Paint mixture ratio in drum 2

Proportion of paint A in Drum 1: $f_1 = \frac{18}{18+7} = \frac{18}{25}$. Proportion of paint A in final mixture: $f_m = \frac{13}{13+7} = \frac{13}{20}$. Let $f_2$ be the proportion of paint A in Drum 2.

The mixtures are mixed in ratio 3 : 4, so: $3(f_1 - f_m) = 4(f_m - f_2)$ $3\left(\frac{18}{25} - \frac{13}{20}\right) = 4\left(\frac{13}{20} - f_2\right)$ $3\left(\frac{72 - 65}{100}\right) = 4\left(\frac{13}{20} - f_2\right)$ $\frac{21}{100} = \frac{52}{20} - 4f_2 \implies 4f_2 = \frac{260 - 21}{100} = \frac{239}{100}$ $f_2 = \frac{239}{400}$

So in drum 2, A is 239 parts out of 400 total. B is $400 - 239 = 161$ parts. Therefore, the ratio of A to B in drum 2 is $239 : 161$.