IIFT 2010 Question Paper

Let the number of people who watched exactly one movie be 'a', the number of people who watched exactly two movies be 'b' and the number of people who watched exactly three movies be 'c'
Then, a + b + c = 100 - 3 = 97 ...............(i)
also, a + 2b + 3c = 61 + 46 + 29 = 136.......................(ii)
On subtracting (i) from (ii), we get
b + 2c = 39
It is given that, b = 25
So, c = 7
Therefore, 7% people watched all the three movies.
Hence, option D is the correct answer.

Let AC = $$2^k$$
AC = 2 * AB
So, AB = $$2^{k - 1}$$
Only 256 and 512 satisfy this condition as the third side is 295 and we have to ensure that sum of the two sides of the triangle is greater than the third side.
Thus, the perimeter = (256 + 512 + 295) = 1063.
Hence, option C is the correct answer.

It is given that sum of the first $$(x - 1)$$ numbers is equal to sum of the numbers from $$(x + 1)$$ to 49
or, sum of $$(x - 1)$$ numbers = sum of first 49 numbers - sum of first $$x$$ numbers
$$\dfrac{x(x - 1)}{2} = \dfrac{49 * 50}{2} - \dfrac{x(x + 1)}{2}$$
On solving, we get $$x$$ = 35
Hence, option C is the correct answer.

Let the interest on the second part be $$r$$ %
Then, the rate on the first part = ($$r$$ + 10)%It is given that, 
$$30000(1 + \dfrac{r + 10}{100})^2 - 30000(1 + \dfrac{r}{100})^2 = 7500$$
On solving, we get $$r$$ = 20%
Let the first part be Rs. $$a$$
Then, the second part = Rs. (60000 - $$a$$)
$$a(1 + \dfrac{20 + 10}{100})^2 - (60000 - a)(1 + \dfrac{20}{100})^2$$ = 38800
On solving, we get $$a$$ = Rs. 40000

Hence, option C is the correct answer.

We can write $$(1 - x^{6})^{4}$$ = $$4C0(1)^4(x^6)^0$$ - $$4C1(1)^3(x^6)^1$$ + $$4C2(1)^2(x^6)^2$$ - $$4C3(1)^1(x^6)^3$$+ $$4C4(1)^0(x^6)^4$$

$$\Rightarrow$$ $$(1 - x^{6})^{4}= (1-4x^6+6x^{12}-4x^{18}+x^{24})$$

Therefore, we can say

$$(1 - x^{6})^{4}(1 - x)^{-4}=(1-4x^6+6x^{12}-4x^{18}+x^{24})*(1 - x)^{-4}$$

We have to find out coefficient of $$x^{12}$$, $$x^6$$, $$x^0$$ in $$(1 - x)^{-4}$$.

We can use binomial expansion for negative coefficients. Therefore, coefficient of $$x^{12}$$ in $$(1 - x)^{-4}$$

$$\Rightarrow$$ $$\dfrac{(-4)*(-4-1)*(-4-2)* ... *(-4-11)}{12!}$$

$$\Rightarrow$$ $$\dfrac{15!}{12!*3!}$$

$$\Rightarrow$$ $$\dfrac{15*14*13}{3*2*1}$$

$$\Rightarrow$$ $$455$$

Similarly, coefficient of $$x^6$$ in $$(1 - x)^{-4}$$

$$\Rightarrow$$ $$\dfrac{(-4)*(-4-1)*(-4-2)* ... *(-4-5)}{6!}$$

$$\Rightarrow$$ $$\dfrac{9!}{6!*3!}$$

$$\Rightarrow$$ $$\dfrac{7*8*9}{3*2*1}$$

$$\Rightarrow$$ $$84$$

Coefficient of $$x^0$$ in $$(1 - x)^{-4}$$ is 1. 

Therefore, we can say that the coefficient of $$x^{12}$$ in the expansion of $$(1 - x^{6})^{4}(1 - x)^{-4}$$ = 455+(-4*84)+(1*6) = 125. Hence, option C is the correct answer. 

Alternative Solution:

$$(1-x^6)^4 = (1-x)^4(1+x+x^2+x^3+x^4+x^5)^4$$

$$\Rightarrow (1-x^6)^4(1-x)^{-4} = (1+x+x^2+x^3+x^4+x^5)^4$$

Hence we need to find coeff of  $$x^{12}$$ in $$(1+x+x^2+x^3+x^4+x^5)^4$$ =  $$(1+x+x^2+x^3+x^4+x^5)\times$$$$(1+x+x^2+x^3+x^4+x^5)\times$$$$(1+x+x^2+x^3+x^4+x^5)\times$$$$(1+x+x^2+x^3+x^4+x^5)$$

This will be equal to number of integral solutions for a + b + c + d = 12, 0<=a,b,c,d<=4

a is the power of x from the first expression, b is the power of x 

Lets find the set of values for (a,b,c,d)

(5,5,2,0) => Number of ways of arranging = 4!/2! = 12

(5,5,1,1) => Number of ways of arranging = 4!/(2!*2!) = 6

(5,4,3,0) => Number of ways of arranging = 4! = 24

(5,4,2,1) => Number of ways of arranging = 4! = 24

(5,3,2,2) => Number of ways of arranging = 4!/2! = 12

(5,3,3,1) => Number of ways of arranging = 4!/2! = 12

(4,4,4,0) => Number of ways of arranging = 4!/3! = 4

(4,4,3,1) => Number of ways of arranging = 4!/2! = 12

(4,4,2,2) => Number of ways of arranging = 4!/(2!*2!) = 6

(4,3,3,2) => Number of ways of arranging = 4!/2! = 12

(3,3,3,3) => Number of ways of arranging = 4!/4! = 1

Hence the coeff of $$x^{12}$$ = 24*2 + 12*5 + 6*2 + 4 + 1 = 125

Let us assume that Mukesh and Suresh started on bike from Delhi to Mathura and Dinesh started walking at the same time. Mukesh drops Suresh in a mid-way such that Suresh will take same time to reach Mathura with Mukesh and Dinesh. In the same time Mukesh will turn back and will pick up Dinesh. 

Let us assume that '4D' is the distance covered by by Mukesh and Suresh. Then Suresh must have covered remaining 300 - 4D kms by walking at 15 km/hr. When Mukesh dropped Suresh by that time Dinesh have already covered D kms and now Mukesh and Dinesh and would have 3D kms apart. Since Mukesh is 4 times faster than Dinesh, Mukesh would have covered (4/5) of 3D kms. 

In addition to this, Mukesh and Dinesh have to cover 2.4D+300-4D kms to reach Mathura at 60 km/hr.

By equating the time taken by Mukesh and Suresh to reach Mathura after the point when Mukesh dropped Suresh,

$$\dfrac{300-4D}{15}=\dfrac{2.4D+2.4D+300-4D}{60}$$

$$\Rightarrow$$ 16.8*D = 900

$$\Rightarrow$$ D = $$\dfrac{375}{7}$$ kms

Therefore, minimum time taken by them to finish the journey = $$\dfrac{4D}{60}+\dfrac{300-4D}{15}$$

$$\Rightarrow$$ $$20 - \dfrac{D}{5}$$

$$\Rightarrow$$ $$20 - \dfrac{375}{7*5}$$

$$\Rightarrow$$ $$20 - \dfrac{75}{7}$$

$$\Rightarrow$$ $$\dfrac{65}{7}$$ or $$9\dfrac{2}{7}$$ hours. Hence, option B is the correct answer. 

Let us draw an appropriate diagram. Let us assume that 'R' is the radius of the hole. 

$$\dfrac{1}{2}$$*(Total volume - Volume of the hole) = Volume of the hole

Volume of the hole = $$\dfrac{1}{3}*\text{Total volume of the rocket}$$

$$\pi*R^2*(8/3)x$$ = $$\dfrac{1}{3}*(\pi*8^2*5x+\dfrac{1}{3}*\pi*8^2*3x)$$

$$R^2*(8/3)$$ = $$\dfrac{1}{3}*(8^2*5+\dfrac{1}{3}*8^2*3)$$

$$R^2 = 48$$

$$R$$ = $$4\sqrt{3}$$ cm. Hence, option A is the correct answer.

Let the production cost be Rs. 100
The, Selling Price = Rs. 120
Price of component A = Rs. 10
Price of component B = Rs. 20
Price of other components = Rs. (100 - 10 - 20) = Rs. 70

After increase in prices,

Price of component A = Rs. 12
Price of component B = Rs. 28
Price of other components = Rs. 70
Total Cost of production = Rs. (12 + 28 + 70) = Rs. 110
Selling price = Rs. (1.15 * 120) = Rs. 138
Profit = Rs. 28
Profit % = 25.45%

Hence, option B is the correct answer.

Applying the cosine rule , we get $$c^{2} = a^{2} + b^{2} - 2ab cos θ$$

Hence, option A is the correct answer.

Total number of subsets that can be formed with the left of 11 elements = $$2^{11}$$ = 2048
The number of subsets which contain only odd numbers(1, 3, 5, 7, 9, 11} = $$2^{6}$$ = 64
Number of subsets which contain at least one even integer = Total number of subsets -  the number of subsets which contain only odd numbers
$$\Rightarrow$$ 2048 - 64 = 1984. Hence, option C is the correct answer.