Question

Little is known of the personal life of the Greek Mathematician Diaphantus , who lived in Alexandria, Egypt, in the third century A.D .Legend has it that the following epitaph marked on Diaphantus's grave . Diaphantus passed one-sixth of his life in his childhood, one twelvth in youth and one seventh as a bachelor.Five years after his marriage was born a son who died four years before his father at one half his father's (final age). What was Diaphantus's age at death?

Answer 1

Hi sir

Let his total life years be x.
(1/6+1/12+1/7+5) of x + x/2 + 4 = x

Let me explain you

1/6*x = childhood
1/12*x = youth
1/7*x = bachelor
5 yrs = married life until son was born
x/2 = his age at his son's death
4 yrs = son died four years before his father at one half his father's (final age). 

∴ (1/6+1/12+1/7+5) of x + x/2 + 4 = his final age at death.
1/6x + 1/12x + 1/7x + 5 + x/2 + 4 

So take LCM, which is 84

(14/84 + 7/84 + 12/84 + 42/84)x + 5 + 4.
x= (14+7+12+42/84)x +9
x= 75/84 x + 9.

1x can also be written as 84/84x in order to solve this problem

∴ 84/84x = 75/84 x + 9
84/84x - 75/84x = 9
84-75/84x = 9
9/84x = 9
x = 9 * 84/9
(Cancel 9 on denominator and numerator)

x = 84.

∴ He was 84 yrs old when he died.

Hope this helps you ^_^

Answer 2

Let Diaphantus 's age at death = x years

a/ c to question ,
passed life in childhood = x/6 years
passed life in youth = x/12 years
passed life in bachelor = x/7 years

agiain
he married when his age
= ( x/6 + x/12 + x/7 + 5) years

let the son's age = P years
a/c to question ,
P = x/2 - 4
so, Diaphantus alived after his marriage = x/2 + 4

hence,
final age at death = ( x/6 + x/12 + x/7 + 5) + (x/2 +4)

x = ( x/6 + x/12 + x/7+ 5) +x/2 + 4

x = (1/6 +1/12 +1/7 +1/2)x + 9

x = 25x/28 + 9

3x/28 = 9

x = 84

hence, Diaphantus's age at death = 84 years