10 years ago, average age of a family of 4 members was 24 years. Two children having been born (with

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$10$ years ago, the average age of a family of $4$ members was $24$ $years.$ Two children having been born (with age difference of $2\, years$), the present average age of the family is the same. The present age of the youngest child is (in $years$)

A

$1$

B

$2$

C

$3$

D

$5$

Solution

Average age of $4$ members ( $10 \,years$ ago) $=24$ years

$\therefore$ Average age of $4$ members at present $=34$ years

Let the age of children be $x$ and $x+2$ years

Present average age of family $=\frac{34 \times 4+x+x+2}{6}=24$

$\frac{136+2+2 x}{6}=24$

$138+2 x=144$

$2 x=6$

$x=3$ years

$x+2=5$ years

Age of youngest child $=3$ years

Standard 13

Quantitative Aptitude

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