Question

A natural number $x$ is chosen at random from the first $100$ natural numbers. The probability that $x+\frac{100}{x}>50$ is

• A. $\frac{1}{20}$
• B. $\frac{11}{50}$
• C. $\frac{1}{10}$
• D. $\frac{11}{20}$

Answer 1

Answer: (D)

$\frac{11}{20}$

Given equation is,

$x+\frac{100}{x}>50$

$\Rightarrow x^2-50x+100>0$

$\Rightarrow x^2-50x+100+525-525>0$

$\Rightarrow x^2-50x+625>525$

$\Rightarrow {(x-25)}^2>525$

$\Rightarrow x-25<-\sqrt{525}orx-25>\sqrt{525}$

$\Rightarrow x<25-\sqrt{525}orx>25+\sqrt{525}$

As $x$ is a positive integer and $\sqrt{525}=22.91$, we must have

$x<25-22.91$ or $x>25+22.91$

$x<2.09$ or $x>47.91$

As only integer values are considered. We can say that, $x\le 2$ or $x\ge 48$

Let $E$ be the event for favourable cases and $S$ be the sample space.

$E=\{1,2,48,49,50,...100\}$

$\therefore n\left(E\right)=55$ and $n\left(S\right)=100$

Hence the required probability $P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{55}{100}=\frac{11}{20}$