The following relationships hold among the ages of the members of a family of four. (All ages are integral).

(a) The mother is three times as old as the daughter was when the father was the same age as the mother is now.

(b) When the daughter reaches half the age the mother is now, the son will be half as old as the father was when the mother was twice the age the daughter is now.

(c) When the father reaches twice the age the mother was when the daughter was the same age as the son is now, the daughter will be four times as old as the son is now.

Given that one of their ages is a perfect square, what are the four ages ?

Let the ages of the Father, Mother, Son and Daughter be F, M, S and D respectively.

The father was the same age as the mother is now (F-M) years ago at which time the daughter was D-(F-M). So

M = 3(D-(F-M))

The mother was twice the age the daughter is now (M-2D) years ago, at which time the father was F-(M-2D). Also, the daughter will reach half the age the mother is now in M/2 - D years, at which time the son will be S + M/2 + D, so

S + M/2 + D = (F-(M-2D))/2

The daughter was the same age the son is now D-S years ago, at which time the mother was M-(D-S). The father will reach twice that age 2(M-(D-S))-F years from now, at which time the daughter will be D+2(M-(D-S))-F, so

D+2(M-(D-S))-F = 4S

Simplifying:

3F - 2M -3D = 0 (A)
F/2 - M - S + 2D = 0 (B)
F - 2M + 2S + D = 0 (C)

Combining (B) and (C), we get

S = 3D/4

Combining (A) and (C) we get

2F - 2S - 4D = 0,

and given the relation of S and D,

F = 11D/4

Using these we get

M = 21D/8

So D must be a multiple of 8:

If D=8, F=22, M=21 and S = 6. None of these is a perfect square.

D=16 is already a perfect square, and then F=44, M=42 and S=12, which is the solution, as the next multiples of 8,22,21,6 do not include perfect squares until unreasonable ages.

Father = 44
Mother = 42
Son = 12
Daughter = 16

Posted by Charlie on 2003-04-25 03:56:11