IIFT 2014 Question Paper

For Food and Beverage, IT and Telecom, Diversified,  Financial Services and Other services satisfy the condition that the merger activity was more in the first three years than in the last three years.

The turbulence for the Financial Service sector is much higher when compared to the other sectors.

Turbulence for the Financial service sector can be calculated as (201-194)+(201-160)+(160-116)+(193-116)+(193-177)= 185

Similarly, turbulence for the IT and Telecom sector = 108

Turbulence for Food and Beverage = 50

Turbulence for other sectors = 62

The number of acquisitions in the period 2005-06 is more than that of the previous year for the seven sectors, and it is the highest.

The total imports from various sources for the year 2010 are as follows:

25% of 540(1.15) + 50% of 290(1.1) + 505 of 80(1.05) + 5% of 50(1.1) = 360 MTOE (approx)

Import of natural gas for the year 2010 = 50% of 80(1.05) = 42

Import of natural gas for 2012 = 30% of 80(1.05)(1.1)(1.05) = 29 (approx)

Hence the required difference = 42 - 29 = 13

Domestic production of crude oil in the year 2011 = 55% of 290(1.1)(1.1) = 193 MTOE (approx)

Consumption of energy in the given country for the year 2012 is as follows:

540(1.15)(1.1)(1.15) + 290(1.1)(1.1)(1.15) + 80(1.05)(1.1)(1.05) + 20(1.2)(1.15)(1.1) + 0(1.1)(1.05)(1.1) + 20(1.15)(1.15)(1.1)

= 786 + 404 + 97 + 30 + 64 + 29 =  1410 MTOE

Hence the required percentage is $$\frac{786}{1410}\times100=\ 56\%\ $$

Domestic production of nuclear energy and hydroelectricity in the year 2011 are as follows:

20(1.2)(1.15) + 95% of 50(1.1)(1.05) = 82 MTOE

Let us draw the diagram according to the information given,

There will be infinite shaded areas as shown in the figure. 

Area of circle P$$_{1}$$ = $$\pi*R^2$$

Area of square Q$$_{1}$$ = $$(\sqrt{2}*R)^2$$ = $$2R^2$$

Area of circle P$$_{2}$$ = $$\pi*(\dfrac{R}{\sqrt{2}})^2=\pi*R^2/2$$

Area of square Q$$_{2}$$ = $$R^2$$

Area of circle P$$_{3}$$ = $$\pi*(\dfrac{R}{2})^2=\pi*R^2/4$$

Therefore, S$$_{N}$$ = [$$2R^2-\pi*\dfrac{R^2}{2}$$]+[$$R^2-\pi*\dfrac{R^2}{4}$$]+... 

S$$_{N}$$ = $$(2R^2+R^2+\dfrac{R^2}{2}+... )$$- $$(\pi*\dfrac{R^2}{2}+\pi*\dfrac{R^2}{4}+\pi*\dfrac{R^2}{8}...)$$

S$$_{N}$$ = $$4R^2-\pi*R^2$$

Therefore, $$\dfrac{S_{N}}{Q_{1}}$$ = $$\dfrac{4R^2-\pi*R^2}{2R^2}$$ = $$\dfrac{4 - \pi}{2}$$. Hence, option A is the correct answer.

1+2+3+4+..................+n < 1200

and 

1+2+3+4+..................+n+ (n+1) > 1200

We need to calculate value of n.

Using first and second inequality,

$$\frac{n (n+1)}{2}$$ < 1200

N = 48

Total number of boxes of sweets distributed is: 48*49/2 = 1176

So, roll no 24 got doubles number of boxes.