The average age of 5 members is 21 text{ year | vedclass

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(C) Total age of $5$ members $= 21 	imes 5 = 105 	ext{ years}$.The youngest member is $5 	ext{ years}$ old,which means $5 	ext{ years}$ ago,the yo

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$20$

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(C) Total age of $5$ members $= 21 	imes 5 = 105 	ext{ years}$.The youngest member is $5 	ext{ years}$ old,which means $5 	ext{ years}$ ago,the youngest member was just born.At that time ($5 	ext{ years}$ ago),each of the $5$ members was $5 	ext{ years}$ younger. So,the total age of the family was $105 - (5 	imes 5) = 105 - 25 = 80 	ext{ years}$.At the time of the birth of the youngest member,the family consisted of only $4$ members (excluding the youngest one).Therefore,the average age of the $4$ members at that time $= \frac{80}{4} = 20 	ext{ years}$.

The average age of $5$ members is $21 \text{ years}$. If the age of the youngest member is $5 \text{ years}$,find out the average age of the family at the birth of the youngest member (in $\text{years}$).

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