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?
Verified answer
Hi sirLet 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 ^_^
Verified answer
Let Diaphantus 's age at death = x yearsa/ 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