IIFT 2018 QA | Previous Year IIFT Paper

Answer: Option B

Explanation :

Let the initial amount (in Rs.) with companies X, Y and Z be x, y and z respectively.

X has given an amount equivalent to 10% of the amount originally with Y. Hence, X has given Rs. (0.1y) to Y.

∴ Amount now with Y = y + 0.1y = Rs. (1.1y)

The amount with Y is twice the amount with Z. Hence, the amount with Z is Rs. (0.55y).

Z gave Rs. 10,000 i.e. 5% of his funds to the Trust.

∴ 0.05 × 0.55y = 10000

Amount given by Y = 0.05y

= 0.05 × (10000)/(0.55 × 0.05)

= 10000/0.55

= 18181 i.e. approx. Rs. 18,000

Hence, option (b).

Answer: Option B

Explanation :

The radius of the bucket is 3 cm and the radius of the original water level is (3 + 1) = 4 cm

Original volume = 200 cc

Let the original height of water = h cm

Using the formula for volume of frustum of cone:

$200=\frac{h}{3}\pi [4^2+3^2+(4\left)\right(3\left)\right]$

∴ h = 600/37π

Now, the height of the liquid increases by 0.3 times itself i.e. it becomes 1.3h

In a frustum of cone, ratio of radii is the same as ratio of heights (applying the concepts of similarity).

∴ New radius = 4.3 cm

∴ Volume of ball = new volume – original volume

​​​​​​​$=\frac{1.3h}{3}\pi [4.3^2+3^2+(4.3\left)\right(3\left)\right]-200$

$=\frac{1.3h}{3}\times \pi \times \frac{600}{37\pi }\times 40.4-200$

= 283.9 - 200 = 83.9 cc

The closest value in the options is 80 cc.

Hence, option (b).

Answer: Option D

Explanation :

The base has to be on one of the four lines.

Hence, the base can first be chosen in 4 ways.

Now, on this base, two points need to chosen from five to form the actual base. This can be done in ⁵C₂ ways i.e. 10 ways

For the third point, one of three lines has to be chosen and then one of five points from each line has to be chosen.

∴ Number of possible triangles = 4 × 10 × 3 × 5 = 600

Hence, option (d).

Answer: Option A

Explanation :

Since S is the midpoint of QR, A(∆PSR)

= A(∆PQR)/2

= 5.8/2 = 2.9 sq.cm

Now A(∆PSR) = A(∆PSX) + A(∆SXR)

Also, A(∆RTX) = A(∆TSX) + A(∆SXR)

Now, triangles PSX and TSX lie within the same parallel lines – SX and PT – and hence, have the same area.

∴ A(∆PSX) = A(∆TSX)

∴ A(∆PSR) = A(∆RTX) = 2.9 sq.cm

Hence, option (a).

Answer: Option B

Explanation :

Q(9, 4) = (9)² – (9)(4) + (4)²

= 81 – 36 + 16 = 61

∴ P(7,Q(9,4)) = P(7,61)

= (7)² + (7)(61) + (61)²

= 49 + 427 + 3721 = 4197

Hence, option (b).

Answer: Option C

Explanation :

PA = QB and angles A and B are right angles.

Hence, PABQ is a rectangle.

∴ Required area

= A(rectangle PABQ) – A(segment O-PRQ)

= A(rectangle PABQ) – [A(sector O-PRQ) – A(∆OPQ)]

= A(rectangle PABQ) – A(sector O-PRQ) + A(∆OPQ)

Now, PQ = AB = 25√2 cm and ∠POQ = 90°

∴ Radius of sector = PO = OQ = PQ/√2 = 25 

Now, A(∆OPQ) = (1/2) × OP × OQ
= (1/2) × 25 × 25 = 625/2 sq.cm

A(sector O-PRQ) = (90/360) × π × (OP)²
= (π/4) × (25)² = (625π/4)

Now, height of ∆OPQ = 25/√2

Distance from O to AB is 30 cm

∴ PA = QB = 30 – (25/√2) cm

∴ A(rectangle PABQ) = PA × AB

= [30 – (25/√2)] × (25√2) = 750√2 – 625 sq.cm

∴ Required area = $(750\sqrt{2}-625)$ - $\frac{625\mathrm{\pi }}{4}$ + $\frac{625}{2}$

= 750√2 - 625/2 - 625π/4

= 750√2 - 625$\left(\frac{1}{2}+\frac{\mathrm{\pi }}{4}\right)$

Hence, option (c).

Answer: Option B

Explanation :

The roots of the equation y² – 8y + 14 = 0 are α and β.

∴ Sum of roots = α + β = −(−8)/1 = 8 and Product of roots = αβ = (14)/1 = 14

∴ (1 + α + β²) (1 + β + α²) = 1 + α + β² + β + αβ + β³ + α² + α³ + α²β² 

= 1 + (α + β) + (αβ) + (αβ)² + (α² + β²) + (α³ + β³)

= 1 + (α + β) + (αβ) + (αβ)² + (α + β)² – (2αβ) + (α + β)³ – (3αβ)(α + β)

= 1 + 8 + 14 + (14)2 + (8)² – 2(14) + (8)³ – 3(14)(8)

= 23 + 196 + 64 – 28 + 512 – 336 = 431

Hence, option (b).

Answer: Option B

Explanation :

Since there is no condition imposed on x, y and z, assume x = y = z = 10

Consider log_xyz + 1.

When x = y = z = 10; log_xyz + 1

= log₁₀(10 × 10) + 1

= log₁₀(10)² + 1 = 2 log₁₀(10) + 1

= 2(1) + 1 = 3

Similarly, log_yxz + 1 = log_zxy + 1 = 3

∴ Required value = (1/3) + (1/3) + (1/3) = 1

Hence, option (b).

Answer: Option C

Explanation :

Let the total work be 180 units (LCM of 9, 12 and 15) and let Ram, Ravi and Ratan do a, b and c units of work per day.

∴ a = 180/9 = 20; b = 180/12 = 15 and c

= 180/15 = 12

Ram works alone on Monday, Ravi works alone on Tuesday and Ratan works alone on Wednesday.

∴ Work done in three days = a + b + c

= 20 + 15 + 12 = 47

∴ Work done in nine days = 3(47)

= 141 units

On the 10th day, Ram does 20 units, thereby taking the completed total work to 141 + 20 = 161 units.

On the 11th day, Ravi does 15 units, hence, taking the completed total work to 176 units.

The work gets completed on the 12th day.

Hence, Ravi has done work equivalent to four days when the work is completed.

∴ Ravi’s total work = 4(15) = 60 units

∴ Required proportion = 60/180 = 1/3

Hence, option (c).

Answer: Option C

Explanation :

Here, the area of each square and each circle will keep on increasing. Hence, each A_n will keep on increasing.

Hence, if the value of n is not known, the ratio will not be found.

Here, the value of n is unknown. Hence, the value of the ratio cannot be determined.

Hence, option (c).

Answer: Option B

Explanation :

Since Joseph takes less time while crossing the path diametrically rather than in a semi-circular manner, the distance travelled along the diameter is less.

Let the radius of the path be r m.

∴ Difference in distance = πr – 2r

= (π – 2)r m

Also, difference in distance = difference in time × constant speed

∴ (π – 2)r = 48 × (50/60)

∴ 1.14r = 40 i.e. r = 40/(8/7) = 35 m

∴ Diameter = 2r = 70 m

Hence, option (b).

Answer: Option C

Explanation :

Let Garima initially have (2n) notes of Rs. 2,000.

After the first transaction (in which she gives away half her notes), Garima has n notes of Rs. 2,000.

Now, Garima gets 30 notes of Rs. 2,000 which increases her number of Rs. 2,000 notes by three-fourth.

∴ n + 30 = n + (3n/4)

∴ (3n/4) = 30 i.e. n = 40

∴ Initial number of Rs. 2,000 notes = 2n = 80

This eliminates options 1 and 2.

Let Garima initially have x notes of Rs. 200.

She then gets n notes of Rs. 200, thereby taking the total notes of Rs. 200 to 50.

∴ x + n = 50

∴ x + 40 = 50 i.e. x = 10

∴ Initial number of Rs. 200 notes = x = 10

Hence, option (c).

Answer: Option D

Explanation :

Volume of original solid = volume of cylinder + volume of cone

Radius of original cylinder = radius of cone = 5 cm

Let height of the original cylinder be 3x cm. Hence, height of cone = 2x cm

Volume of solid = $\pi \left(5^2\right)\left(3x\right)+\frac{\pi }{3}\left(5^2\right)\left(2x\right)$

$=25\pi \times \frac{11x}{3}=\frac{275\pi x}{3}$

Let the radius of the cylindrical hole be r cm.

Also, height of cylindrical hole = (2/3)(5x) = (10x/3) cm

∴ Volume of hole = $\pi \left(r^2\right)\frac{10x}{3}$

Hence, as per the given condition:

$\frac{\pi \left(r^2\right){\displaystyle \frac{10x}{3}}}{{\displaystyle \frac{275x}{3}}-\pi \left(r^2\right){\displaystyle \frac{10x}{3}}}=\frac{1}{3}$

$\therefore \frac{10r^2}{275-10r^2}=\frac{1}{3}$

∴ 275 – 10r² = 30r²
∴ 40r² = 245

$\therefore r^2=\frac{245}{40}=\frac{55}{8}$

$\therefore r=\sqrt{\left(\frac{55}{8}\right)}$

Hence, option (d).

Answer: Option C

Explanation :

Tab B takes 1.5 times more than tap A to fill the tank.

A takes 12 hours to fill the tank alone.

Hence, time taken by tap B alone = 1.5 × 12 = 18 hours

Let the total capacity of the tank be 180 litres.

Let taps A and B fill a and b litres per hour and let tap C empty c litres per hour.

∴ a = 180/12 = 15; b = 180/18 = 10 and c = 180/36 = 5

For the first two hours, only A and C are working; for the next four hours, all three of A, B and C are working and for the next n hours, only A and B are working.

∴ 2(a – c) + 4(a + b – c) + n(a + b) = 180

∴ 2(15 – 5) + 4(15 + 10 – 5) + n(15 + 10) = 180

∴ 2(10) + 4(20) + 25n = 180

∴ 25n = 180 – 20 – 80 = 80

∴ n = 80/25 = 3.2 hours

∴ Total time = 6 + 3.2 = 9.2 hours

i.e. 9 hours and 12 minutes

Hence, option (c).

Answer: Option A

Explanation :

Let height of the mountain = x

From the figure, AD = distance travelled by man = 100 m

Also, DE = AD × sin 30° = 100 × 0.5 = 50 m

Also, AE = AD × cos 30° = 100 × (√3/2) = 50√3 m

Now, ∠CAB = 45°

Hence, ABC is a 45-45-90 triangle and AB = BC = x

Also, EB = AB – AE = (x − 50√3) m

∴ DF = EB = (x − 50√3) m

Similarly, DE = FB = 50 m

∴ CF = BC – FB = (x – 50) m

Now, in ∆CDF, tan 60° = CF/DF

∴ √3 = (x – 50)/(x − 50√3)

∴ √3x – 50(3) = x − 50

∴ (√3 – 1)x = 150 – 50 = 100

∴ x = 100/((√3 – 1) = 50(√3 + 1) [Rationalising]

Hence, option (a).

Answer: Option B

Explanation :

Total cost incurred = total cost incurred due to petrol + total cost incurred due to driver charges

= (total litres consumed × cost per litre) + (total driver hours × cost per hour)

Total petrol consumed in 1000 km = 1000 × $\frac{1}{400}\left\{\frac{1000}{x}+x\right\}=2.5\left\{\frac{1000}{x}+x\right\}$

∴ Total petrol cost = $2.5\left\{\frac{1000}{x}+x\right\}\times 65$

$=162.5\times \left\{\frac{1000}{x}+x\right\}$

Time taken to travel 1000 km = 1000/x

∴ Total driver cost = (1000/x) × 50 = Rs. (50000/x)

∴ Total cost = 162.5 × $\left\{\frac{1000}{x}+x\right\}+\left(\frac{50000}{x}\right)$

$=\left(\frac{1000}{x}\right)(162.5+50)+162.5x$

$=\frac{212500}{x}+162.5x$

You can now substitute options and verify that the least cost is for x = 36.

Hence, option (b).

Answer: Option A

Explanation :

Let there be 100 people surveyed in all.

Let a, b, c correspond to people who watched only Star Plus, only Sony and only SAB TV.

Let d, e, f correspond to people who watched only Star Plus and Sony, only Sony and SAB TV and only SAB TV and Star Plus.

Let g correspond to people who watched all three channels.

73% people i.e. 73 people watched atleast one channel.

∴ a + b + c + d + e + f + g = 73 … (i)

Number of people watching Star Plus, Sony and SAB TV is 38%, 39% and 23% respectively i.e. 38, 39 and 23.

∴ a + d + f + g = 38 … (ii)

b + d + e + g = 39 … (iii)

c + e + f + g = 23 … (iv)

Adding (i), (ii) and (iii):

(a + b + c) + 2(d + e + f) + 3g = 39 + 38 + 23 = 100 … (v)

Solving (i) and (v), we get:

(d + e + f) + 2g = 100 – 73 = 27

Number of people surveyed who watch more than one channel = (d + e + f) + g

Also, number of people who watch all channels = g = 11

∴ (d + e + f) + g + 11 = 27

∴ (d + e + f) + g = 16

∴ Required % = (16/100) × 100 = 16%

Hence, option (a).

Answer: Option C

Explanation :

P(playing on artificial turf) = 0.4

∴ P(playing on grass) = 1 – P(playing on artificial turf) = 1 – 0.4 = 0.6

P(injury/artificial turf) = 0.42

Also, P(injury/artificial turf) = 1.5 × P(injury/grass)

∴ P(injury/grass) = 0.42/1.5 = 0.28

Now, P(injury) = P(injury/grass) × P(playing on grass) + P(injury/artificial turf) × P(playing on artificial turf)

= (0.28 × 0.6) + (0.42 × 0.4)

= 0.168 + 0.168 = 0.336

Also, P(injury/grass) × P(playing on grass) = P(grass/injury) × P(injury)

∴ P(grass/injury) = 0.168/0.336 = 0.5

Hence, option (c).

Answer: Option A

Explanation :

Instead of solving algebraically, solve the question by first substituting x = 1. You will get one original value and three values in the options. Square each value from the options and match it with the original value of the expression.

Original expression:$1+x^2+\sqrt{(1+x^2+x^4)}$

$=1+{\left(1\right)}^2+\sqrt{[1+{\left(1\right)}^2+{\left(1\right)}^4]}$

$=1+1+\sqrt{3}=2+\sqrt{3}$

Option 1:

$\frac{1}{\sqrt{2}}\left[\sqrt{1+x+x^2}+\sqrt{1-x+x^2}\right]$

$=\frac{1}{\sqrt{2}}\left[\sqrt{1+1+1^2}+\sqrt{1-1+1^2}\right]$

$=\frac{\sqrt{3}+1}{\sqrt{2}}$

Now, square this value.

$=\frac{3+1+2\sqrt{3}}{2}=\frac{4+2\sqrt{3}}{2}=2+\sqrt{3}$

Hence, option (a).

Answer: Option B

Explanation :

log₂x × log(x/64)² = log(x/16)²

$\therefore \frac{\log x}{\log 2}\times \frac{\log 2}{\log \left({\displaystyle \frac{x}{64}}\right)}=\frac{\log 2}{\log \left({\displaystyle \frac{x}{16}}\right)}$

It is clear that x has to be a power of 2 to solve the equation i.e. x ≠ 12.

Hence, option (c) is eliminated.

If x = 16, then the base of the RHS (i.e. x/16) becomes 1, which is not possible for a logarithm i.e. x ≠ 16. Hence, option (d) is eliminated.

If x = 2, then you get (x/64) = (x/16); which is not possible.

Hence, x = 4

Hence, option (b).