IIFT 2012 Question Paper

We are given that C+B = 10
Thus, the cases are C=3 and B = 7
C = 4 and B = 6
C = 5 and B = 5
C = 6 and B = 4
C = 7 and B = 3
C = 8 and B = 2
C = 9 and B = 1
C = 10 and B = 0
Our required case = C = 9 and B = 1
Thus, the required probability = $$\dfrac{^9C_3}{^{10}C_{3}+^9C_3+^8C_3+^7C_3+^6C_3+^5C_3+^4C_3+^3C_3}$$
= $$\dfrac{14}{55}$$
Hence, option C is the correct answer.

Let us start by putting the value of x.
When x = 1,
$$7^{x - 1} + 11^{x - 1} = 1+1 = 2$$
When x=2,
$$7^{x - 1} + 11^{x - 1} = 7+11 = 17$$
When x=3,
$$7^{x - 1} + 11^{x - 1} = 49+121 = 170$$
When x=4,
$$7^{x - 1} + 11^{x - 1} = 243+1331 = 1574$$
As we can see $$7^{x - 1} + 11^{x - 1}$$ is an increasing function and the only value that satisfies $$7^{x - 1} + 11^{x - 1} = 170$$ is x=3
Thus, the given equation has only 1 solution.
Hence, option B is the correct answer.

Let 100 be the value in 2008.
In 2009 it will become 100*1.18
In 2010 it will become 100*1.18*1.18
In 2011 it will become 100*1.18*1.18*0.88
In 2012 it will become 100*1.18*1.18*0.88*1.18 $$\approx$$ 145
thus, the effect = 45%
Hence, option C is the correct answer.

If $$log{3}, log(3^{x} - 2)$$ and $$log (3^{x}+ 4)$$ are in arithmetic progression
Then, $$2*log(3^{x} - 2) = log{3}+log (3^{x}+ 4)$$
Thus, $$log{(3^{x} - 2)^2} = log{3(3^x+4)}$$
Thus, $$(3^{x} - 2)^2 = 3(3^x+4)$$
=> $$3^{2x} - 4*3^x +4 = 3*3^x + 12$$
=> $$3^{2x} - 7*3^x - 8 = 0$$
=> $$(3^x+1)*(3^x-8) = 0$$
But $$3^x+1 \neq 0$$
Thus, $$3^x = 8$$
Hence, $$x = log_{3}{8}$$
Hence, option B is the correct answer.

There are 3 ways in which she can answer the questions:-
1. 3 from group 1 and 3 from group 2
She can select 3 questions from each group in $$^5C_3$$ ways and thus the total number of ways = $$^5C_3*^5C_3$$ = 100
2. 4 questions from group 1 and 2 questions from group 2.
She can select 4 questions from group 1 in 5 ways and 2 questions from group 2 in 10 ways and thus, total number of ways = 50
Same will be the case for 4 questions from group 2 and 2 questions from group 1
Thus, the total number of ways = 100+50+50 = 200
Hence, option C is the correct answer.

Each of the 5 papers can be checked by any of the 9 professors and thus the total number of way = 9$$^5$$
Selecting exactly 2 of the 9 professors can be done in $$^9C_2$$ way = 36
They can correct the 5 papers in 2$$^5$$ - 2(all the 5 papers checked by the same professor) = 30  ways.
Thus, the total number of ways = 36*30
Hence, the required probability = $$\dfrac{36*30}{9^5}$$ = $$\dfrac{40}{2187}$$
Hence, option B is the correct answer.

Let the amount invested in X = x
Thus, the amount invested in Y = 25000-x
The interest incurred = 3518 Rs, thus the total amount at the end of the 2nd year = 28518 Rs.
Thus, $$x*(1.06)^2 + (25000-x)*(1.08)^2 = 28518$$
=> $$x*1.1236 + 25000*1.1664 - x*1.1664 = 28518$$
=> $$-0.0428x + 29160 = 28518$$
=> $$ 642 = 0.0428x$$
Hence, $$x = 15000 Rs$$
Hence, option C is the correct answer.

The probability that the LPG will last atmost 90 days = 0.6
Thus, the probability that the LPG will last more than 90 days = 0.4
The probability that the LPG will last more than 60 days = 0.8 = The probability that the LPG will last 60 to 90 days + The probability that the LPG will last more than 90 days
Hence, 0.4 + The probability that the LPG will last 60 to 90 days = 0.8
Hence, The probability that the LPG will last 60 to 90 days = 0.8- 0.4 = 0.4
Hence, option A is the correct answer.

Total amount allotted in 2011 = 4.5*10$$^7$$
Total number of departments = 200
Thus, the amount received by each department = 225000
Total amount allotted in 2012 = 60000000
The amount received by each department = 300000
Thus the excess of amount each department is getting = 300000 - 225000 = 750000  = Rs. 7.5*10$$^4$$ 
Hence, option C is the correct answer.

The area covered by the rectangular tank = 130*110 = 14300 m$$^2$$
The total area of the field = 20350+14300 = 34650 m$$^2$$
We know, 34550 m$$^2$$ = $$\pi*r^2$$
Thus, $$r^2 = \frac{34650*7}{22}$$
Thus, $$r = 105$$ m
Hence, option C is the correct answer.