A3. The goldsmith has two alloys. The first alloy contains $90 \%$ gold, the second $54 \%$ gold. The goldsmith mixed $320 \mathrm{~g}$ of the first alloy and $160 \mathrm{~g}$ of the second alloy to obtain a new alloy. What percentage of gold does the new alloy contain?
(A) 33
(B) 48
(C) 65
(D) 72
(E) 78
A3. In $320 \mathrm{~g}$ of the first alloy, there is $320 \cdot \frac{90}{100}=32 \cdot 9=288 \mathrm{~g}$ of gold, and in $160 \mathrm{~g}$ of the second alloy, there is $160 \cdot \frac{54}{100}=16 \cdot \frac{27}{5} \mathrm{~g}$ of gold. The proportion of gold in the new alloy is therefore

$$
\frac{32 \cdot 9+16 \cdot \frac{27}{5}}{320+160}=\frac{16 \cdot\left(18+\frac{27}{5}\right)}{16 \cdot(20+10)}=\frac{\frac{117}{5}}{30}=\frac{117}{150}=\frac{39}{50}=\frac{78}{100}
$$

which amounts to $78 \%$.