Question

If a number $x$ is selected from the first $100$ natural numbers at random, then the probability that
$x+\frac{100}{x}>50$ is

• A. $\frac{9}{20}$
• B. $\frac{1}{2}$
• C. $\frac{11}{20}$
• D. $\frac{5}{20}$

Answer 1

Answer: (C)

$\frac{11}{20}$

The given inequality reduces to $x^2-50x+100>0$
(we can multiply with x without changing the sign since we know that x is positive).
Thus, we have: $x>\frac{50+10\sqrt{21}}{2}orx<\frac{50-10\sqrt{21}}{2}i.e.x>25+5\sqrt{21}orx<25-5\sqrt{21}$
$\Rightarrow$, $x>47.9orx<2.08$
Thus, among the first 100 natural numbers, we have the numbers $1,2,48,49,...100$ i.e. $55$ numbers.
Therefore, probability = $\frac{55}{100}=\frac{11}{20}$
Hence, (c) is correct.