10 years ago,the average age of a family of 4 | vedclass

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(C) The average age of $4$ members $10$ years ago was $24$ years.Therefore,the sum of their ages $10$ years ago was $24 	imes 4 = 96$ years.At presen

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$3$

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(C) The average age of $4$ members $10$ years ago was $24$ years.Therefore,the sum of their ages $10$ years ago was $24 	imes 4 = 96$ years.At present,each of the $4$ members has aged by $10$ years,so the sum of their present ages is $96 + (4 	imes 10) = 96 + 40 = 136$ years.Let the ages of the two children be $x$ and $x+2$ years,where $x$ is the age of the youngest child.The total number of family members is now $4 + 2 = 6$.The present average age of the family is given as $24$ years.Thus,the sum of the ages of all $6$ members is $24 	imes 6 = 144$ years.We can write the equation: $136 + x + (x + 2) = 144$.$138 + 2x = 144$.$2x = 144 - 138$.$2x = 6$.$x = 3$.Therefore,the present age of the youngest child is $3$ years.

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

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