A natural number ‘n’ is selected at random from the set of first 100 natural numbers. The probability that $n+\frac{100}{n}≤50$ is equal to
Answer & explanation
Correct answer: option 3
$n+\frac{100}{n} \leq 50 $
$\Rightarrow n^2-50 n+100 \leq 0$
$\Rightarrow 25-5 \sqrt{21} \leq n \leq 25+5 \sqrt{21}$
⇒ n = 3, 4, 5...., 47
Thus, favourable number of ways
= (47 – 3 + 1) = 45
Thus, required probability = $\frac{45}{100}=\frac{9}{20}$