Question

A natural numbers is selected from $1$ to $100$ so that the probability, if it satisfy the number $\frac{x^2-60x+800}{x-30}<0$ is

• A. $\frac{7}{25}$
• B. $\frac{4}{25}$
• C. $\frac{2}{25}$
• D. $\frac{8}{25}$

Answer 1

The correct option is A $\frac{7}{25}$

$x\in [1,100]$

$\frac{x^2-60x+800}{(x-30)}<0$

$\Rightarrow \frac{x^2-60x+900-100}{(x-30)}$

$\Rightarrow \frac{(x-30{)}^2-{10}^2}{(x-30)}<0$

$\Rightarrow \frac{(x-40)(x-20)}{(x-30)}<0$

case (i): Case (ii): $(x-40)(x-20)>0$

$(x-40)(x-20)<0$; $(x-30)<0$

$(x-30)>0$ $x\in (100,40)\cup (20,1)$

$x\in (40,20)$ $x\in (100,30)$ $x\in (1,30)$

$\Rightarrow x\in (40,30)$ $x\in [1,20)$

$\Rightarrow x=\{31,32,.....39\}$ $x=\{1,2,.....19\}$

$\therefore x=\{1,2,......19\}\cup \{31,32,...\dots ..39\}$

$n\left(c\right)=19+9=28$

$\therefore$ Probability $=\frac{n\left(x\right)}{100}=\frac{28}{100}$

$=\frac{7}{25}$.