The problem states that the mean age of $R$ men in a club is 50 years. Two men, aged 55 and 63, leave the club. As a result, the mean age of the remaining men reduces by 1 year. We need to find the value of $R$.

1. Problem Description

The problem states that the mean age of $R$ men in a club is 50 years. Two men, aged 55 and 63, leave the club. As a result, the mean age of the remaining men reduces by 1 year. We need to find the value of $R$.

2. Solution Steps

Let $S$ be the sum of the ages of the $R$ men in the club.

The mean age is given by:

$\frac{S}{R} = 50$

$S = 50R$

Two men, aged 55 and 63, leave the club. The sum of their ages is $55 + 63 = 118$.

The new sum of ages is $S - 118 = 50R - 118$.

The number of men remaining is $R - 2$.

The new mean age is $50 - 1 = 49$.

So, we have:

$\frac{50R - 118}{R - 2} = 49$

$50R - 118 = 49(R - 2)$

$50R - 118 = 49R - 98$

$50R - 49R = 118 - 98$

$R = 20$

3. Final Answer

The value of R is

2

0. So, the answer is B. 20.