Question 1002899: A man like be his life for x years. When he was in his midlife his first child was born. When he was in his 2/3 his second child was born. When the man died the sum of the children's ages is 60. How long does the man live.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the man dies at x years of age.
he has his first child at 1/2 * x years of age.
he has his second child at 2/3 * x years of age.
when the man dies, his first child is x - 1/2 * x = 1/2 * x years of age.
when the man dies, his second child is x - 2/3 * x = 1/3 * x years of age.
the sum of the children's ages when the man dies is 60.
you get:
1/2 * x + 1/3 * x = 60
multiply both sides of this equation by 6 to get:
3 * x + 2 * x = 6 * 60 which becomes:
5 * x = 360
divide both sides of this equaton by 5 to get:
x = 72
the man dies when he is 72 years of age.
his first child is born when he is 1/2 * 72 = 36 years old.
his second child is born when he is 2/3 * 72 = 48 years old.
when he dies, his first child is 72 - 36 = 36 years old.
when he dies, his second child is 72 - 48 = 24 years old.
when he dies, the sum of his children's ages is 36 + 24 = 60.
solution is confirmed to be correct.
your solution is that the man dies when he is 72 years old.
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