The average age of group of eight members is the same as it was 3 years ago when a young member is substituted for an old member. The incoming member is younger to the outgoing member by

11

years

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24

years

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28

years

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16

years

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Solution

The correct option is B $24$ years
Let the average age of eight members be $¯ x$

Let the ages be $a_{1}, a_{2}, . . ., a_{8}$

Therefore, $¯ x = \frac{a_{1} + a_{2} + . . . . + a_{8}}{8}$

$a_{1} + a_{2} + . . . . + a_{8} = 8 ¯ x$

$a_{2} + . . . . + a_{8} = 8 ¯ x - a_{1}$ -------(1)

Let the old member age be $a_{1}$ and
the young member age be $a_{9}$

Given that 3 years ago, young member is substituted for old member.

Given that, average is same after substitution.

Therefore, $¯ x = \frac{(a_{9} - 3) + (a_{2} - 3) + . . . . + (a_{8} - 3)}{8}$

$a_{9} + a_{2} + . . . . + a_{8} - 3 (8) = 8 ¯ x$

$a_{9} + a_{2} + . . . . + a_{8} - 24 = 8 ¯ x$ -------(2)

substituting (1) in (2) we get

$a_{9} + 8 ¯ x - a_{1} - 24 = 8 ¯ x$

$⟹ a_{9} - a_{1} = 24$

Therefore, the incoming member is $24$ years younger to the outgoing member.

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