# IIFT 2017 QA

Paper year paper questions for IIFT 2017 QA

**1. IIFT 2017 QA | Arithmetic - Simple & Compound Interest**

Swarn, an SME enterprise, borrowed a sum of money from a nationalised bank at 10% simple interest per annum and the same amount at 8% simple interest per annum from a microfinance firm for the same period. It cleared the first loan 6 months before the scheduled date of repayment and repaid the second loan just at the end of scheduled period. If, in each case, it had to pay Rs. 62,100 as amount, then how much money and for what time period did it borrow?

- A.
Rs. 55,750, 2 years

- B.
Rs. 52,500, 2 years

- C.
Rs. 51,750, 2.5 years

- D.
Rs. 55,750, 2.5 years

Answer: Option C

**Explanation** :

Let the amount borrowed from each source be Rs. P for n months.

Since Swarn pays Rs. 62,100 to each firm, it pays the same simple interest to each firm.

Since it repays the bank 6 months before repayment, its effective loan tenure is (n − 6) months; while for the loan with the microfinance firm, its tenure is n months.

Considering same interest: [P × 10 × (n − 6)]/(12 × 100) = [P × 8 × n]/(12 × 100)

∴ 5(n − 6) = 4n i.e. n = 30

Hence, time period of borrowing = 30 months = 2.5 years. Hence, options 1 and 2 are eliminated.

Considering the loan with the microfinance firm:

62100 − P = (P × 8 × 2.5)/100

∴ 62100 = 1.2P i.e. P = Rs. 51,750

Hence, option 3.

Alternatively,

Consider the loan with the microfinance firm. Let the amount borrowed be Rs. P for n years.

∴ 62100 − P = (P × 8 × n)/100

∴ 62100 = P(1 + 0.08n)

Now, observe that there are only two values of n (2 and 2.5) in the options). Substitute each value in the above equation and check if the Principal value given in that option is obtained.

When n = 2; P = 62100/1.16 = Rs. 53,535 (approximately). SInce this value is not in the options, this case is invalid.

When n = 2.5; P = 62100/1.2 = Rs. 51,750

Hence, option 3.

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**2. IIFT 2017 QA | Geometry - Mensuration**

A rectangular plank √10 m wide, is placed symmetrically along the diagonal of a square of side 10 m as shown in the figure. The area of the plank is:

- A.
10(√20 − 1) sq.m

- B.
10(√5 − 1) sq.m

- C.
10√20 − 1 sq.m

- D.
None of the above

Answer: Option A

**Explanation** :

Consider the figure given in the question.

The two small triangles formed in the region between the vertex of the square and the width of the plank (top left corner) will be congruent to each other. Also, because they are congruent and because the diagonal of the square is the angle bisector of the angle at the vertex, each of these two triangles will be a 45-45-90 triangle.

∴ Height of small triangle = width of small triangle/2 = √10/2

By symmetry, this will also be the height of the other small triangle (bottom right corner).

Diagonal of the square = 10√2 m

∴ Length of plank = 10√2 − (√10/2) − (√10/2) = 10√2 − √10 = √10(√20 − 1) m

Width of plank = √10 m

∴ Area of plank = √10 × √10(√20 − 1) = 10(√20 − 1) sq.m

Hence, option 1.

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**3. IIFT 2017 QA | Arithmetic - Time & Work**

Somesh, Tarun and Nikhil can complete a work separately in 45, 60 and 75 days. They started the work together but Nikhil left 5 days after the start and Somesh left 2 days before the completion of the work. In how many days will the work be completed?

- A.
$25\frac{1}{7}$

- B.
$50\frac{1}{7}$

- C.
$35\frac{5}{7}$

- D.
$40\frac{5}{7}$

Answer: Option A

**Explanation** :

Let the total work be equivalent to 900 units and let Somesh, Tarun and Nikhil respectively complete a, b and c units of work per day.

∴ a = 900/45 = 20; b = 15 and c = 12

Let the work be completed in n days.

For the first five days, all three people were working. For the last two days, only Tarun was working.

Hence, for the remaining n − 5 − 2 i.e. n − 7 days in between, both Somesh and Tarun were working.

Hence, considering amount of work done in each day:

5(20 + 15 + 12) + (n − 7)(20 + 15) + 2(15) = 900

∴ 235 + 35n − 245 + 30 = 900

∴ 35n = 880 i.e. n = 25.14 days

Hence, option 1.

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**4. IIFT 2017 QA | Algebra - Logarithms**

$(1+5){\mathrm{log}}_{e}3+\frac{(1+{5}^{2})}{2!}{\left({\mathrm{log}}_{e}3\right)}^{2}+\frac{\left(1+{5}^{3}\right)}{3!}{\left({\mathrm{log}}_{e}3\right)}^{3}+...$

- A.
12

- B.
244

- C.
243

- D.
245

Answer: Option B

**Explanation** :

Consider log_{e}3 = x.

Hence, the series can also be written as

$\left(x+\frac{{x}^{2}}{2!}+\frac{{x}^{3}}{3!}+\frac{{x}^{4}}{4!}+...\right)+\left(5x+\frac{{\left(5x\right)}^{2}}{2!}+\frac{{\left(5x\right)}^{3}}{3!}+...\right)$

Now, each bracket is a standard expression that can be expressed as a power of e. Hence, the expression becomes:

$({e}^{x}-1)+({e}^{5x}-1)={e}^{{\mathrm{log}}_{e}3}-1+{e}^{5{\mathrm{log}}_{e}3}-1$

= 3 – 1 + 3^{5} – 1 = 1 + 243 = 244

Hence, option 2.

**Note:**

****This is one of the rare questions where pure engineering maths based concepts have been used. In the exam, you are advised to leave such a question if you are not conversant with such series.

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**5. IIFT 2017 QA | Algebra - Functions & Graphs**

f f(x) = 1/(1 + x), then find the value of f[f{f(x)}] at x = 5.

- A.
7/9

- B.
7/13

- C.
5/13

- D.
5/9

Answer: Option B

**Explanation** :

f(5) = 1/(1 + 5) = 1/6

f[f(5)] = f(1/6) = 1/[1 + (1/6)] = 6/7

f[f{f(5)}] = f(6/7) = 1/[1 + (6/7)] = 7/13

Hence, option 2.

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**6. IIFT 2017 QA | Arithmetic - Mixture, Alligation, Removal & Replacement**

A pharmaceutical company produces two chemicals X and Y, such that X consists of 5% salt A and 10% salt B and Y consists of 10% salt A and 6% salt B. For producing the chemicals X and Y, the company requires at least 7 gm of Salt A and at least 7 gm of salt B. If chemicals X costs Rs. 10.50 per gm and chemical Y costs Rs. 7.80 per gm, what is the minimum cost at which the company can meet the requirement by using a combination of both types of chemicals?

- A.
Rs. 810

- B.
Rs. 850

- C.
Rs. 537

- D.
None

Answer: Option A

**Explanation** :

Let the total quantity of chemicals A and B be a and b respectively.

∴ Quantity of salt A required = (0.05a + 0.1b) and Quantity of salt B required = (0.1a + 0.06b)

Atleast 7 gms of each salt need to be present.

∴ 0.05a + 0.1b ≥ 7 and 0.1a + 0.06b ≥ 7

On solving, b ≥ 50 and a ≥ 40

Consider the minimum quantities required of A and B as cost is to be minimized.

∴ Minimum cost = (10.5)(40) + (7.8)(50)

= 420 + 390 = Rs. 810

Hence, option 1.

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**7. IIFT 2017 QA | Arithmetic - Ratio, Proportion & Variation**

Suntex Company plans to manufacture a new product line of Razor next year and sell it at a price of Rs. 12 per unit. The variable costs per unit in each production run is estimated to be 50% of the selling price, and the fixed costs for each production run is estimated to be Rs. 50,400. Based om their estimated costs how many units of the new product will company Suntex need to manufacture and sell in order for their revenue to be equal to their total costs for each production run?

- A.
5400

- B.
4200

- C.
8400

- D.
2100

Answer: Option C

**Explanation** :

Let n units be sold.

Variable cost per unit = 0.5 × 12 = Rs. 6 per unit

∴ Total cost for one production run = Rs. (50400 + 6n)

Total revenue by selling n units = Rs. (12n)

When cost = revenue; 50400 + 6n = 12n

∴ 6n = 50400 i.e. n = 8400

Hence, option 3.

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**8. IIFT 2017 QA | Algebra - Progressions**

In a certain sequence the term x_{n} is given by formula ${x}_{n}=5{x}_{n-1}-\frac{3}{4}{x}_{n-\mathrm{2\; for\; n\; \ge \; 2.}}$

What is the value of x_{3} if x_{0} = 4 and x_{1} = 2?

- A.
67/2

- B.
37/2

- C.
123/4

- D.
None

Answer: Option A

**Explanation** :

*x*_{2} = 5*x*_{1} − (3/4)(*x*_{0}) = 5(2) − (3/4)(4) = 7

∴ *x*_{3} = 5*x*_{2} − (3/4)(*x*_{1}) = 5(7) − (3/4)(2) = 67/2

Hence, option 1.

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**9. IIFT 2017 QA | Arithmetic - Percentage**

A mobile company that sells two models ACN-I and ACN-II of mobile, reported that revenues from ACN-I in 2016 were down 12% from 2015 and revenue from ACN-II sales in 2016 were up by 9% from 2015. If the total revenues from sales of both the mobile models ACN-I and ACN-II in 2016 were up by 3% from 2015, what is the ratio of revenue from ACN-I sales in 2015 to revenue from ACN-II sales in 2015?

- A.
5 : 2

- B.
2 : 5

- C.
3 : 4

- D.
None of the above

Answer: Option B

**Explanation** :

Let the revenues of ACN-I and ACN-II in 2015 be a and b respectively.

∴ Total revenues in 2015 = (a + b)

Since total revenues grew by 3% in 2016, total revenue in 2016 = 1.03(a + b).

Also, revenue of ACN-I dropped by 12% while those of ACN-II increased by 9%

∴ Revenue of ACN-I in 2016 = 0.88a and Revenue of ACN-II in 2016 = 1.09b

∴ Total revenue in 2016 = (0.88a + 1.09b)

∴ 0.88a + 1.09b = 1.03(a + b)

∴ 0.06b = 0.15a i.e. a : b = 2 : 5

Hence, option 2.

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**10. IIFT 2017 QA | Algebra - Number System**

If 10^{67} – 87 is written as an integer in base 10 notation, what is the sum of digits in that integer?

- A.
683

- B.
489

- C.
583

- D.
589

Answer: Option D

**Explanation** :

10^{67} is nothing but 10...00000 (67 times) i.e. 1 followed by 67 0s.

When a two-digit number is subtracted from it, the last two 0s get replaced by (100 − that two-digit number number) while the remaining 0s get replaced by 9s due to carry of the subtraction.

Hence, when 87 is subtracted from a number having 67 0s, we get 67 − 2 = 65 9s and the last two digits as 100 − 87 = 13.

∴ Sum of digits = 65(9) + 1 + 3 = 589

Hence, option 4.

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**11. IIFT 2017 QA | Geometry - Trigonometry**

A flag pole on the top of a mall building is 75 m high. The height of the mall building is 325 m. To an observer at height of 400 m, the mall building and the pole subtend an equal angle Ө. If the horizontal distance of the observer from the pole is x, what is the value of x?

- A.
20√10 m

- B.
30√10 m

- C.
25√5 m

- D.
None

Answer: Option B

**Explanation** :

Consider the layout of the mall, flag pole and oberver as shown below.

Let the horizontal distance from the oberver to the pole be *d* m.

**Considering pole: ** tan *θ* = 75/*d* ... (i)

**Considering mall building:** tan 2*θ* = 400/*d* ... (ii)

∴ (2tan*θ*) / (1 − tan^{2}*θ*) = 400/*d*

∴ 2(75/*d*) = (400/*d*) × [1 − (75/*d*)^{2}]

∴ 1 − (5625/*d*^{2}) = 3/8

∴ (5625/*d*^{2}) = 5/8

∴ *d*^{2} = 5625 × (8/5) = 9000

∴ *d* = 900 × 10 = 30√10 m

Hence, option 2.

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**12. IIFT 2017 QA | Modern Math - Probability**

Writex Brown, an E-commerce company gives home delivery of its valuable products, after receiving final order in their website, by different modes of transportation like bike, scooter, tempo and truck. The probabilities of using bike, scooter, tempo and truck are respectively 2/9, 1/9, 4/9, and 2/9. The probabilities of delivering the product late to the destination by using these modes of transport are 3/5, 2/5, 1/5, and 4/5. If the product reaches the destination in time, find the probability that the company has used scooter to reach the destination.

- A.
1/10

- B.
4/25

- C.
3/25

- D.
None of these

Answer: Option C

**Explanation** :

This is a case of conditional probability.

The product being delivered on time implies no delay in delivering the product.

P(Scooter | No Delay) = [P(Scooter and No Delay Using Scooter] / [P(No Delay)]

[P(Scooter and No Delay] = P(Scooter) × P(No Delay Using Scooter)

= (1/9) × [1 − (2/5)] = (1/9) × (3/5) = 3/45

P(No delay) = [(2/9) × (2/5)] + [(1/9) × (3/5)] + [(4/9) × (4/5)] + [(2/9) × (1/5)]

= (4/45) + (3/45) + (16/45) + (2/45) = 25/45

∴ Required probability = (3/45) / (25/45) = 3/25

Hence, option 3.

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**13. IIFT 2017 QA | Geometry - Mensuration**

A pest control person uses a particular machine for his job. It moves along the circumference of a circular hall of radius 49 meters in 148 minutes to finish the pest control. How many minutes more will it take him to move along the perimeter of a hexagon of side 54 meters?

- A.
7.69 minutes

- B.
14.36 minutes

- C.
14.00 minutes

- D.
4.28 minutes

Answer: Option A

**Explanation** :

Circumference of circular hall = 2 × (22/7) × (49) = 308 m

∴ Speed of machine = (308/148) = (77/37) m/min

Perimeter of hexagon = 54 × 6 = 324 m

∴ Time taken to move along hexagon = (324) / (77/37)

= (324 × 37) / 77 = 155.69 minutes

∴ Extra time required = 155.69 − 148 = 7.69 minutes

Hence, option 1.

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**14. IIFT 2017 QA | Venn Diagram**

A premier B-school, which is in process of getting an AACSB accreditation, has 360 second year students. To incorporate sustainability into their curriculum, it has offered 3 new elective subjects in the second year namely Green Supply Chain, Global Climate Change & Business and Corporate Governance. Twelve students have taken all the three electives, and 120 students study Green Supply Chain. There are twice as many students who study Green Supply Chain and Corporate Governance but not Global Climate Change and Business, as those who study both Green Supply Chain and Global Climate Change & Business but not Corporate Governance, and 4 times as many who study all the three. 124 students study Corporate Governance. There are 72 students who could not muster up the courage to take up any of these subjects. The group of students who study both Green Supply Chain and Corporate Governance but not global Climate Change & Business is exactly the same as the group made up to the students who study both Global Climate Change & Business and Corporate Governance. How many students study Global Climate Change & Business only?

- A.
176

- B.
104

- C.
152

- D.
188

Answer: Option B

**Explanation** :

The number of students who study each combination of subjects (based on the direct data) given is as shown below:

It is given that: (GSC and CG but not GCCB) = 4 times (all three electives)

∴ 2x = 4(12) i.e. x = 24

Also: (GSC and CG but not GCCB) = (all three electives) + (GCCB and CG but not GSC)

∴ (GCCB and CG but not GSC) = 2x − 12 = 2(24) − 12 = 36

So, the figure becomes:

Now, CG only = 124 − (48 + 12 + 36) = 28

∴ GCCB alone = 360 − 120 − 36 − 28 − 72 = 104

Hence, option 2.

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**15. IIFT 2017 QA | Arithmetic - Time, Speed & Distance**

Ramesh and Sohan start walking away from each other from a point P at an angle of 120°. Ramesh walks at a speed of 3 kmph while Sohan walks at a speed of 4 kmph. What is the distance between them after 90 minutes?

- A.
9.89 km

- B.
10.56 km

- C.
9.12 km

- D.
12.42 km

Answer: Option C

**Explanation** :

Let Ramesh and Sohan reach points R and S respectively after 90 minutes.

∴ PR = distance travelled by Ramesh in 90 minutes = 1.5 × 3 = 4.5 km

Similarly, PS = 1.5 × 4 = 6 km

Now, PRS forms a triangle, where two sides (PR and PS) as well as the included angle (RPS = 120°) are known. Hence, the third side can be found using the cosine rule.

∴ Cos 120 = [(4.5)^{2} + (6)^{2} − (RS)^{2}] / (2 × 4.5 × 6)

∴ −0.5 = [20.25 + 36 − (RS)^{2}] / 54

∴ 56.25 − (RS)^{2 }= −27

∴ (RS)^{2 }= 83.25 i.e. RS = 9.12

Hence, they are 9.12 km apart.

Hence, option 3.

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**16. IIFT 2017 QA | Geometry - Circles**

A chord AB of length 24 cm is drawn in a circle of radius 13 cm. Find the area of the shaded portion APB.

- A.
13π x cm

^{2} - B.
$\frac{13\pi x}{180}$cm

^{2} - C.
$\frac{169\pi x}{360}-60c{m}^{2}$

- D.
$\frac{169\pi x}{180}-60c{m}^{2}$

Answer: Option C

**Explanation** :

Area of sector O-APB = (x/360) × π × (13)^{2} = 169π*x*/360 cm^{2}

Let the perpendicular from O to AB meet AB at M. Hence, OM is the perpendicular bisector of AB.

Hence, AM = 24/2 = 12 cm and OM = 5 cm (5, 12, 13 form a pythagorean triplet in right triangle OAM).

∴ Area of triangle OAB = (1/2) × OM × AB = (1/2) × 5 × 24 = 60 cm^{2}

∴ Required area = [(169π*x*/360) − 60] cm^{2}

Hence, option 3.

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**17. IIFT 2017 QA | Geometry - Circles**

Two tangents are drawn from a point P on the circle with centre at O, touching the circle at point Q and T respectively. Another tangent AB touches the circle at point S. If angle QPT=55°, find the angle AOB=?

- A.
125°

- B.
62.5°

- C.
97.5°

- D.
95°

Answer: Option B

**Explanation** :

In quadrilateral PQOT, ∠QOT = 360 − ∠QPT − ∠PQO − ∠PTO

Since PQ and PT are tangents to the circle at Q and T, ∠PQO = ∠PTO = 90°

∴ ∠QOT = 360 − 55 − 90 − 90 = 125°

∴ ∠QOS + ∠SOT = ∠QOT = 125°

Now, when two tangents are drawn from a point to the circle, the line joining the external point and the centre of the circle is the angle bisector of the central angle subtended by the two tangential points at the centre.

For instance, when PQ and PT are tangents from P to the circle with centre O, then line PO is the angle bisector of ∠QOT.

Similarly, AQ and AS are tangents from A to the circle with centre O. Hence, AO is the angle bisector of ∠QOS.

∴ ∠QOA = ∠AOS = (∠QOS)/2

Similarly, for quadrilateral BSOT, ∠TOB = ∠BOS = (∠SOT)/2

Now, ∠AOB = ∠AOS + ∠BOS

= (∠QOS)/2 + (∠SOT)/2 = (∠QOS + ∠SOT)/2

= 125/2 = 62.5°

Hence, option 2.

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**18. IIFT 2017 QA | Geometry - Coordinate Geometry**

The coordinates of a triangle ABC are A (1, 5), B (−2, 3), and C(0, −4). What is the equation of the median AD?

- A.
7x − 3y + 8 = 0

- B.
5x − 4y + 15 = 0

- C.
x + 3y − 16 = 0

- D.
11x − 4y + 9 = 0

Answer: Option D

**Explanation** :

The co-ordinates of D are {[(−2 + 0)/2], [(3 − 4)/2]} = (−1, − 0.5)

Using the two point form, the equation of median AD is:

(y − 5) / (x − 1) = (−0.5 − 5) / (−1 − 1)

∴ (y − 5) / (x − 1) = 5.5 / 2 = 11 / 4

∴ 4y − 20 = 11x − 11

∴ 11x − 4y + 9 = 0

Hence, option 4.

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**19. IIFT 2017 QA | Arithmetic - Mixture, Alligation, Removal & Replacement**

The Drizzle Pvt. Ltd., a squash company has 2 cans of juice. The first contains 25% water and the rest is fruit pulp. The second contains 50% water and rest is fruit pulp. How much juice should be mixed from each of the containers so as to get 12 litres of juice such that the ratio of water to fruit pulp is 3 : 5?

- A.
6 litres, 6 litres

- B.
4 litres, 8 litres

- C.
5 litres, 7 litres

- D.
9 litres, 3 litres

Answer: Option A

**Explanation** :

Quantity of water in 12 litres of juice = (3/8) × 12 = 4.5 litres

Quantity of fruit pulp in 12 litres = 12 − 4.5 = 7.5 litres

Let x litres of the first can and y litres of the second can be used.

∴ Quantity of water used in mixture = (0.25x + 0.5y) and quantity of pulp used = (0.75x + 0.5y)

Comparing the expressions for quantity of water and pulp in the mixtures:

0.25x + 0.5y = 4.5 ... (i)

0.75x + 0.5y = 7.5 ... (ii)

On solving, x = 6 and y = 6

Hence, option 1.

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**20. IIFT 2017 QA | Arithmetic - Time & Work**

An overhead tank, which supplies water to a settlement, is filled by three bore wells. The first two bore wells operating together fill the tank in the same time as taken by the third bore well to fill it. The second bore well fills the tank 10 hours faster than the first one and 8 hours slower than the third one. The time required by the third bore well to fill the tank alone is:

- A.
9 hours

- B.
12 hours

- C.
18 hours

- D.
20 hours

Answer: Option B

**Explanation** :

Let the first bore well fill the tank in x hours.

Hence, the second fills it in (x − 10) hours and the third fills it in (x − 10) − 8 = (x − 18) hours.

Let the total capacity of the tank to be filled be (x)(x − 10)(x − 18) and let the individual quantity filled by the three respective bore wells be a, b and c.

∴ a = (x − 10)(x − 18); b = (x)(x − 18); c = (x)(x − 10)

Since the first two bore wells fill the tank in the same time as the third bore well alone, amount of work done by first two bore wells per hour is the same as amount of work done by the third bore well per hour.

i.e. a + b = c

∴ (x − 10)(x − 18) + (x)(x − 18) = (x)(x − 10)

∴ (x − 18)(2x − 10) = (x)(x − 10)

∴ 2x^{2} − 46x + 180 = x^{2} − 10x

∴ x^{2} − 36x + 180 = 0

∴ x = 30 or x = 6

Since (x − 18) is number of hours, this value has to be positive. Hence, x > 18 i.e. x = 30

∴ Time taken by third borewell = (x − 18) = 12 hours

Hence, option 2.

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