Mike bought two books from the Canadian Super Mathematics Company. He paid full price for a $\$ 33$ book and he received $50 \%$ off the full price of a second book. In total, he saved $20 \%$ on his purchase. In dollars, how much did he save?
Solution 1

Suppose that the full price of the second book was $\$ x$.

Since $50 \%$ is equivalent to $\frac{1}{2}$, then the amount that Mike paid, in dollars, for the two books was $33+\frac{1}{2} x$.

The combined full price, in dollars, of the two books was $33+x$.

Since Mike saved a total of $20 \%$ on his purchase, then he paid $80 \%$ of the combined full cost. Since $80 \%$ is equivalent to $\frac{4}{5}$, then, in dollars, Mike paid $\frac{4}{5}(33+x)$.

Therefore, we obtain the equivalent equations

$$
\begin{aligned}
\frac{4}{5}(33+x) & =33+\frac{1}{2} x \\
\frac{4}{5}(33)+\frac{4}{5} x & =33+\frac{1}{2} x \\
\frac{3}{10} x & =\frac{1}{5}(33) \\
x & =\frac{10}{3} \cdot \frac{1}{5}(33) \\
x & =2 \cdot 11=22
\end{aligned}
$$

Therefore, the full cost was $\$(33+x)=\$(33+22)=\$ 55$.

Mike saved $20 \%$ of the full cost or $\frac{1}{5}(\$ 55)=\$ 11$.

Solution 2

Suppose that the full price of the second book was $\$ x$.

The combined full price, in dollars, of the two books was $33+x$.

Since $50 \%$ is equivalent to $\frac{1}{2}$, then the amount that Mike saved, in dollars, on the second book was $\frac{1}{2} x$.

Since $20 \%$ is equivalent to $\frac{1}{5}$, then, in dollars, Mike saved a total of $\frac{1}{5}(33+x)$ on the purchase of both books.

Therefore, we obtain the equivalent equations

$$
\begin{aligned}
\frac{1}{5}(33+x) & =\frac{1}{2} x \\
2(33+x) & =5 x \\
66+2 x & =5 x \\
3 x & =66 \\
x & =22
\end{aligned}
$$

Therefore, the full cost of the second book was $\$ 22$.

Mike saved $50 \%$ of the full cost or $\frac{1}{2}(\$ 22)=\$ 11$.