IIFT 2012 Question Paper

The volume of the hot water in the hemispherical bowl = $$\frac{2*\pi*r^3}{3}$$
Given that Radius of the hemispherical bowl and that of the cylindrical vessel is the same.
Also, radius is 50% more than its height.
Thus, h = $$\frac{2r}{3}$$
Thus, the volume of the cylindrical vessel = $$\frac{2*\pi*r^3}{3}$$
Hence, the volume of the hot water in the cylindrical vessel is 100% of the cylindrical vessel.
Hence, option C is the correct answer.

The given situation is as shown:-

Speed of train 1 = 60 km/hr = 50/3 m/s
Speed of train 2 = 48 km/hr = 40/3 m/s
Thus, their relative speed = 90/3 = 30 m/s
Let, Length of train 1 = l
Thus, the length of train 2 will be 0.5 l
Time taken by them to cross each other = 15 s
Thus, $$ 15*30 = 1.5l$$
Thus, $$ l = 300 $$
Train 1 takes 51s to cross a bridge
Let the length of the bridge be b
Thus, $$\frac{50*51}{3} = 300+b $$
Hence, $$ 850 = 300+b $$
Hence, $$b = 550$$ m
Thus, option A is the correct answer.

12 men can complete a work in ten days. Thus, 1 men will take 120 days.
In 1 day 1 men will complete $$\frac{1}{120}$$ of the work.
20 women can complete the same work in twelve days.
Thus, 1 women will take 240 days to complete the work.
Thus, In 1 day 1 women will complete $$\frac{1}{240}$$ of the work.
8 men and 4 women work for 9 days completing $$\frac{8*9}{120}$$+ $$\frac{4*9}{240}$$ = $$\frac{3}{4}$$ of the work.
Thus, $$\frac{1}{4}$$ of the work is remaining.
Now 8 men and 14 women work completing $$\frac{8}{240}$$+$$\frac{14}{240}$$ = $$\frac{1}{8}$$ of the work in 1 day.
To complete $$\frac{1}{4}$$ they'll take 2 days.
Hence, option A is the correct answer.

Given that, $$(45-x) \ltimes \sqrt{n} $$
=> $$(45-x) = k*\sqrt{n} $$
Where x is the speed of the train, k is the proportionality constant and n is the number of wagon wheels attached.
Now given that when n = 9, x= 30
Thus, $$ 15 = k*3 $$
thus, k = 5
Thus, $$ 45-x = 5\sqrt{n} $$
Thus, when n = 81 x = 0
Thus, for the train to move the maximum number of wagon wheels = 80
Hence, option C is the correct answer.

Let the total fees = T, the fixed fees = F, the number of students = n and let x be the varying fees.
Thus, T = F + nx
Given, when  n = 20, T = 8000
Thus, 8000 = F + 20x
Also, when n = 40, T = 12000
Thus, 12000 = F + 40x
Solving we get F = 4000 and x = 200
Thus, the average rent when 80 students are staying = 4000/80 + 200 = 250 Rs
Hence, option A is the correct answer.

Let the CP of 1 soap = S
Thus, the CP of 1 toothbrush = 0.6S
Given that, SP of 1 soap = 1.15 S and SP of 1 toothbrush = 0.6S+20
Also, 15*1.15*S + 8*(0.6S+20) - 15S - 8*0.6*S = 385
Thus, solving we get S = 100
Hence, Total CP of 20 soaps and 12 toothbrush = 20*100 + 12*60 = 2720
SP of 15 soaps and 8 toothbrush = 15*1.15*100 + 8*80 = 2365
Thus, the overall loss = 2365 - 2720 = 355 Rs.
Hence, option A is the correct answer.

Let, $$\sqrt{7+\sqrt{7-\sqrt{7+\sqrt{7-.....\infty}}}}$$ = $$x$$
Thus, $$\sqrt{7+\sqrt{7-x}}$$ = $$x$$
=> $$7+\sqrt{7-x} = x^2$$
=> $$7-x = (x^2-7)^2$$
Putting options we get,
x=1 => 6$$\neq(-6)^2$$
x=2 => 5$$\neq(-3)^2$$
x=3 => 4=$$(9-7)^2$$
x=4 => 3$$\neq(9)^2$$
Hence, option C is the correct answer.

The unit digit in the product of $$(8267)^{153} \times (341)^{72}$$ is the same as the unit digit in the product of $$(7)^{153} \times (1)^{72}$$
1 raised to anything is 1.
7 has a cyclicity  of 4. Thus, $$7^{153} = 7^1 = 7$$
Hence, option C is the correct answer.

It is given that Z = 31!
X = 31! + 1
X+1 = 31!+2 this is divisible by 2
X+2 = 31!+3 this is divisible by 3
X+3 = 31!+4 this is divisible by 4
.
.
.
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X+29 = 31!+30 this is divisible by 30
X+30 = 31!+31 this is divisible by 31
Hence, none of X + 1, X + 2, ..., X + 29, X + 30 is a prime number.
Hence, option D is the correct answer.