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10 years ago, the average age of a family of five members was 38 years. Now, two new members join, whose age difference is 8 years. If the present average age of the family is the same as it was 10 years ago, What is the age (in years) of the new younger member? 

10 years ago, the average age of a family of five members was 38 years. Now, two new members join, whose age difference is 8 years. If the present average age of the family is the same as it was 10 years ago, What is the age (in years) of the new younger member?  Correct Answer 9

Total age of family 10 years ago = 38 × 5 = 190 years

Total age at present of the family = 38 × 7 = 266 years

Age of 5 members increased = 5 × 10 = 50 years

Difference of total age then and now = 266 – (190 + 50) = 26 (sum of ages of new members)

Let the two members be x and y.

X + y = 26

X – y = 8

On solving two equations,

X = 17 and y = 9

Age of younger member = 9

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