question for math whizzes?

Question

A man passed 1/6 of his life in childhood, 1/12 in youth, 1/7 in childless marraige. After 5 years of marraige, the man had a child. Alas! late born wretches child; after attaining half her father's life, cruel fate overtook her, leaving the man to spend the last four years of his life lonely. What was the mans final age?????

Show ur sollution.....

Answer 1

He was 84 years old and here are two solutions

1) Hard way
x/6+x/12+x/7 + 5 + 4 = x/2

x/2 -(x/6+x/12+x/7) = 9

(42-14-7-12)x/84 = 9
9x/84 = 9
x = 84

2) Easier way. Since the equation has denominators like 1/6,1/7,1/12 find the lcm for these numbers. this not surprisingly is 84. Once you have 84, you can easily verify that this is the right answer as follows

1/6 of life in childhood = 14 years
1/12 of life in youth = 7 years
1/7 of childless marriage = 12 years
further then 5 years of further marriage
Child dies after reaching half of fathers age = 42 years
Last 4 lonely years = 4 years

If you add all of these you get 84 which is what you were looking for

Answer 2

(1/7)x=5
x=5*7
x=35
the man spent a seventh of his life in childless marriage and the question says after 5 years of mrriage he had a child, so 1/7 of the mans life is 5 years