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Explanation:

i) Let the ages of the five members at present be a, b, c, d & e years.

And the age of the new member be f years.

ii) So the new average of five members’ age = (a + b + c + d + f)/5 ——- (1)

iii) Their corresponding ages 3 years ago = (a-3), (b-3), (c-3), (d-3) & (e-3) years

So their average age 3 years ago = (a + b + c + d + e – 15)/5 = x —– (2)

==> a + b + c + d + e = 5x + 15

==> a + b + c + d = 5x + 15 – e —— (3)

iv) Substituting this value of a + b + c + d = 5x + 15 – e in (1) above,

The new average is: (5x + 15 – e + f)/5

Equating this to the average age of x years, 3 yrs, ago as in (2) above,

(5x + 15 – e + f)/5 = x

==> (5x + 15 – e + f) = 5x

Solving e – f = 15 years.

Thus the difference of ages between replaced and new member = 15 years.