Question

A natural number is selected at random from the set $X=x:1\le x\le 100$. The probability that the number satisfies the inequation $x^2-13x\le 30$, is

• A. $\frac{9}{50}$
• B. $\frac{3}{20}$
• C. $\frac{2}{11}$
• D. $\frac{7}{9}$

Answer 1

The correct option is B $\frac{3}{20}$

$x^2-13x-30\le 0$
$(x-15)(x+2)\le 0$
Since $xϵ[1,100]$
Hence $x\le 15$
Therefore the values of x which satisfies the above inequality is $[1,15]$
Hence, first 15 natural numbers.
Therefore the required probability is
$\frac{15}{100}$

$=\frac{3}{20}$