IIFT 2021 Slot 1 Question Paper (23rd Dec)

Since in first case we are compounding half-yearly and in second case we are compounding yearly and the amount received and the principal invested in both the cases is same so interest rate in the first case is lower than the interest rate in the second case.

There are 48 alphabets out of which we need 3

So, no. of ways of selecting 3 alphabets from 48 alphabets is $$^{48}C_3$$

Since we need to arrange these alphabets in ascending order so there is only one possible way for every three alphabets.

The first digit is 3 more than the third digit.

For a particular first digit there will be a particular third digit but the 2nd digit can be any of the numbers

So, no. of ways of selecting first digit= 15

No. of ways of selecting 2nd digit = 18

Therefore, total no. of ways = $$^{48}C_3\times\ 15\times\ 18$$ = 4669920

Increase in circumference is 8% which means that the increase in radius is also 8%

Increase in area = 8+8+ (8*8)/100 = 16+0.64 = 16.64%

$$P\left(A\ U\ B\ U\ C\right)=P\left(A\right)+P\left(B\right)+P\left(C\right)-\left\{P\left(A∩B\right)+P\left(B∩C\right)+P\left(C∩A\right)\right\}+P\left(A∩B∩C\right)$$

Let the no. of people who watch all three be x

78= 36+48+32-(14+20+12)+x

or x= 8

No. of people who watch times now only = 48 - (14+20)+8 = 22

No. of people who watch republic only = 32 - (20+12) + 8 = 8

Ratio = 22/8 = 11:4

$$f\left(x\right)=\left(1-\alpha\ \right)x^2-6\alpha x+8\alpha\ =0$$

Now roots are greater than 2 therefore,

$$-\frac{b}{2a}>2$$

$$f\left(2\right)>0$$

D>0

$$-\frac{b}{2a}>0$$

$$\frac{6\alpha}{2\left(1-\alpha\ \right)}>0$$

$$\frac{\alpha}{\alpha\ -1}<0$$

$$\alpha\ \in\ \left(0,1\right)$$

f(2)>0

$$\left(1-\alpha\ \right)4-12\alpha\ +8\alpha\ >0$$

$$4-8\alpha\ >0$$

$$\alpha\ <\frac{1}{2}$$

D>0

$$36\alpha\ ^2-32\alpha\ \left(1-\alpha\ \right)>0$$

$$68\alpha^2-32\alpha\ >0$$

$$\alpha\ \left(68\alpha\ -32\right)>0$$

$$\alpha\ \in\ \left(-\infty\ ,0\right)U\left(\frac{32}{68},\infty\ \right)$$

Taking the intersection of all we get $$\alpha\ \in\ \left(\frac{32}{68},\frac{1}{2}\right)$$

$$\cos\left(A\right)\ =\sin\left(\frac{\pi}{2}-A\right)$$

$$\sin A=\cos\left(\frac{\pi}{2}-A\right)$$

$$\cos\left(\frac{\pi}{32}\right)=\sin\left(\frac{\pi}{2}-\frac{\pi}{32}\right)=\sin\left(\frac{15\pi}{32}\right)$$

$$\sin^2\left(A\right)+\cos^2\left(A\right)=1$$

So the series simplifies to (1+1+1+1)-(1+1+1+1/2)

therefore value of series is 1/2

Let the CP be x

Now as per question, $$600-x=20\left(x-175\right)$$

21x = 4100

x= 195.24

Therefore, to make a profit of 30% selling price of second article should be = $$1.3\times\ 195.24\ =\ 253.81$$

We need a third 6 in the 7th throw which means that in the first 6 throws we should get 6 exactly twice.

No. of ways of getting 6 exactly twice in first 6 throws= $$^6C_2$$

Probability of getting 6 for the third time in the seventh throw= $$\frac{\left(^6C_2\times\ 5^4\right)}{6^7}$$

A, B, C, D, E, F, G all lie of a single line

H, I, J all lie on another single line

K, L lie on another single line

If we are given three points which are not collinear than we can draw a circle from these three points

The difference between the numbers xyz and zyx will be a multiple of 99.

therefore, 99(x-z)= 99*7

x-z =7

So, there are two possibilities for x and z.

i) x = 9 and z = 2,

ii) x = 8 and z = 1

To maximize the LCM, the first case will be preferred.

Now, for maximum LCM, y will be maximum.

Let us check for y = 8.

LCM = 8*9 = 72

For y = 7, LCM = 7*2*9 = 126.

Therefore, maximum possible LCM=126