Question

A natural number x is chosen at random from the first 100 natural numbers. Then the probability, for the equation $x+\frac{100}{x}>50$ to be true is

• A. $\frac{1}{20}$
• B. $\frac{11}{20}$
• C. $\frac{1}{3}$
• D. $\frac{3}{20}$

Answer 1

Answer: (B)

Given equation
$x+\frac{100}{x}>50\Rightarrow x^2-50x+100>0\Rightarrow {\left(x-25\right)}^2>525$
$\Rightarrow x-25<-\sqrt{\left(525\right)}$ or $x-25>\sqrt{\left(525\right)}$
$\Rightarrow x<25-\sqrt{\left(525\right)}$ or $x>25+\sqrt{\left(525\right)}$
As x is positive integer and $\sqrt{\left(525\right)}=22.91$, we must have $x\le 2$ or $x\ge 48$
Let E be the event for favourable cases and S be the sample space.
$\therefore E=\left\{1,2,48,49,......100\right\}$
$\therefore n\left(E\right)=55$ and $n\left(S\right)=100$
Hence the required probability $P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{55}{100}=\frac{11}{20}$