IIFT 2013 Question Paper

Let $$x$$(in km) be the distance from the home to the clinic
Thus, the time taken to reach the clinic = $$\dfrac{x}{30}$$ hours
They stay at the clinic for $$\dfrac{48}{60} = \dfrac{4}{5}$$ hours
The time taken to reach home from clinic = $$\dfrac{x}{12}$$ hours
Total time = 5 hours
Thus, $$\dfrac{x}{30}+\dfrac{4}{5}+\dfrac{x}{12}$$ = $$5$$
=> $$\dfrac{2x}{60}+\dfrac{48}{60}+\dfrac{5x}{60}$$ = $$5$$
Thus, $$300 = 7x+48$$
Thus, $$x=36$$
Hence, the doctor's clinic is 36 kms away.
Hence, option C is the correct answer.

Let x and y be the percentage of aluminum in the 8 kg and 16 kg piece.
Let us take a kg away from both.
Thus, we will have xa and ya amount of aluminum in both the pieces.
In the remaining 8-a and 16-a pieces we will have (8-a)x and (16-a)y amount of aluminum.
Given that, $$\dfrac{(8-a)x+ay}{8} = \dfrac{(16-a)y + ax}{16}$$
Thus, $$2*(8x-ax+ay) = 16y-ay+ax$$
=> $$16x-2ax+2ay = 16y-ay+ax$$
Thus, $$16(x-y) = 3a(x-y)$$
Thus, $$a = \dfrac{16}{3} = 5.33$$
Hence, option C is the correct answer.

Hi total gain = 5%
Thus, the amount at the end of 3 years = 105 lakh Rupees
The amount he gets from the bank = $$30(1.12)^3 = 42.14784$$ lakh rupees
Let x be the percentage at which he sells the assets of worth 70 lakhs
Thus, the amount he gets = 0.7x lakhs
Thus, 70x + 42.1478 = 105
Thus, 70x = 62.8525
Thus, x is closest to 0.90= 90%
Hence, option C is the correct answer.

$$log_{13} log_{21} (\sqrt{x+21}+ \sqrt{x} ) =0 $$
Thus, $$log_{21} (\sqrt{x+21}+ \sqrt{x} ) = 1 $$
Thus, $$(\sqrt{x+21}+ \sqrt{x} ) = 21 $$
Let, $$\sqrt{x} = t$$
Thus, $$x = t^2$$
Thus, $$x+21 = t^2+21$$
Thus, $$\sqrt{t^2+21}+t = 21$$
Thus, $$(t^2+21) = (21-t)^2$$
=> $$t^2 + 21 = 441 - 42t + t^2$$
=> $$42t = 420$$
Hence, $$t = 10$$
Hence, option D is the correct answer.

$$3x^{2} - 4x + 7$$ is the given quadratic equation
The smallest value of a given quadratic equation $$ax^2 + bx + c = 0$$ occurs at $$ x = \dfrac{-b}{2a}$$
Thus, the smallest value of the given expression will occur at  $$ x = \dfrac{-(-4)}{2*3} = \dfrac{2}{3}$$
Thus, the smallest vale of the given expression is $$3*(\dfrac{2}{3})^2 - 4*(\dfrac{2}{3}) + 7 = \dfrac{17}{3}$$
Hence, option D is the correct answer.

Let the 7 consecutive numbers be a-3, a-2, a-1, a, a+1, a+2 and a+3.
Sum of the numbers = 7a and the average of these numbers = a
If next 3 numbers a+4, 4+5 and a+6 are also added then the average of these 10  numbers = $$\dfrac{7a+a+4+a+5+a+6}{10} = a+1.5$$
Thus, the average increases by 1.5
Hence, option C is the correct answer.

Let D be the distance, T be the time, V be the velocity, N be the number of carriages and X be the liters of Diesel required.
Given T$$\ltimes \dfrac{D}{V}$$
Also given that, V $$\ltimes \dfrac{\sqrt{\frac{X}{D}}}{N}$$
Thus, T = $$\dfrac{k*N*D^{\frac{3}{2}}}{\sqrt{X}}$$
Given when in a journey of 70 km, in 45 minutes with 15 carriages, 10 litres of diesel is required
Thus, 45 = $$\dfrac{k*15*70^{\frac{3}{2}}}{\sqrt{10}}$$
Thus, k = $$\dfrac{3}{70\sqrt{7}}$$
In a journey of 50 km, in half an hour with 18 carriages
30 = $$\dfrac{3}{70\sqrt{7}}*\dfrac{50\sqrt{50}*18}{\sqrt{X}}$$
Thus, $$\sqrt{X} = \dfrac{9*\sqrt{50}}{7*\sqrt{7}}$$
Thus, X = 11.8
Hence, option B is the correct answer.

We are given that 1.6Y = X
Thus, $$\dfrac{1}{1.6Y}+\dfrac{1}{Y} = \dfrac{1}{40}$$
=> $$104/1.6 = Y$$
Thus, Y = 65 hours.
Hence, option B is the correct answer.