IIFT 2016 QA | Previous Year IIFT Paper
Previous year paper questions for IIFT 2016 QA
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1.
IIFT 2016 QA
| Algebra - Inequalities & Modulus
The smallest integer x for which the inequality
Answer: Option
D
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Explanation :
The LHS is of the form: (x – 7) / (x + 9) (x – 4)
For x > 7, the expression will be positive.
Similarly, for any negative integer,
[(x – 7)/(x – 4)] will always be positive.
For the expression to be positive,
(x + 9) > 0 i.e. x > −9
Hence, the smallest integer that satisfies this inequality is x = −8
Hence, option (d)
2.
IIFT 2016 QA
| Algebra - Progressions
A child, playing at the balcony of his multi-storied apartment, drops a ball from a height of 350 m. Each time the ball rebounds, it rises 4/5th of the height it has fallen through. The total distance travelled by the ball before it comes to rest is
- A.
2530 m
- B.
2800 m
- C.
3150 m
- D.
3500 m
Answer: Option
C
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Explanation :
Distance covered by ball when it touches ground for the first time = 350 m.
Distance travelled when it touches for the second time = 2 × (4/5) × 350 m
Distance travelled when it touches for the third time = 2 × (4/5) × (4/5) × 350 m
∴ Total distance = 350 + [2 × (4/5) × 350] + [2 × (4/5) × (4/5) × 350] + …= 350 + [2 × (4/5) × 350][(1 + (4/5) + (4/5)2 + …]
The last term above is an infinite G.P. with a = 1 and r = 4/5
The sum of such an infinite G.P. = a/(1 – r) = 1/([1 – (4/5)] = 1/(1/5) = 5
∴ Total distance = 350 + [2 × (4/5) × 350](5) = 350 + 8(350) = 9(350) = 3150 m
Hence, option (c).
3.
IIFT 2016 QA
| Algebra - Logarithms
Find the value of x which satisfies the following equation
4 log7 (x - 8) = log3 81
- A.
8
- B.
18
- C.
20
- D.
None of the above
Answer: Option
D
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Explanation :
4 log7(x – 8) = log3 81
∴ 4 log7(x – 8) = log3(34) = 4
∴ log7(x – 8) = 1
∴ x – 8 = 7 i.e. x = 15
Hence, option (d).
4.
IIFT 2016 QA
| Modern Math - Permutation & Combination
A playschool contains 4 boys and y girls. On every Wednesday during winter, five students, of which at least three are boys, go to Zoological Garden, a different group being sent every week. At the Zoological Garden, each boy in the group is given a ball. If the total number of balls distributed is 368, then the value of y is
Answer: Option
D
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Explanation :
Since the group has to have atleast three boys, number of ways in which a group of five can be formed = (4C3 × yC2) + (4C4 × yC1) = [4 × (y)(y – 1)/2] + [(1)(y)] = 2y(y – 1) + y
Since each boy in the group gets a ball, total balls distributed = 3[2y(y – 1)] + 4y = 6y2 – 2y
Since total balls distributed= 368; 6y2 – 2y = 368
∴ 3y2 – y – 184 = 0
On solving this, y = 8
Hence, option (d).
5.
IIFT 2016 QA
| Modern Math - Permutation & Combination
Which of the following statements regarding arrangement of the word ‘RIYADH’ is/are true:
i. Two vowels can be arranged together in 120 ways
ii. Vowels do not occur together in 240 ways
Which of the above statements are true?
Answer: Option
D
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Explanation :
(i) Two vowels can be arranged together in RIYADH in 5! × 2! = 240 ways
Hence, statement (i) is not true.
(ii) Total arrangements where vowels are not together = total arrangements possible – arrangements where vowels are together = 6! – 240 = 720 – 240 = 480
Hence, statement (ii) is also not true.
Hence, option (d).
6.
IIFT 2016 QA
| Arithmetic - Average
Two farmers were cultivating wheat on their respective agricultural land in a village. Farmer A had an average production of 20 bushels from a hectare. Farmer B, who had 15 hectares of more land dedicated to wheat cultivation, had and output of 30 bushels of wheat from a hectare. If farmer B harvested 530 bushels of wheat more than farmer A, how many bushels of wheat did farmer A cultivate?
- A.
50
- B.
80
- C.
160
- D.
200
Answer: Option
C
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Explanation :
If farmer A had x hectares, farmer B would have had (x + 15) hectares.
Average production of farmers A and B is 20 bushels and 30 bushels respectively.
∴ Total production of A = 20x and total production of B = 30(x + 15)
Since total production of B exceeds that of A by 530 bushels; 30(x + 15) – 20x = (10x + 450) = 530
∴ 10x = 80 i.e. x = 8
∴ Total cultivation of farmer A = 20x
= 160 bushels
Hence, option (c).
7.
IIFT 2016 QA
| Arithmetic - Time, Speed & Distance
Shruti and Krishna left Delhi for Noida at the same time. While Shruti was driving her car, Krishna, an environmentalist by profession, was traveling on his bicycle. Having reached Noida, Shruti turned back and met Krishna an hour after they started. Krishna continued his journey to Noida after the meeting, while Shruti turned back and also headed for Noida. Having reached Noida, Shruti again turned back and met Krishna 30 minutes after their first meeting. The time taken by Krishna to cover the distance between Delhi and Noida is
- A.
2 hours
- B.
2.5 hours
- C.
3 hours
- D.
None of the above
Answer: Option
A
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Explanation :
Shruti and Krishna leave for the first time. Their first meeting is after one hour, the second is after an additional half an hour.
So, the interval between subsequent meetings keeps halving every time.
∴ Total time = 60 + 30 + 15 + ….
This is an infinite G.P. with a = 60 and r = 1/2
∴ Sum of terms = a / (1 – r) = 60 /[1 – (1/2)] = 120 minutes = 2 hours
Hence, option (a).
8.
IIFT 2016 QA
| Arithmetic - Profit & Loss
In a local shop, as part of promotional measures, the shop owner sells three different varieties of soap, one at a loss of 13 percent, another at a profit of 23 percent and the third one at a loss of 26 percent. Assuming that the shop owner sells all three varieties of soap at the same price, the approximate percentage by which average cost price is lower or higher than the selling price is
- A.
10.5 higher
- B.
12.5 lower
- C.
14.5 lower
- D.
8.5 higher
Answer: Option
A
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Explanation :
Let the S.P. of each type be Rs. 100
C.P. of soap sold at 13% loss = 100/0.87 ≅ Rs. 115
C.P. of soap sold at 23% profit = 100/1.23 ≅ Rs. 81
C.P. of soap sold at 26% loss = 100/0.74 ≅ Rs. 135
∴ Average C.P. = (115 + 81 + 135)/3 = Rs. 110.3
∴ % by which average C.P. is higher = (10.3/100) × 100 = 10.3%
The closest value in the options is 10.5%
Hence, option (a).
9.
IIFT 2016 QA
| Arithmetic - Time & Work
In the marketing management course of an MBA programme, you and your roommate can complete an assignment in 30 days. If you are twice as efficient as your roommate, the time required by each to complete the assignment individually is
- A.
45 days and 90 days
- B.
30 days and 60 days
- C.
40 days and 120 days
- D.
45 days and 135 days
Answer: Option
A
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Explanation :
If “you” are twice as efficient as your roommate, the roommate will take double the time taken by you i.e. the number of days for each will be in the ratio 1 : 2.
Hence, options 3 and 4 can be eliminated.
Let the total work correspond to 90 units.
If your roommate does x units of work per day, you can do 2x units of work.
∴ Total work done per day when both of you work together = x + 2x = 3x units
Since both of you can together complete the work in 30 days, work done per day = 90/30 = 3 units
∴ 3x = 3 i.e. x = 1
Hence, you and your roommate can individually do 2 units and 1 unit per day.
Hence, time taken by you = 90/2
= 45 days and time taken by roommate
= 90/1 = 90 days
Hence, option (a).
10.
IIFT 2016 QA
| Geometry - Quadrilaterals & Polygons
Let PQRSTU be a regular hexagon. The ratio of the area of the triangle PRT to that of the hexagon PQRSTU is
- A.
0.3
- B.
0.5
- C.
1
- D.
None of the above
Answer: Option
B
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Explanation :
A regular hexagon comprises six equilateral triangles.
∴ Area of hexagon = 6 × area of equilateral triangle = area of six equilateral triangles
P, R and T are alternate vertices of the hexagon. The triangle formed by joining them splits each equilateral triangle into half. Six half-equilateral triangles is equivalent to three equivalent triangles.
∴ Area of triangle PRT = area of three equilateral triangles.
∴ Required ratio = (area of three equilateral triangles) / (area of six equilateral triangles) = 0.5
Hence, option (b).
11.
IIFT 2016 QA
| Venn Diagram
290 students of MBA (International Business) in a reputed Business School have to study foreign language in Trimesters IV and V. Suppose the following information are given
i. 120 students study Spanish
ii. 100 students study Mandarin
iii. At least 80 students, who study a foreign language, study neither Spanish nor Mandarin
Then the number of students who study Spanish but not Mandarin could be any number from
- A.
80 to 170
- B.
80 to 100
- C.
50 to 80
- D.
20 to 110
Answer: Option
D
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Explanation :
Atleast 80 students study neither Spanish nor Mandarin.
Hence, maximum number of students who study atleast one language = 290 – 80 = 210
Minimum number of students who study both languages = 100 + 120 – 210 = 10
∴ Maximum number of students who study Spanish but not Mandarin = 120 – 10 = 110
Maximum number of students who study both languages = smaller value of 100 and 120 = 100
∴ Minimum number of students who study Spanish but not Mandarin = 120 – 100 = 20
Hence, the range could be any number from 20 to 110.
Hence, option (d).
12.
IIFT 2016 QA
| Geometry - Mensuration
A right circular cylinder has a height of 15 and a radius of 7. A rectangular solid with a height of 12 and a square base, is placed in the cylinder such that each of the corners of the solid is tangent to the cylinder wall. Liquid is then poured into the cylinder such that it reaches the rim. The volume of the liquid is
- A.
147(2π - 8)
- B.
180(π - 5)
- C.
49(5π - 24)
- D.
49(15π - 8)
Answer: Option
A
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Explanation :
Volume of liquid = volume of cylinder – volume of rectangular solid
Volume of cylinder = šr2h = (72)(15)= 735š
Since each corner of the solid with a square base is tangential to the cylinder wall, the diagonal of the square base is equal to the diameter of the base of the cylinder.
∴ Diagonal of square base = 2(7) = 14
∴ Side of square = 14/√2 = 7√2
Volume of rectangular solid = (7√2) × (7√2) × 12 = 1176
∴ Volume of liquid = 735š – 1176
= 147(5š – 8)
Hence, option (a).
13.
IIFT 2016 QA
| Modern Math - Permutation & Combination
A reputed paint company plans to award prizes to its top three salespersons, with the highest prize going to the top salesperson, the next highest prize to the next salesperson and a smaller prize to the third-ranking salesperson. If the company has 15 salespersons, how many different arrangements of winners are possible (Assume there are no ties)?
- A.
1728
- B.
2730
- C.
3856
- D.
1320
Answer: Option
B
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Explanation :
This is a case of first selecting the top three salesmen out of 15 and then arranging them as first, second and third i.e. 15P3āāāāāāā
∴ Number of arrangements = 15C3 × 3! = 2730
Hence, option (b).
14.
IIFT 2016 QA
| Algebra - Progressions
What is the sum of integers 54 through 196 inclusive?
- A.
28,820
- B.
24,535
- C.
20,250
- D.
17,875
Answer: Option
D
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Explanation :
Sum of integers from 54 to 196 = (sum of integers from 1 to 196)– (sum of integers from 1 to 53)
Sum from 1 to 196 = (196)(197)/2 = 98 × 197 = 19306
Observe that this value is already less than three out of four options. Hence, it will further decrease when the sum from 1 to 53 is subtracted from it.
Hence, the answer should be less than 19306. Only 17875 satisfies this condition.
Hence, option (d).
15.
IIFT 2016 QA
| Modern Math - Probability
The student mess committee of a reputed Engineering College has n members. Let P be the event that the Committee has students of both sexes and let Q be the event that there is at most one female student in the committee. Assuming that each committee member has probability 0.5 of being female, the value of n for which the events A and B are independent is
- A.
2
- B.
3
- C.
4
- D.
None of the above
Answer: Option
B
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Explanation :
P = event that both males and females are present in the committee = 1/n + 2/n +....+ (n – 1)/n
= [1 + 2 + 3 + … + (n – 1)]/n
= [(n)(n − 1)/2n] = (n – 1)/2
Q = probability that at most one female is present in the committee = 1/n
Probability of PāQ = P × Q
P ā Q = 1/n as “only one female in the committee is the intersection of the two events”.
∴ (1/n) = [(n − 1)/2] × (1/n)
∴ n – 1 = 2 i.e. n = 3
Hence, option (b).
16.
IIFT 2016 QA
| Algebra - Progressions
A multi-storied office building has a total of 17 rows of parking spaces. There are 20 parking spaces in the first row and 21 parking spaces in the second row. In each subsequent row, there are 2 more parking spaces than in the previous row. The total number of parking spaces in the office building is
- A.
380
- B.
464
- C.
596
- D.
712
Answer: Option
C
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Explanation :
Observe that the last 16 rows form an A.P. with a = 21, d = 2 and n = 16.
The term before that is 20.
Hence, sum = (19 + 1) + A.P. of 16 terms
Hence, this can even be considered as an A.P. of 17 terms with a = 19 and d = 2
∴ Total parking spaces = 1 + (17/2) × [2(19) + 16(2)] = 1 + 17(19 + 16) = 596
Hence, option (c).
17.
IIFT 2016 QA
| Algebra - Surds & Indices
The highest number amongst āāāāāāā and
- A.
- B.
- C.
- D.
None of the above
Answer: Option
B
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Explanation :
Since the powers are (1/2), (1/3), (1/4), raise them to their LCM i.e. 12.
Hence, the terms become (26)1/12; (34)1/12 and (43)1/12 i.e. (64)1/12; (81)1/12 and (64)1/12
Since 81 > 64, the largest value would be ā3.
Hence, option (b).
18.
IIFT 2016 QA
| Modern Math - Permutation & Combination
In an MBA entrance examination, a minimum is to be secured in each of the 6 sections to qualify the cut-offs. In how many ways can a candidate fail to secure the cut-offs?
Answer: Option
D
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Explanation :
Number of ways in which the candidate fails to secure the cut-offs in no section out of 6 sections = 6Co = 1
Number of ways in which the candidate may/may not secure the cut-offs
= 6Co+ 6C1 + 6C2 + 6C3 + 6C4 + 6C5 + 6C6
= 26 = 64
Thus, the number of ways in which he fails to secure the cut offs in at least one section = 64 − 1 = 63
Hence, option (d).
19.
IIFT 2016 QA
| Algebra - Progressions
The sum of 4 + 44 + 444 + …. upto n terms in
- A.
- B.
- C.
- D.
Answer: Option
C
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Explanation :
A simple way to solve this is to take the value of n as a suitable integer and tally the options with the sum.
e.g. when n = 1, sum = 4
Only the expression in option 3 gives this value.
Hence,option 3.
Note:
You can verify option 3 by putting n = 2.
20.
IIFT 2016 QA
| Geometry - Coordinate Geometry
Suppose the two sides of a square are along the straight lines 6x - 8y = 15 and 4y - 3x = 2. Then the area of the square is
- A.
2.52 sq.units
- B.
3.61 sq.units
- C.
4.33 sq.units
- D.
None of the above
Answer: Option
B
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Explanation :
The two lines are 4y – 3x = 2 and 4y – 3x = −7.5
As a1 / a2 = b1 / b2 ≠ c1 / c2; thetwo lines are parallel to each other.
The distance between the two parallel lines would be the side of the square
i.e. |c1 – c2| / (√a2 + b2)
Here a2 + b2 = (−3)2 + (4)2 = 25 and |c1 – c2| = |2 – (−15/2)| = 19/2
∴ Side of square = (19/2)/5 = 19/10
= 1.9 ∴ Area of square = (1.9)2
= 3.61 sq. units
Hence, option (b).
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