Question

A natural number is randomly chosen from first 56 natural numbers. The probability that $x+\frac{56}{x}>28$ is

• A. $\frac{1}{28}$
• B. $\frac{33}{56}$
• C. $\frac{31}{56}$
• D. None of these

Answer 1

The correct option is B $\frac{33}{56}$

$x+\frac{56}{x}>28$ $\Rightarrow x^2-28x+56>0$ $\Rightarrow {(x-14)}^2-196+56>0$ $\Rightarrow {(x-14)}^2-{\left(\sqrt{140}\right)}^2>0$ $\Rightarrow (x-14-\sqrt{140})(x-14+\sqrt{140})>0$ $\Rightarrow (x-14-11.8)(x-14+11.8)>0$ $\Rightarrow (x-25.8)(x-2.2)>0$ $\Rightarrow x\le 2$ or $x>25.8$ $\Rightarrow x\le 2orx\ge 26$
Now $n\left(A\right)=2+$number $x$ such that $26\le x\le 56=2+31=33$
$\therefore P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{33}{56}$ $(\because n\left(S\right)=56)$