IIFT 2011 Question Paper

$$\log_{10}{3}+\log_{10}(4x+1)=\log_{10}(x+1)+1$$ can be written as

$$\log_{10}{3}+\log_{10}(4x+1)=\log_{10}(x+1)+\log_{10}{10}$$

We know that $$\log_{10}{a}+\log_{10}{b}=\log_{10}{ab}$$

$$\log_{10}{3*(4x+1)}=\log_{10}{(x+1)*10}$$

$$12x+3=10x+10$$

$$x=7/2$$. Hence, option B is the correct answer. 

i. Volume of the parallelepiped of length 5 cm, breadth 3 cm and height 4 cm = 3*4*5 = 60 cm$$^3$$
ii. Volume  of the cube of each side 4 cm = 4^3 = 64 cm$$^3$$
iii.Volume  of the cylinder of radius 3 cm and length 3 cm = $$\pi*3^2*3$$ = 84.82 cm$$^3$$
iv. Volume  of the  sphere of radius 3 cm = $$4/3*\pi*3^3$$ = 113.09 cm$$^3$$

Therefore, we can say that volumes of the objects in decreasing order = iv,iii,ii,i.

Hence, option A is the correct answer. 

Given that  $$|x - 1| + |x - 2| + |x - 3| \geq 6$$.

Case 1: When x > 3

$$(x - 1) + (x - 2) + (x - 3) \geq 6$$

$$x \geq 4$$

Therefore, the value of x $$\in$$ [4, $$\infty$$) 

Case 2: When 2 < x < 3

$$(x - 1) + (x - 2) - (x - 3) \geq 6$$

$$x \geq 6$$

Therefore, no possible value of x in this domain.

Case 3: When 1 < x < 2

$$(x - 1) - (x - 2) - (x - 3) \geq 6$$

$$x \leq -2$$

Therefore, no possible value of x in this domain.

Case 3: When x < 1

$$-(x - 1) - (x - 2) - (x - 3) \geq 6$$

$$x \leq 0$$

Therefore, the value of x $$\in$$ (-$$\infty$$, 0]

Therefore, the value of x that will satisfy this inequality: x $$\in$$ (-$$\infty$$, 0] $$\cup$$ [4, $$\infty$$).

Hence, option B is the correct answer.

For a number to be divisible by 3, the sum of it's digit should be a multiple of 3. This can be done in two ways. When the number are formed by using either {1, 2, 3, 4, 5} or {0, 1, 2, 4, 5}.

Total 5 digit numbers that can be formed by {1, 2, 3, 4, 5} without repetition = 5*4*3*2*1 = 120 

Total 5 digit numbers that can be formed by {0, 1, 2, 4, 5} without repetition = 4*4*3*2*1 = 96

Therefore, total such number = 120 + 96 = 216. Hence, option D is the correct answer. 

Given that 2, a, b, c, d, e, f and 65 are in an AP.

65 = 2 + (8-1)d

d = 9. 

Therefore, e = 2+(6-1)*9 = 2+45 = 47. Therefore, option B is the correct answer. 

Let 'W' be the amount of work. It is given that,

$$\dfrac{2}{5}*W = 104*8*30$$ ... (1)

Let 'X' be the number of additional men required to finish the work on time. 

$$W - \dfrac{2}{5}*W = (104+x)*9*(56-30)$$ ... (2)

By equation (1) and (2) we can say that, 

$$\dfrac{2}{3} = \dfrac{104*8*30}{(104+x)*9*(56-30)}$$

$$\Rightarrow$$ $$(104+x)*9*26 = 12*104*30$$

$$\Rightarrow$$ $$(104+x)=160$$

$$\Rightarrow$$ $$x=56$$.

Hence, option A is the correct answer.

5 balls are drawn at random, hence the probability that 3 of them are red and 2 are blue = $$\dfrac{8c3*6c2}{14c5}$$ = $$\dfrac{60}{143}$$

Hence, option D is the correct answer. 

Let ABC be the arc as shown in the figure. It is known that the diameter of the circle = 84 cm. Hence, the radius of the circle = 42 cm. 

We can see that both CO and BO are equivalent to the radius of the circle. Hence, CO = 42 cm and BD = BO - OD = 42 - 21 = 21 cm. 

In triangle ODC, $$cosDOC = \dfrac{21}{42}$$

Hence, $$\angle$$ DOC = 60°

In triangle BOC, OB = OC. Hence, we can say that $$\angle$$ OCB = $$\angle$$ OBC = 60°.

Therefore, OBC is an equilateral triangle. Hence, OB = OC = BC = 42 cm. 

Hence, we can say that the length of the chord of ‘half of the arc’ = 42 cm. Therefore, option C is the correct answer.

Mr. and Mrs. Gupta have three children - Pratik, Writtik and Kajol, all of whom were born in different cities. Pratik is 2 years elder to Writtik. Mr. Gupta was 30 years of age when Kajol was born in Hyderabad, while Mrs. Gupta was 28 years of age when Writtik was born in Bangalore. If Kajol was 5 years of age when Pratik was born in Mumbai, then what were the ages of Mr. and Mrs. Gupta respectively at the time of Pratik’s birth.

It is given that Pratik is 2 years elder to Writtik and Kajol was 5 years of age when Pratik was born in Mumbai. Hence, we can say that Kajol is the eldest and Writtik is the youngest. 

Mrs. Gupta was 28 years of age when Writtik was born in Bangalore. Hence, at the time of Pratik's birth Mrs. Gupta would have been two years younger i.e. 28 - 2 = 26 years old.

Mr. Gupta was 30 years of age when Kajol was born. Hence, at the time of Pratik's birth Mr. Gupta would have been 5 years older i.e. 30 + 5 = 35 years old.

Hence, option A is the correct answer.

He purchased a new vehicle with 40 percent of the money received. He then gave 20 percent of the remaining amount to each of his two sons for investing in their business. Thereafter, Mr. Sinha spent half of the remaining amount for renovation of his house. One-fourth of the remaining amount was then used for purchasing a LCD TV and the remaining amount - Rs. 1,35,000/- was deposited in a bank. What was the amount of his cash prize?

Let '100x' be the amount won by Mr. Sinha in the lottery contest. 

The amount left with him after purchasing the new vehicle = 0.60*100x = 60x.

The amount left with him after distributing his two sons for investing in their business = (1-2*0.20)*60x = 36x.

The amount left with him after renovating his house = 0.50*36x = 18x.

The amount left with him after purchasing the LCD TV = 0.75*18x = 13.5x.

It is given that, 13.5x = 135000

Therefore, 100x = 10,00,000. Hence, option A is the correct answer.