imc

Let Claudia have $x$ 5-cent coins and $\left(12-x\right)$ 10-cent coins. It is easily observed that any multiple of $5$ between $5$ and $5x+10(12-x)=120-5x$ inclusive can be obtained by a combination of coins. Thus, $24-x=17$ combinations can be made, so $x=7$. But the answer is not $7,$ because we are asked for the number of 10-cent coins, which is $12-7=\boxed{5}$