After replacing an old member by a new member, it was found that the average age of five members of a club is the same as it was 3 years ago. What is the difference between the ages of the replaced and the new member ?
Solution
Correct Answer: Option D
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.
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