IIFT 2018

It is given that 73% of those surveyed viewed at least one of the three Channels; this implies 27% viewed none of the channels
a + b + c + d + e + f + g = 73
It is given, g = 11
a + b + c + d + e + f = 62 ......... (1)
a + b + c + 2(d+e+f) + 3g = 38 + 39 + 23
a + b + c + 2(d+e+f) + 3g = 100
a + b + c + 2(d + e + f) = 67 .......... (2)
Subtracting (1) from (2), we get
d + e + f = 67 - 62 = 5
The percentage of those surveyed, viewed more than one of the three TV channels = d + e + f + g = 5 + 11 = 16
The answer is option A.

Probability of playing on grass = 0.6
Probability of playing on Artificial turf = 0.4
Probability of injury on artificial turf = $$\frac{3}{2}$$(probability of injury on grass)
0.42 = $$\frac{3}{2}$$(probability of injury on grass)
Probability of injury on grass = 0.28
Applying Baye's theorem,
Probability of injury on grass = $$\ \frac{\ 0.6\times\ 0.28}{0.6\times\ 0.28+0.4\times\ 0.42}\approx\ 0.5$$
The answer is option C.

$$1+x^2+\sqrt{\ 1+x^2+x^4}$$

=$$\frac{1}{2}\left(2+2x^2+2\sqrt{\ 1+x^2+x^4}\right)$$

=$$\frac{1}{2}\left(1+x+x^2+1-x+x^2+2\sqrt{\left(1+x+x^2\right)\left(1-x+x^2\right)}\right)$$

=$$\frac{1}{2}\left(\sqrt{\ 1+x+x^2}+\sqrt{\ 1-x+x^2}\right)^2$$

$$\sqrt{\ 1+x^2+\sqrt{\ 1+x^2+x^4}}=\sqrt{\frac{1}{2}\left(\sqrt{\ 1+x+x^2}+\sqrt{\ 1-x+x^2}\right)^2\ }$$

$$\sqrt{\ 1+x^2+\sqrt{\ 1+x^2+x^4}}=\frac{1}{\sqrt{\ 2}}\left(\sqrt{1+x+x^2}+\sqrt{\ 1-x+x^2}\ \right)$$

The answer is option A.

$$\log_ba\ =\ \frac{1}{\log_ab}$$

$$\log_2 x.\log_{\frac{x}{64}}2$$ = $$\frac{\log_2x}{\log_2\left(\frac{x}{64}\right)}=\frac{\log_2x}{\log_2x-6}$$

Similarly,

$$\log_{\frac{x}{16}}2$$=$$\frac{1}{\log_2x-4}$$

Let $$\log_2x=t$$ => $$\frac{t}{t-6}=\frac{1}{t-4}$$ => $$t^2-5t+6=0$$ => $$\left(t-2\right)\left(t-3\right)=0$$ => t=2,3

.'. $$\log_2x=2\ or\ 3\ =>\ x=4\ or\ 8.$$

Hence, B is correct.