jmc

Let $c$ be the number of questions correct, $w$ be the number of questions wrong, and $b$ be the number of questions left blank. We are given that $$\begin{array}{rl}c+w+b& =20\\ 5c-2w& =48\end{array}$$ Adding equation $(2)$ to double equation $(1)$, we get $$7c+2b=88$$ Since we want to maximize the value of $c$, we try to find the largest multiple of $7$ less than $88$. This is $84=7\times 12$, so let $c=12$. Then we have $$7(12)+2b=88\Rightarrow b=2$$ Finally, we have $w=20-12-2=6$. We want $c$, so the answer is $12$, or $\boxed{12}$.