Math Problem

[Jim Brier]

A man spends one-sixth of his life as a child, one-twelfth as an adolescent, and one-seventh as a bachelor. Five years after he is married, he has a son who dies 4 years before his own death and at one-half his own final age. How long was the son alive?

[BruceZ]

42 years.

Man lives 84 years.
child 84/6 = 14 years
adolescent 84/12 = 7 years until age 21
bachelor 84/7 = 12 years until age 33
son born at age 38
son dies when man is 80 at age 42
man dies 4 years later at age 84

Now, if a hen and a half lays an egg and a half in a day and a half, how long does it take a hen to lay a dozen eggs?

[ronniejames]

One hen lays 12 eggs in 18 days. (Three hens lay 6 eggs in 3 days, 3 hens lay 12 eggs in 6 days, 1 hen lays 4 eggs in 6 days, etc.)

[Jim Brier]

Thanks Bruce! I couldn't have answered it better myself.

[SittingBull]

The # of hens is directly proportional to the # of eggs being produced and inversely proportional to the length of time it takes the hens to lay the eggs.
Hence H=K(E/T) where K is a constant,E=is the # of eggs produced,T is the length of time it takes to produce the eggs and H is the # of hens laying the eggs.

Happy pokering,Bruce!
Sitting Bull

[Mike Haven]

you give the answer, and you prove the answer is correct in conjunction with logical ages for childhood, etc, but how did you get the answer in the first place, without random guesses?

i would be interested to see your workings

thanks

[ripdog]

It looks to me like you take the 1/6, 1/12, and 1/7 numbers and simply find the LCD, which is 84. Divide 84 by 2 and you get the age of the son. Sounds good to me.

Simple algebra:

s = final age of son, m = final age of man

s = .5m = m - 4 - [(1/6 +1/12 + 1/7)m + 5]

m = 84, s = 42

[BruceZ]

[Bob T.]

36

Good luck,
Play well,

Bob T.