S is the set of all prime numbers that satisfy the inequality

Got a bit lost in the wordings there. From what I understand, x should be a prime number that satisfies the inequality, and S is the set of all prime numbers x.


\(\frac{1}{|10 - x|} \leq \frac{5}{x^2}\) can be written as \(\frac{x^2}{|10 - x|} \leq 5\)

When x = 2, we have \(\frac{2^2}{|10 - 2|} = \frac{4}{8}\), which is \(\leq\) 5

When x = 3, we have \(\frac{3^2}{|10 - 7|} = \frac{9}{7}\), which is \(\leq\) 5

When x = 5, we have \(\frac{5^2}{|10 - 5|} = \frac{25}{5}\), which is \(\leq\) 5

Any other prime number greater than 5 will have the fractional value greater than 5.


S = {2, 3, 5}

The arithmetic mean = \(\frac{2 + 3 + 5}{3} = \frac{10}{3}\). The numerator is 10.


Option C

Arun Kumar