice, K is not at 3, which is a required condition. 79. [1] If V is at 5 and K is at 3 then the only positions for S and T are 1 and 2 since they have to be together. 80. [4] KTS are at positions 3, 4 and 5. Hence V cannot be at 5. 81.

From statement I, we get 1, 3, 5, 7, 9. From statement II, all numbers which are multiples of 3. Combining both, we get the numbers 3 and 9. Hence, we cannot decide.

82.

Both the statements are needed. Apply a sinA

83.

=

b sinB

Both the statements are needed. First statement says x > or x < 1

1

2

2

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84.

Second statement indicates x > 0. Cannot be determined. The second statement indicates that Raju can be a liar or can be truthful.

85.

Cannot be determined. The length of the race has to be known.

For questions 86 to 88: The possible combinations that meet all the conditions Mon Tue Wed Thurs Fri Sat Sun a) N M R O/P S Q P/O b) M N R O/P S P/O Q On the basis of these possibilities, we can solve the questions easily. 86. [3] 87. [1] 88. [3] For questions 89 to 92: For solving these questions, we use the diagram given below:

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CA xdazbcy B

A = 40, B = 50 and

C = 30

A + B + C = 120 n(A∪B∪C) = 100 – 20 = 80 Number that play exactly two games (x + y + z = 20) Thus, we get the following equations (a + b + c) + d = 80 – 20 = 60 (a + b + c) + 3d = 120 – 2 (20) = 80 ∴ d = 10 = n (A∩C) 89.

Number of students playing at least 2 games = 20 + 10 = 30.

90.

Number of students playing exactly one game = 80 – 30 = 50.

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91.

Number of students playing exactly one game = 100 – 20 = 80

92.

Number of students playing all 3 games =d = 10

93.

In alternative [1], V and Z are not supposed to sit together. Alternative [2], is correct. In alternative [3], Y sits at the head of table. In alternative [4], X sits at the head of table.

94.

Since V and Y sit together, W and T sit near Z leaving X next to T.

95.

U

X

Z W

Y 96.

V U

V

T Y Z

97.

W

T

X

U

1

2 Z

Y

V

X

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For questions 98 to 100: Use diagrams to arrive at the correct answers. 98.

[1]

99.

[3]

100.

[3]

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SECTION IV 151. [2] The passage is about the Victorians’ idea of invention, which resulted in a number of useless inventions. 152. [2] It is stated that inventions should facilitate life, i.e., utility should over-ride the mere act of inventing. 153. [3] It is the author’s view of inventions that triumphs. 154. [3] “numerous plans and inventions.” 155. [1] The discussion about the mousetrap shows that Alice regards utility as more important. 156. [1] The knight wants to do things in a particular way even if they are useless and serve no purpose. Hence the best choice to describe the Victorian era is a). 157. [1] a see brackets in the passage. 158. [2] From the Victorian era, when some inventions did not serve any purpose, it moved to the Industrial Revolution, during which all inventions were put to commercial use. 159. [2] It was the Industrial era that produced wealth. 160. [2] The passage talks about the art and utility of inventions, hence the best choice is b). 161. [1] “is to make sure that their site name is top-of-mind”, i.e., awareness. 162. [3] Third paragraph: “providing a tangible cue about the service always helps.” 163. [1] Fourth paragraph: “single most important benefit that is sought by a consumer.” P r e p ato r I nwww.manyagroup.com Logon

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164. [3] Fifth paragraph: unexpected positive service 165. [4] Spends on billboards are going down because advertisers are looking at other options, according to the passage. 166. [1] In a bid to fill up empty coffers, municipal corporations are giving permissions to put up hoardings without any limit. 167. [2] The companies are holding out and are making average profits, as stated. 168. [2] Only television is mentioned in the passage. 169. [4] The article does not make any future predictions. 170. [4] The article is about dotcom advertising but later talks of hoardings and outdoor advertising. Only e) reflects both these themes. 171. [3] Antipode: an exact opposite. 172. [4] Moribund: dying, stagnant, obsolescent 173. [1] Pristine: immaculate, unspoiled, original state. 174. [4]Insidious: gradual and harmful 175. [2] Ingenuity: inventiveness 176. [3] P r e p ato r I nwww.manyagroup.com Logon

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177. [1] 178. [3] 179. [3] Lurk and wait are synonyms, so are abscond and depart. 180. [4] Exhort is to urge; goad is to prod. 181. [1] Look for units of measurement 182. [3] We use needle to knit, loom to weave. 183. [2] Zoology is the study of animals, botany is the study of plants. 184.

‘Equilibrium’ means a state of composure.

185.

Synopsis is a brief review of a subject.

186.

Partisan is to take parts or sides.

187.

Metamorphosis is to change form completely.

188.

Controversy is a dispute, argument or debate.

189. [1] 190. [1]

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191. [4] 192. [4] 193. [4] success….. accomplishment 194. [3] frugal: economical in manner 195. [4] truculent - fierce or cruel. A ferocious manner will alienate her colleagues 196. [1] A person should not be categorized, or labeled as…… 197. [1] acerbic - bitter 198. [3] to take exception to - were in opposition to 199. [1] unprecedented - without precedence 200. [2] cursory- careless or without details

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