Question

A number $x$ is selected from first $100$ natural numbers. Find the probability that $x$ satisfies the condition $x+\frac{100}{x}>50$

• A. $\frac{55}{100}$
• B. $\frac{45}{100}$
• C. $1$
• D. $\frac{50}{100}$

Answer 1

Answer: (A)

$\frac{55}{100}$

$x+\frac{100}{x}>50⟹x^2+100>50x⟹x^2-50x+100+(25{)}^2-(25{)}^2>0⟹x^2+(25{)}^2-50x-525>0⟹(x-25{)}^2>525⟹x-25>\pm \sqrt{525}⟹x>25+22.91,x<25-22.91⟹x>47.91,x<2.09⟹x\ge 48,x\le 2$

As we have to select from 1 to 100 and x is greater than or equal to 48. Hence we have to select from 49 to 100 which is 53 numbers and favorable cases are 1, 2.

Total favorable case = 53+2=55.

Total number of cases = 100

Hence the required probability, $=\frac{55}{100}$