Question

A natural number x is chosen at random from the first one hundred natural numbers. The probability that $\frac{(x-20)(x-40)}{x-30}<0$ is

• A. 1/50
• B. 3/50
• C. 3/25
• D. 7/25

Answer 1

The correct option is D

7/25

Let $E=\frac{(x-20)(x-40)}{(x-30)}=\frac{(x-20)(x-30)(x-40)}{(x-30{)}^2}$
Sign of E is same as that of sign of (x-20)(x-30)(x-40)=F(say).
Note that F<0 if and only if
0<x<20 or 30<x<40.
$\therefore E<0in(0,20)\cup (30,40)$
Thus, E is negative for x=1, 2, ...., 19, 31, 32, ....., 39, that is E<0 for 28 natural numbers.
$\therefore$ Required probability =$\frac{28}{100}=\frac{7}{25}.$