Question

A natural number $x$ is chosen at random from the first 100 natural numbers. The probability for the equation $x+\frac{100}{x}>50$ is $\frac{10+k}{20}$.Find $k$ ?

Answer 1

Given $x+\frac{100}{x}>50\Rightarrow x^2-50x+100>0$

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

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

As $x$ is a positive integer and $\sqrt{525}=22.91$, we must have $x\le 2$ or $x\ge 48$

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

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

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

Hence the required probability

$P\left(E\right)=\frac{n\left(E\right)}{n\left(s\right)}=\frac{55}{100}=\frac{11}{20}$

by comparing it with $\frac{10+k}{20}$ we get $10+k=11$

$\therefore k=1$