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A boy is half as old as the girl will be when the boy’s age is twice the sum of their ages when the boy was the girl’s age.

How many times older than the girl is the boy at their present age?

This is a logical problem sum.

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sorry could you use semicolons to understand sentences correctly – dato datuashvili Aug 9 '12 at 7:33
i am trying to understand problem ,but could not it,please fix sentences – dato datuashvili Aug 9 '12 at 7:38
by the way,you could denote age of boy's as $x$ and age of girl as $y$,you know that $y=k*x$ where $k$ is some natural number – dato datuashvili Aug 9 '12 at 7:46
@dato: Why must $k$ be a natural number? – Cameron Buie Aug 9 '12 at 8:27
nu if consider proportional coefficient as floating number,then $k$ have not be necessary natural – dato datuashvili Aug 9 '12 at 8:51

If $x$ is the boy's age and $y$ is the girl's age, then when the boy was the girl's current age, her age was $2y-x$. So "twice the sum of their ages when the boy was the girl's age" is $2(3y-x)=6y-2x$. The boy will reach this age after a further $6y-3x$ years, at which point the girl will be $7y-3x$. We are told that $x$ is half of this; so $2x=7y-3x$, which means that $x=\frac{7}{5}y$.

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Let the boy's age (in years) be $b$ and the girl's age be $g$.

When the boy was the girl's age, the girl was $g+(g-b)=2g-b$, so the sum of their ages then was $3g-b$, and twice that sum is $6g-2b$. The boy will be that age in $6g-3b$ years, at which point the girl will be $g+(6g-3b)=7g-3b$.

The boy is half that old now, so $b=\frac12(7g-3b),$ from which we see (multiplying by $2$ and rearranging) that $5b=7g$.

Thus, $b=\frac75g$, so the boy is $\frac75$ times the girl's age at present.

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