Question 118140: a man passed 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. Alas! late born with child after attaining half her father's life, cruel fate overtook her, leaving the man to spend his last 4 years solving MSA's Math problems. What was the man's final age?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! a man passed 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. Alas! late born with child after attaining half her father's life, cruel fate overtook her, leaving the man to spend his last 4 years solving MSA's Math problems. What was the man's final age?
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Let x = man's final age
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a man passed one-sixth of his life in childhood,
 x
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one-twelfth in youth
 x
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and one-seventh in childless marriage.
 x
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after 5 years of marriage, the man had a child.
x = 5
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child after attaining half her father's life, cruel fate overtook her,
 x
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leaving the man to spend his last 4 years solving What was the man's final age?
x = x + x + x + x + 4
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We know that x = 5:
x = x + x + 5 + x + 4
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x = x + x + x + 9
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Multiply equation by 12 to get rid of the denominators, resulting in:
12x = 2x + x + 6x + 108
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12x = 9x + 108
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12x - 9x = 108
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3x = 108
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x = 
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x = 36 yrs old
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Check our solutions
child years = 6; one sixth
youth years = 3; one twelveth
childless marriage = 5 years
child's final age 18; half his final age
years alone = 4 given
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Final age total = 36
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There are flaws in this, that *36 is not quite 5, (5.14285) and being married at age 9 is somewhat far-fetched, then having a child at age 14???
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Hope this helps you anyway.
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