$10\frac{1}{2}$

From the three equations we see that $(abc)^2=2\times 24\times 3=144$ and so, since $abc$ is positive, $abc=12$. Then the third equation tells us that $a=1/2$, the second that $b=4$ and the first that $c=6$. Therefore $a+b+c=10\frac{1}{2}$.

*This problem is taken from the UKMT Mathematical Challenges.*

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