Answer Key & Explanations - IIFT 2016 (All Sets)
Answer Key & Explanations - IIFT 2016 (All Sets)
Using options, we can check that only option (D) satisfies the above equation.
There are two vowels I and A in RIYADH
(i) Let these two vowels IA as one unit.
∴ No. of ways in which 2 vowels can be arranged together = 5! × 2! = 240
5. 108. 66. 30. D Hence statement (i) is false
(ii) Total no. of arrangements = 6! = 720.
No. of ways in which vowels do not occur together = 720 – 240 = 480
Hence statement (ii) is false.
Let farmer A has ‘x’ hectare land. ∴ Total production of A = 20x
Farmer B has x + 7 hectare land ∴ Total production of B = (x + 15) × 30
6. 109. 67. 31. C
Given that (x + 15) × 30 – 20x = 530 ⇒ 30x + 450 – 20x = 530 ⇒ 10x = 80 ⇒ x = 8.
∴ Production of farmer A = 20x = 20 × 8 = 160 bushels.
A B
d d
Delhi Noida
2
Let the distance between Delhi and Noida is x km.
Let they first meet at point A after one hour.
Distance covered by Shruti = 2x – d
7. 110. 68. 32. A Distance covered by Krishna = d
∴ Ratio of speeds of Shruti and Krishna is 2x- d : d …………. (i)
Let next they meet at B after half an hour. As Krishna covered distance ‘d’ in one hour, so
' d' d
he will cover distance in half an hour.∴ AB =
2 2
3d 5d
Distance covered by Shruti in second meeting =x – d + x − =2x-
2 2
IIFT - 2016
Answer key &
Explanations
5d d
∴ratio of speeds of Shruti and Krishna is 2x - : ……………… (ii)
2 2
5d
2x −
2x − d 2 ⇒ 2x –d = 4x – 5d ⇒ 2x = 4d ⇒ x = 2d.
=
d d
2
Now Krishna covered distance‘d’ in one hour, so he will cover distance x in 2 hours.
Let SP of each article be Rs. 100
100 100
Thus, CP 1 will be × 100 = 115, CP 2 = × 100 = 81.3
87 123
100
8. 111. 69. 33. A CP 3 = × 100 = 135.1
74
Hence, total CP = 331.4.
331.4 − 300
% by which CP is lower/higher than SP = × 100 = 10.5% higher.
300
If I do 2 units per day, my roommate will do 1 unit per day.
Together we do 3 units per day.
9. 112. 70. 34. A Since, we take 30 days to finish the complete work, so total work must be 30 × 3 = 90
units. Now working @ 2 units/day, I will take 45 days for 90 units and my roommate will
take 90 days.
Let the side of hexagon PQRSTU is ‘a’ Q
3
∴ Area of hexagon = 6× ×a2
4 a
a 60° 60°
30°
P R
3a 3a
2 2
10. 113. 71. 35. B
∴ Area of ∆PRT =
4
3
( )
× 3a =
2 3 3 2
4
a .
S
U
3 3 2 4 3 1
∴ Reqd. ratio is a × = = = 0.5 T
4 6 3a 2 6 2
Total students = 290. Let 80 students do not study either Spanish or Mandarin.
∴ No. of students who study Spanish or Mandarin or both = 290 – 80 = 210.
11. 114. 72. 36. D ∴ n(S∪M) = n(S) + n (M) – n(S∩M) ⇒ 210 = 120 +100 – n(S∩M) ⇒ n(S∩M) = 10
∴ Number of students who study Spanish but not Mandarin = 120 -10 = 110.
As 110 is given in option D, hence it is the answer.
The Volume of Cylinder = 15 × 49π
The rectangle solid is placed in cylinder such that each of the corners of solid is
tangent to walls of cylinder. Hence the diameter of cylinder will be diagonal to the
square base.
As the diameter of cylinder is 14, so diagonal of square is 14 and hence side of
square is 7√2.
12. 115. 73. 37. A
14 7√2
1 1
P(P) = 1 − n
( when all males) − n ( when all females)
2 2
1 n ×1 1
P(Q) = n ( when all males) + × n −1 ( when one female and rest males)
2 2 2
P(P ∩ Q) = exactly one female
1 1
15. 118. 76. 40. B =n× ×
2 2 n −1
For independent events
P(P ∩ Q) = P (P) × P (Q)
n 1 1 1 n
= 1 − n − n n + n
2 2 2
n
2 2
Solving we get n = 3
The number of parking spaces
16
16. 119. 77. 41. C = 20 +21 + 23 + ……… = 20 + 2 [2 × 21 + 15 × 2]
= 20 + [8[72]] = 20 + 576 = 596
We have
1 1 1
,
2 3 4
2 3 ,4
1 1 1
17. 120. 78. 42. B ×12 ×12 ×12
=2 2
,3 3
,4 4
6 4 3
= 2 , 3 , 4 = 64, 81, 64. As 81 is the largest number among the above numbers
1
so 3 3 is the highest number.
Number of ways in which a candidate can fail to secure cut offs.
18. 121. 79. 43. D 6 6 6 6 6
= C 0 + C 1 + C 2 + ........ C 5 = 2 – 1 = 63.
4 + 44 + 444 + -------------- n terms
= 4 (1 + 11 + 111 + -------------- n terms)
4
= (9 + 99 + 999 + -------------- n terms)
9
4
= [(10-1) + (100-1) + (1000-1) +-------------- + n terms]
19. 122. 80. 44. C 9
=
4 4 10 10 n − 1
[10 + 100 + 1000 + -------------- - n]=
− n
( )
9 9 9
=
40 n
81
(
10 − 1 −
4n
9
)
The two sides of the square are
6x – 8y = 15 and 4y – 3x = 2
or 6x – 8y = 15 (1)
and 6x – 8y = - 4 (2)
These two lines are parallel. So the distance between these lines is the side of
20. 123. 81. 45. B 15 − (− 4 ) 19
the square. ∴ side of square = =
62 + 82 10
2
19 361
∴ Area of square = = = 3.61 sq.units
10 100
CDB is 2 triangles ahead of GHF in clockwise order and both GHF, CDB have alphabets in
21. 1. 82. 46. A clockwise order in their respective triangles. So following this, we get the answer as A
option.
By observation, HNP & DLP are vertices of 2 different triangles. We are moving ACW from
HNP to DLP. This means we have to move ACW from PDA. While in option PHE we are
22. 2. 83. 47. B
moving clockwise from PDA. So PHE is not possible. Also answer has to start from P.
Answer can’t be PME because it’s a straight line. Hence answer is PJG.
23. 3. 84. 48. C By observation, line IO comes after AK in ACW direction. Thus CL must be after EM in
IIFT - 2016
Answer key &
Explanations
ACW direction. By observing closely the 2 options starting with EM, we get the answer as
EMDL.
By observation, BPM is not a triangle. Thus correct answer will not form a triangle. Thus
24. 4. 85. 49. A option D is eliminated. Further BPM is almost opposite to GN. So FP is almost opposite to
AK. Hence answer is FPO.
a) If the Maternal grandmother is from tribe A, then mother will be from tribe A and the
female in question is from tribe A. As given that the female is from tribe B, so
statement a is false.
25. 5. 86. 50. B
b) If paternal grandmother is from A, then father is from tribe A and after marriage, he will
become member of tribe B. His daughter, the female in question will be of tribe B.
Hence statement b is true.
We will check the options one by one.
a) If the boy is born in tribe B then he will marry in tribe A and his daughter will be in tribe
A. Hence option (a) is incorrect.
b) If the boy is born in tribe B, then he will marry in tribe A. His son will be in tribe A. So
26. 6. 87. 51. C his daughter in law will be from tribe B. Hence option (b) is incorrect.
c) If the boy is born in tribe B, then his mother’s brother can be from tribe B and his
father’s brother can be from tribe A. Hence option (c) is correct.
d) If the boy is born in tribe B, then he will marry in tribe A and his divorced son will be in
tribe A. Hence option (d) is incorrect.
a) Any widower will return to his tribe. So he can marry his wife’s sister which is from
other tribe. Hence this marriage is permissible.
b) This marriage is also permissible as the divorced husband will return to his tribe.
Hence the mother can marry the divorced husband of her daughter.
27. 7. 88. 52. C
c) The mother’s brother will be of same tribe as that of girl. Hence the girl cannot marry
him. Hence this marriage cannot take place.
d) Any widower will return to his own tribe A. His brother’s widow will be of tribe B. So he
can marry his brother’s widow.
This is debatable set and there are two interpretations possible for statement 5. A number
of students wasted quite a lot of time on this set in the exam because of this ambiguity.
Some institutes have given a key without considering this possibility. We have highlighted
both scenarios in our solution and we sincerely hope this clarifies the given problem set.
Figure (i)
Carpenter/ Tailor /
Ana Belle
Plumber/
Cook/ Diana Ferida
Hairdresser/ Washerwoman/
28- 8- 89- 53- Elsa Cinderella
31. 11. 92. 56.
The correct seating arrangement is given above : (One should be careful in the
th
interpretation of 5 condition according to which the correct arrangement is Cook-
Carpenter-Belle)
Tailor /
Figure (ii) Cook/ Ana Diana
Plumber/
Carpenter/ Ferida
Elsa
Hairdresser/ Washerwoman/
Belle Cinderella
IIFT - 2016
Answer key &
Explanations
28. 8. 89. 53. * None of option*
29. 9. 90. 54. * None of option*
30. 10. 91. 55. * None of option*
31. 11. 92. 56. C
From the given information
We can infer that
32- 12- 93- 57- Admin → E & G (female)
34. 14. 95. 59. Finance → C, A and one of B & E
Logistics → H and one of B & E.
Order of income → G > H > A > F, B, E > C
32. 12. 93. 57. * Finance department will have 3 people.
33. 13. 94. 58. A B earns less than A and H.
nd
34. 14. 95. 59. A H is at 2 position in descending order of income.
All fathers are males and some doctors are males as well as fathers. Also some doctors
35. 15. 96. 60. C rd
are females. Hence 3 option.
(2 + 6) × (15 – 5) = 80
36. 16. 97. 61. B (7 + 6) × (9 - 4) = 65
So answer will be (16 + 8) × (13 - 11) = 24 × 2 = 48.
10, 26, 74, 218, 654
10 × 3 – 4 = 26
26 × 3 – 4 = 74
37. 17. 98. 62. A
74 × 3 – 4 = 218
218 × 3 – 74 = 650
Hence 654 is wrong and should be replaced by 650.
st
Sum of alphabets position in 1 row = 1 + 4 + 1 + 3 + 2 + 2 + 4 + 3 + 3 = 23
nd
Similarly, sum of digits in 2 row = 1 + 3 + 1 + 2 + 4 + 2 = 13
rd
Since, these are prime numbers, so sum of all given and unknown alphabets in 3 row
38. 18. 99. 63. D
should also be a prime no.
rd
Thus, 3 row = 1 + 2 + 3 + 4 + 4 + 3 + 1 + 1 = 19 (taking D option as correct).
No other option gives a prime value in such manner, hence correct answer is (D)
L
F
B
39. 19. 100. 64. B