IIFT 2017 QA | Previous Year IIFT Paper

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 (c). 

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 (c).

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 (a).

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 (a).

Answer: Option B

Explanation :

Consider log_e3 = 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:

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

= 3 – 1 + 3⁵ – 1 = 1 + 243 = 244

Hence, option (b).

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.

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 (b).

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 (a).

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 (c).

Answer: Option A

Explanation :

x₂ = 5x₁ − (3/4)(x₀) = 5(2) − (3/4)(4) = 7 

∴ x₃ = 5x₂ − (3/4)(x₁) = 5(7) − (3/4)(2) = 67/2

Hence, option (a).

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 (b).

Answer: Option D

Explanation :

10⁶⁷ 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 (d).

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 2θ = 400/d ... (ii)

∴ (2tanθ) / (1 − tan²θ) = 400/d 

∴ 2(75/d) = (400/d) × [1 − (75/d)²] 

∴ 1 − (5625/d²) = 3/8

∴ (5625/d²) = 5/8 

∴ d² = 5625 × (8/5) = 9000 

∴ d = 900 × 10 = 30√10 m

Hence, option (b).

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 (c).

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 (a).

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 (b).

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 (c).

Answer: Option C

Explanation :

Area of sector O-APB = (x/360) × π × (13)² = 169πx/360 cm²

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 cm²  

∴ Required area = [(169πx/360) − 60] cm²

Hence, option (c).

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 (b).

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 (d).

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 (a).  

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)

∴ 2x² − 46x + 180 = x² − 10x

∴ x² − 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 (b).