Mathematics?

Question:If a man pass one-sixth of his life in childhood, one-twelfth in youth, and one-seventh in childless marriage. After 5 years of marriage, the man had a child, but after attaining half of her father's life, cruel faith overtook her, leaving the man to spend his last 4 years in charity works. What was the man's final age?

Answers:
Let X be the final age.

First part of the man's life until he has a daugther will be
((1/6)X + (1/12)X + (1/7)X + 5)

Since her daugther died when she is half the man's age, the man's age will be
2((1/6)X + (1/12)X + (1/7)X + 5)

Hence, the man final age can be solve by the equation:
2((1/6)X + (1/12)X + (1/7)X + 5) + 4 = X
(1/3)X + (1/6)X + (2/7)X + 10 + 4 = X
(2/6)X + (1/6)X + (2/7)X + 14 = X
(3/6)X + (2/7)X + 14 = X
(1/2)X + (2/7)X + 14 = X
(7/14)X + (4/14)X + 14 = X
(11/14)X + 14 = X
(3/14)X = 14
X = 196 / 3
X = 65 (1/3)

The man's final age is 65 yrs 4 mths.
to old?
84
If 1/7 of his life was spent in a childless marriage and he had a child after 5 years of marriage then 5 years is one seventh of his life. This makes his life 35 years long. NOT 84.



Tancy assumes that the man goes straight into marriage after youth and childhood. Which is not the case.

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