IIFT 2008 Question Paper

Distance between the top of ladder and foot of wall = $$\sqrt{25^2 - 7^2}$$ = 24 m. Let us assume that C is moved by a distance x m. 

Using, Pythagoras theorem we can say that, 25^2 = 12^2 + (7+x)^2 => x = $$\sqrt{481}$$ - 7 = 14.93 m.

$$2-\frac{\sqrt{6407522209}}{\sqrt{3600840049}}=2-\frac{80047}{60007}$$
=$$2-1.3339610$$
$$=0.666039$$
Therefore, option A is the right answer. 

It has been given that a, b, and c are in an arithmetic progression.
Let a = x-p, b = x, and c = x+p
We know that a, b, and c are real numbers.
Therefore, the arithmetic mean of a,b,c should be greater than or equal to the geometric mean.
$$\frac{a+b+c}{3} \geq \sqrt[3]{abc}$$
$$\frac{a+b+c}{3} \geq \sqrt[3]{4}$$
$$\frac{3x}{3}\geq\sqrt[3]{4}$$

$$x\geq\sqrt[3]{4}$$
We know that $$x$$ = $$b$$.
Therefore,$$b\geq\sqrt[3]{4}$$or $$b\geq 2^{\frac{2}{3}}$$
Therefore, option B is the right answer.

The given equation is $$ax^{2} + bx + c = 0$$
Roots of the given quadratic equation are $$\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$$
We know that one of the 2 roots is double the other. 
Therefore, $$\dfrac{-b+\sqrt{b^2-4ac}}{2a}$$ =$$2*\dfrac{-b-\sqrt{b^2-4ac}}{2a}$$
=> $$-b +\sqrt{b^2-4ac} = -2b -2\sqrt{b^2-4ac}$$
=> $$b = -3\sqrt{b^2-4ac}$$
Squaring on both sides, we get, 
$$b^2 = 9*(b^2-4ac)$$
$$8b^2 = 36 ac$$
$$2b^2 = 9ac$$.
Therefore, option A is the right answer. 

Alternately

Suppose two roots are x, 2x

Sum of the roots= 3x=-b/a

Product of the roots=$$2x^{2}$$ = c/a

Putting the value of x from the first eqn.

 We get $$2b^2 = 9ac$$..

Let us assume the upstream speed of the boat = u and the downstream speed = v

30/u+44/v=10

40/u+55/v=13

Consider 1/u=a, 1/v=b, these equations become

30a+44b=10

40a+55b=13

We get a=1/5 and b=1/11 Hence,  1/u=1/5 and 1/v =1/11 => u=5 and v=11

Assuming speed of boat in still water is x and the speed of stream=y

u=x-y=5

v=x+y=11

From these equations, we get x=8 km/hr

$$Cot^{-1}[\frac{\sqrt{sin^2(a/2)+cos^2(a/2)-2sin(a/2)cos(a/2)}+\sqrt{sin^2(a/2)+cos^2(a/2)+2sin(a/2)cos(a/2)}}{\sqrt{sin^2(a/2)+cos^2(a/2)-2sin(a/2)cos(a/2)}-\sqrt{sin^2(a/2)+cos^2(a/2)+2sin(a/2)cos(a/2)}}]$$

$$Cot^{-1}[\frac{cos(a/2)-sin(a/2) + cos(a/2) + sin(a/2)}{cos(a/2)-sin(a/2) - cos(a/2) - sin(a/2)}]$$

$$Cot^{-1}[\frac{-2cos(a/2)}{2sin(a/2)}]$$

$$Cot^{-1}[Cot(-a/2)]$$

We know that $$Cot^{-1}[-x] = \pi - Cot^{-1}[x]$$

Therefore, $$Cot^{-1}[Cot(-a/2)]$$ = $$\pi - Cot^{-1}[Cot(a/2)]$$ = $$\pi-\frac{1}{2}a$$

Let us construct a diagram based on the given statements.
There are 2 diametrically opposite gates A and B.
A pole is erected on the circumference such that the distance of the pole from one of the gates is 7 m more than the distance of the pole from the other gate. Let the distances be x and x+7 m.

Now, APB is a right-angled triangle (since AB is the diameter).
Applying Pythagoras theorem, we get,
$$13^2 = x^2 + (x+7)^2$$
$$169=x^2+x^2+14x+49$$
$$120 = 2x^2 + 14x$$
$$x^2 + 7x - 60 = 0$$
$$(x+12)(x-5)=0$$
Therefore, $$x$$ can be $$-12$$ or $$5$$.
$$x$$ cannot be negative. Therefore, $$x$$ has to be $$5$$.
The distance of the pole from one of the gates is 5 m. Therefore, option C is the right answer. 

The length of the spiral will be equal to the length of the circular arc portions of the semicircles. 
Length of the circular arc portion of a semicircle = $$\frac{22}{7}*r$$.
Length of the spiral = $$\frac{22}{7}*(0.5+1+1.5+2+...+6.5)$$
 = $$\frac{22}{7}*0.5*[1+2+3+....+13]$$
= $$\frac{11}{7}*\frac{13*14}{2}$$
=$$143$$ cm.
Therefore, option B is the right answer.

Standard deviation is the measure of how much the values in the group differ from the average. 
If the salary of each person is increased by 100, the average salary will also be increased by 100. Therefore, the standard deviation will remain unaltered in this case. 
Every person is given a hike of 20%. Therefore, the standard deviation will also increase by 20%. Therefore, the new standard deviation will be 1.2*400 = 480. Therefore, option B is the right answer. 

1 person in every 100 suffers from the disease.

Probability of a person being healthy = $$\ \frac{\ 99}{100}$$

In a group of 50 people if the test is positive, then he could then test each person’s blood individually otherwise he will consider that the entire group is healthy

The number of tests =50+1 = 51

The probability that all the people in the group are healthy = $$^{50}C_{50}\ \times\ \left(\ \frac{\ 99}{100}\right)^{50}$$

= approx 0.605

So the probability that atleast one person suffers in a group of 50 = 1-0.605= 0.395

Expected number of tests = 51*0.395 + 0.605*1

= 20.145+0.605

=20.75 = 21 tests.

C is the correct answer.