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)

$\frac{11}{20}$

Answer 2

Given equation $x+\frac{100}{x} > 50 \Rightarrow x^{2}-50x+100 > 0 \Rightarrow \left(x-25\right)^{2} > 525$ $\Rightarrow x-25 \sqrt{\left(525\right)}$ $\Rightarrow x < 25-\sqrt{\left(525\right)}$ or $x > 25+\sqrt{\left(525\right)}$ As x is positive integer and $\sqrt{\left(525\right)}=22.91$, we must have $x\le2$ or $x \ge48$ 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 $P\left(E\right) =\frac{n\left(E\right)}{n\left(S\right)}=\frac{55}{100}=\frac{11}{20}$