Question 54522: Diophantus of Alexandria, a third century mathmatician, lived one-sixth of his life in childhood, one-twelfth in his youth, and one-seventh as a bachelor. Then he married and five years later had a son. The son died four years before diophantis at half the diophantis was when he himself died. How long did he live?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Diophantus of Alexandria, a third century mathmatician, lived one-sixth of his life in childhood, one-twelfth in his youth, and one-seventh as a bachelor. Then he married and five years later had a son. The son died four years before diophantis at half the age diophantus was when he himself died. How long did he
live?
:
Let his age at death = x:
:
childhood + youth + young man + 5 yrs + son's age + 4 yrs = His age at death
1/6x + 1/12x + 1/7x + 5 + 1/2x + 4 = x
:
A good common denominator is 84, mult equation by 84 and you get:
14x + 7x + 12x + 420 + 42x + 336 = 84x
75x + 756 = 84x
75x - 84x = - 756
- 9x = - 756
x = -756/-9
x = 84 years old
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