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(D) Let the sum of the ages of the $5$ members $3 \, 	ext{years}$ ago be $S$.Then,the average age $3 \, 	ext{years}$ ago was $\frac{S}{5}$.The curre

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$15$

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(D) Let the sum of the ages of the $5$ members $3 \, 	ext{years}$ ago be $S$.Then,the average age $3 \, 	ext{years}$ ago was $\frac{S}{5}$.The current sum of the ages of the same $5$ members (without replacement) would be $S + (5 	imes 3) = S + 15$.Let the age of the old member be $X$ and the new member be $Y$.After replacing the old member with the new member,the new sum of ages is $(S + 15) - X + Y$.The new average age is $\frac{S + 15 - X + Y}{5}$.According to the problem,this new average is equal to the average age $3 \, 	ext{years}$ ago,which is $\frac{S}{5}$.So,$\frac{S + 15 - X + Y}{5} = \frac{S}{5}$.$S + 15 - X + Y = S$.$15 - X + Y = 0$.$X - Y = 15$.Therefore,the difference between the ages of the replaced member and the new member is $15 \, 	ext{years}$.

After replacing an old member with a new member,it was found that the average age of five members of a club is the same as it was $3 \, \text{years}$ ago. What is the difference between the ages of the replaced member and the new member (in $\text{years}$)?

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