Eleven years earlier,the average age of a fam | vedclass

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(A) $1$. Eleven years ago,the total age of $4$ members was $4 	imes 28 = 112 \, 	ext{years}$.$2$. Presently,these $4$ members have aged by $11$ year

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(A) $1$. Eleven years ago,the total age of $4$ members was $4 	imes 28 = 112 \, 	ext{years}$.$2$. Presently,these $4$ members have aged by $11$ years each,so their total age is $112 + (4 	imes 11) = 112 + 44 = 156 \, 	ext{years}$.$3$. Currently,the family has $6$ members,and their average age is $28 \, 	ext{years}$. Thus,the total age of the $6$ members is $6 	imes 28 = 168 \, 	ext{years}$.$4$. The sum of the ages of the $2$ children is $168 - 156 = 12 \, 	ext{years}$.$5$. Let the present age of the youngest child be $x$. Since the children were born within the last $11$ years,the elder child's age is $x + k$,where $k$ is the age gap.$6$. The problem states that at the time of the birth of the youngest child,the elder child's age was equal to the total number of family members $(6)$.$7$. Let $t$ be the time elapsed since the birth of the youngest child. Then the youngest child's age is $t$,and the elder child's age is $6 + t$. $8$. The sum of their ages is $t + (6 + t) = 12 \Rightarrow 2t + 6 = 12 \Rightarrow 2t = 6 \Rightarrow t = 3$.$9$. Thus,the present age of the youngest member is $3 \, 	ext{years}$.

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