[K Sengupta]

Each of the following relationships hold amongst the ages of the members of a family of four. Each of the four ages is an integer when expressed in years.

(I) The mother is three times as old as the daughter was when the father was the same age as the mother is now, and:

(II) 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, and:

(III) 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 precisely one of their ages (in years) is a perfect square, what are the four ages?

This post has been edited by K Sengupta: 03 November 2009 - 06:03 PM

[ljb]

Spoiler for I got

Son 12, Daughter 16, Mother 42, Father 44

[asser moustafa]

ljb, on 03 November 2009 - 10:19 PM, said:

Spoiler for I got

Son 12, Daughter 16, Mother 42, Father 44

how ??.. it doesn't work

[LMcNair27]

asser moustafa, on 03 November 2009 - 11:01 PM, said:

how ??.. it doesn't work

What doesn't work about it?

Spoiler for explanation

Son 12, Daughter 16, Mother 42, Father 44

(I) The mother is three times as old as the daughter was when the father was the same age as the mother is now:
when the father (who is now 44) was the same age as the mother is now (42), then their ages were Son 10, Daughter 14, Mother 40, Father 42. So the mother's current age (42) is three times the daughters previous age (14). 14x3=42.

(II) 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:
when the daughter reaches 21 (half the mother's current age), then their ages will be Son 17, Daugher 21, Mother 47, Father 49.
When the mother was 32 (twice the daughter's current age), then their ages were Son 2, Daughter 6, Mother 32, Father 34.
So the son's future age (17) will be half the father's previous age (34). 34/2=17.

(III) 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:
The mother's age when the daughter was 12 (son's current age) would have been 38. So when the father reaches 76 (2x38), then their ages will be Son 44, Daughter 48, Mother 74, Father 76.
So the daughter's future age (48) will be four times the son's current age (12). 12x4=48.

Finally, the Daughter's current age is a perfect square. 4^2=16.

[asser moustafa]

sorry sorry,, I forgot something.... but still he didn't tell how he got it ???

[ljb]

asser moustafa, on 03 November 2009 - 05:36 PM, said:

sorry sorry,, I forgot something.... but still he didn't tell how he got it ???

I'm leaving the fun part for you

[asser moustafa]

:S I faaaaaaailed :$
I've solved tens of brain teasers in this site,,, but this >>>> can't break it down

[ljb]

Spoiler for here's my method

Let S = age of son, D = age of daughter, M = age of mother, F = age of father.
Changing the word equations into algebraic ones, I got:
(I) M = 3(D-(F-M)) => simplifies to 3F = 3D+2M
(II) S+M/2-D = (F-(M-2D))/2 => simplifies to 2S+2M = 4D+F
(III) D+2(M-(D-2))-F = 4S => simplifies to 2M = 2S+D+F

Solve (III) for S and substitute in (II) to get 4M = 5D+2F
Solve (I) for M and substitute in above to get 11D = 4F

It's a start - I plugged it into Excel, varying D and seeing the results of the other three (M and S can be found in terms of D and F) and found the one that worked

[jerbil]

Spoiler for I agree with ljb, but ...

Given that Methuselah lived to the age of 969, then there are technically more than one solution to the problem, though I agree that ljb’s solution is the only realistic one.

Genesis 5:25 states that Noah had a son when he was aged 500, but the age of the mother is not mentioned.

Mathematically, S is always a multiple of 6, D a multiple of 8, M a multiple of 21 and F a multiple of 22. S < D < M < F, and ljb’s equations always apply.

Assuming that one believes Genesis and that Methuselah was the longest living male ever, then the possible solutions for (S,D,M,F) are:-

(12, 16, 42, 44), (36, 48, 126, 132), (48, 64, 168, 176), (108, 144, 378, 396),
(126, 168, 441, 462), (132, 176, 462, 484), (144, 192, 504, 528) and (192, 256, 672, 704).

One of these is not an integer multiple of ljb’s solution.