IIFT 2021 Slot 1 Question Paper (5th Dec)

Area of shaded region = area of semicircle on BC + area of semicircle of AC - area of unshaded semicircle regions
Area of unshaded semicircle regions = area of semicircle on AB - area of triangle ABC
Given,
AB = 50, BC = 14
In triangle ABC, $$AC^2+BC^2 = AB^2$$
Solving we get,
AC = 48
Area of unshaded semicircle regions = $$\ \frac{\pi\ \left(25\right)^2}{2}\ -\frac{1}{2}\times\ 48\times\ 14=\frac{625\pi\ }{2}-336$$
Area of shaded region = $$\frac{576\pi\ }{2}\ +\frac{49\pi}{2}\ -\frac{625\pi}{2}\ +336$$ = 336

The answer is option A.

It is given that both the roots of given equation exceeds 2. Therefore, at x = 2, equation will be greater than zero.
$$4 - 8a + 3 - 3a + 4a^2 > 0$$
$$4a^2-11a+7>0$$
(4a-7)(a-1) > 0
a < 1 and a > 7/4
Answer is option D.

Shankar's profit% = $$\frac{P}{C.P}\times\ 100$$ = X
Shah's profit% = $$\frac{P}{S.P}\times\ 100$$ = Y
$$\ \frac{\ S.P}{C.P}=\frac{X}{Y}$$
It is given,
$$Y\ =\ \frac{909}{2000}X$$
$$\frac{X}{Y}\ =\ \frac{2000}{909}$$
$$\frac{S.P}{C.P}\ =\ \frac{2000}{909}$$
$$\frac{P}{C.P}\ =\ \frac{1091}{909}$$
It is given,
C.P = Rs 330
Profit $$P\ =\ \frac{1091}{909}\times\ 330\ \approx\ 396$$
The cost price for Mehrauli Nursery = 396 + 330 = Rs 726

Answer is option A.

Midpoint of AC passes through the line y = 2x + c

Midpoint of AC = (3,2)

2 = 6 + c

c = -4

Line passing through B is y = 2x - 4

(slope of AB)(slope of BC) = -1

$$\left(\ \frac{\ y-1}{x-5}\right)\left(\ \frac{\ y-3}{x-1}\right)=-1$$

Substituting y = 2x - 4 and solving, we get

B(2,0) or B(4,4)

B cannot lie on X-axis. Therefore, co-ordinates of B = (4,4)

Midpoint of BC = $$\left(\frac{9}{2},\frac{5}{2}\right)$$

The answer is option B.

AE is altitude from A on BC. $$\angle\ AEC\ =\ 90^{\circ\ }$$

It is given,

$$\angle\ ABC\ =\ 60^{\circ\ },\ \angle\ ACB\ =\ 45^{\circ\ }$$

In triangle ABE,

AE = $$\sqrt{\ 3}x$$

As $$\angle\ EAC\ =\angle\ ECA$$, EC = $$\sqrt{\ 3}x$$

In triangle AED, $$\angle\ EAD\ =\ 60^{\circ\ }$$

Therefore, CD = $$3x-\sqrt{\ 3}x$$

It is given,

$$BC\ \cdot\ CD\ =\ \frac{3^4.2^3.11}{19}$$

$$x\left(\sqrt{\ 3}+1\right)\sqrt{\ 3}\left(\sqrt{\ 3}-1\right)x\ =\ \frac{3^4.2^3.11}{19}$$

$$x\ =\ \frac{3^2.2.11}{19}$$

AE = 9*2 = 18

$$AE^2$$ = 324

Answer is option B.

"CORONAVIRUS" has 7 distinct alphabets and two pairs of repeated characters ", O" and "R".

There are three possible cases for creating 4 letter words.

1. Two letters are "O" and the other two are "R".

The total number of arrangements = $$\frac{4!}{2!\times2!}=6$$

2. Two of the letters are either "O" or "R" and the others are distinct.

The total number of arrangements = $$^2C_1\times^8C_2\times\frac{4!}{2!}=2\times28\times12=672$$

3. All four letters are distinct.

The total number of arrangements = $$^9C_4\times4!=3024$$

Thus, the total number of four-letter words possible = $$6+672+3024=3702$$.

Hence, the answer is option C.

It is given that,
The mean of 29 observations is 337
Therefore we can say :
$$337\ =\frac{S}{29}$$ where S is sum of observations 
S = (337)(29)    (1) 
The first term is increased by 1 
2nd by 2 .. and so on
New sum = $$S+\frac{29\left(30\right)}{2}$$
Therefore, mean will be $$\frac{\left(S+\frac{29\left(30\right)}{2}\right)}{29}$$
= 337 + 15
= 352

Answer is option C.

$$\log_{3}\frac{16^{x-3}+14}{4^{x-3}+2}=243$$

$$\log_3\frac{16^{x-3}+14}{4^{x-3}+2}=243$$

Let the speed of the train be in the from $$S=72-k\sqrt{\ x}$$(since speed is proportional to the square root of the number of wagons).
Now, when the number of wagons is 4, the speed of the train is 56 km/hr.
Thus, $$56=72-k\sqrt{\ 4}$$
$$56=72-2k$$
$$k=8$$

Thus, $$S=72-8\sqrt{\ x}$$
Since speed cannot be negative,
$$S>0$$
$$72-8\sqrt{\ x}>0$$
$$\ x<81$$
The maximum value of x will be 80.
Thus, the correct option is D.

It is given that,

$$2000\left(1.1\right)^3=3917\left(\frac{100-x}{100}\right)^3$$

$$\left(100-x\right)^3=679601$$

$$88^3=681472$$

Solving we get,

x = 12.xx

Difference = 2%

Answer is option C.