IIFT 2008 Question Paper

The question asks for the thinnest person. Therefore, let us arrange the persons in the decreasing order of their weight.

Eswar is not as fat as Debu.
Therefore, Debu > Eswar.

Anil is fatter than Debu but not as fat as Bibek.
Bibek > Anil > Debu > Eswar.

 Eswar is thinner than Charu, who is thinner than Debu

Bibek > Anil > Debu > Charu > Eswar.

Therefore, Eswar is the thinnest among the five persons and hence, option D is the right answer.

"only brother of the only daughter of my sister’s maternal grandmother"

The daughter of my sister’s maternal grandmother is sister's mother. Mother's brother is maternal uncle of the person.

If Yuvraj is not pointing towards saurav then Saurav is pointing towards Yuvraj.
Thus, Yuvraj is the maternal uncle of Saurav.

Hence, option C is the correct answer.

$$Z$$ is the number of items produced by the company.
Total cost, $$R=\frac{1}{10}Z^{3}-5Z^{2}+10Z+5$$
As we can see, the term '5' does not vary with the number of quantities produced. Therefore, 5 is the fixed cost. 
Variable cost = $$\frac{1}{10}Z^{3}-5Z^{2}+10Z$$
Average variable cost = Total variable cost/ number of quantities.
Average variable cost =$$\frac{1}{10}Z^{2}-5Z+10$$
=$$\frac{Z^2-50Z+100}{10}$$
=$$\frac{(Z^2-50Z+625)- 525}{10}$$
=$$\frac{(Z-25)^2-525}{10}$$
As we can see, the least value of the expression will be obtained at $$Z=25$$
Therefore, option C is the right answer. 

Let us assume that n = 3 and a, b = 2, 6. 

Therefore, the harmonic sequence will be: 2, H$$_1$$, H$$_2$$, H$$_3$$, 6

Hence, H$$_2$$ = $$\dfrac{2*2*6}{2+6}$$ = 3

H$$_1$$ = $$\dfrac{2*2*3}{2+3}$$ = $$\dfrac{12}{5} = 2.4$$

H$$_3$$ = $$\dfrac{2*3*6}{3+6}$$ = $$4$$

Therefore, $$\dfrac{H_{1}+a}{H_{1}-a}+\dfrac{H_{n}+b}{H_{n}-b}$$

$$\dfrac{2.4+2}{2.4-2}+\dfrac{4+6}{4-6}$$

$$\Rightarrow$$ $$11-5$$ = 6. 

Option B: 2n = 2*3 = 6. 

$$(n+2)C_8:(n-2)P_4=57:16$$
$$\Rightarrow$$ $$\dfrac{\dfrac{(n+2)!}{8!*(n-6)!}}{\dfrac{(n-2)!}{(n-6)!}}=\dfrac{57}{16}$$

$$\Rightarrow$$ $$\dfrac{(n+2)(n+1)(n)(n-1)}{8!}=\dfrac{57}{16}$$

$$\Rightarrow$$ $$(n+2)(n+1)(n)(n-1)=57*2520$$

$$\Rightarrow$$ $$(n+2)(n+1)(n)(n-1)=21*20*19*18$$

Therefore, n = 19. 

Let us assume that the common difference of the A.P. is 'd'. 

Then, we can say that a = b - d, c = b + d

It is given that a + b + c = 3/2. i.e. b = 1/2. 

It is given that $$a^{2}, b^{2}$$ and $$c^{2}$$ are in Geometric Progression. Hence, we can say that 

$$b^4 = a^2*c^2$$

$$b^4 = (b - d)^2*(b + d)^2$$

$$b^4 = (b^2 - d^2)^2$$

$$\Rightarrow$$ $$(b^2+d^2-b^2)(b^2-d^2+b^2)=0$$ 

Therefore, $$(b^2 - d^2 + b^2) = 0$$. i.e. $$d = \dfrac{1}{\sqrt{2}}$$

Hence, a = b - d = $$\dfrac{1}{2}$$ - $$\dfrac{1}{\sqrt{2}}$$.

It has been given that the terms $$a, b,$$ and $$c$$ are in harmonic progression.
Therefore, $$\frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b}$$
$$\frac{2}{b}$$ = $$\frac{1}{a}$$ + $$\frac{1}{c}$$
$$\frac{2}{b} = \frac{a+c}{ac}$$
$$b = \frac{2ac}{(a+c)}$$--------------(1)
The given expression is log $$(a+c)$$ + log $$(a-2b+c)$$.
log $$(a+c) + log (a - 2b + c)$$ = log $$((a+c)(a-2b + c))$$
Substituting (1), we get,
log $$(a+c)$$ + log $$(a - 2b + c)$$ = log$$((a+c)(a - \frac{4ac}{(a+c)} +c))$$
 = log ($$a^2 + ac - 4ac + c^2 + ac$$)
 = log $$(a^2 + c^2 - 2ac)$$
 = log $$(c-a)^2$$ [Since c is greater than a]
= 2 log $$(c-a)$$
Therefore, option C is the right answer. 

The given series is $$1^2 - 2^2 + 3^2 - 4^2 +.....+2003^2$$
$$1^2 - 2^2$$ can be written as $$(1+2)(1-2)$$ = $$3*(-1)$$ = $$-3$$
$$3^2 -4^2$$ can be written as $$(3+4)*(3-4)$$ = $$7*(-1)$$ = $$-7$$
$$5^2-6^2$$ can be written as $$(5+6)*(5-6)$$ = $$11*(-1)$$ = $$-11$$
Therefore, all the terms till $$2002^2$$ can be expressed as an AP.
The last term of the AP will be $$(2001+2002)(2001-2002)$$ = $$-4003$$

Therefore, the given expression is reduced to $$-3 - 7 ...-4003 + 2003^2$$
Let is evaluate the value of $$-3 - 7 ...-4003 $$
Number of terms,$$n$$ = $$\frac{4003-3}{4} + 1$$ = $$1001$$
Sum = $$\frac{n}{2} *$$(first term + last term)
       = $$\frac{1001}{2}*(-4006)$$
       = $$ -2005003$$
$$2003^2 = 4012009$$
Value of the given expression = $$4012009 - 2005003 = 2007006$$.

Therefore, option A is the right answer.

There are 9 married couples. Therefore, there will be 9 men and 9 women. 
First, let us select the 2 men. 
2 men can be selected in 9C2 = 36 ways.

Now, the wives of these 2 men cannot be selected. Therefore, we have to select 2 women from the remaining 7 women. 
2 women can be selected in 7C2 = 21 ways.

We have selected 2 men and 2 women. A team should consist of one man and one woman. Therefore, the 2 teams can be formed in 2 ways. 

Therefore, the total number of ways in which the team can be selected is 36*21*2 = 1512.
Therefore, option B is the right answer.

It has been given that the interior angles in a polygon are in an arithmetic progression. 
We know that the sum of all exterior angles of a polygon is 360°.
Exterior angle = 180° - interior angle. 
Since we are subtracting the interior angles from a constant, the exterior angles will also be in an AP.
The starting term of the AP formed by the exterior angles will be 180°-120° = 60° and the common difference will be -5°.

Let the number of sides in the polygon be 'n'. 
=> The number of terms in the series will also be 'n'.

We know that the sum of an AP is equal to 0.5*n*(2a + (n-1)d), where 'a' is the starting term and 'd' is the common difference.
0.5*n*(2*60° + (n-1)*(-5°)) = 360°
120$$n$$ - 5$$n^2$$ + 5$$n$$ = 720
5$$n^2$$ - 125$$n$$ + 720 = 0
$$n^2$$ - 25$$n$$ + 144 =0.
$$(n-9)(n-16) = 0$$

Therefore, $$n$$ can be 9 or 16.
If the number of sides is 16, then the largest external angle will be 60 - 15*5 = -15°. Therefore, we can eliminate this case. 
The number of sides in the polygon must be 9. Therefore, option C is the right answer.