Logical question problem 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.

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.
