[Combinatorics]
Problem: A chemist has three bottles, each containing a mixture of acid and water:

- bottle A contains $40 \mathrm{~g}$ of which $10 \%$ is acid,
- bottle B contains $50 \mathrm{~g}$ of which $20 \%$ is acid, and
- bottle C contains $50 \mathrm{~g}$ of which $30 \%$ is acid.

She uses some of the mixture from each of the bottles to create a mixture with mass $60 \mathrm{~g}$ of which $25 \%$ is acid. Then she mixes the remaining contents of the bottles to create a new mixture. What percentage of the new mixture is acid?

Solution: Bottle A contains $40 \mathrm{~g}$ of which $10 \%$ is acid.

Thus, it contains $0.1 \times 40=4 \mathrm{~g}$ of acid and $40-4=36 \mathrm{~g}$ of water.

Bottle B contains $50 \mathrm{~g}$ of which $20 \%$ is acid.

Thus, it contains $0.2 \times 50=10 \mathrm{~g}$ of acid and $50-10=40 \mathrm{~g}$ of water.

Bottle C contains $50 \mathrm{~g}$ of which $30 \%$ is acid.

Thus, it contains $0.3 \times 50=15 \mathrm{~g}$ of acid and $50-15=35 \mathrm{~g}$ of water.

In total, the three bottles contain $40+50+50=140 \mathrm{~g}$, of which $4+10+15=29 \mathrm{~g}$ is acid and $140-29=111 \mathrm{~g}$ is water.

The new mixture has mass $60 \mathrm{~g}$ of which $25 \%$ is acid.

Thus, it contains $0.25 \times 60=15 \mathrm{~g}$ of acid and $60-15=45 \mathrm{~g}$ of water.

Since the total mass in the three bottles is initially $140 \mathrm{~g}$ and the new mixture has mass $60 \mathrm{~g}$, then the remaining contents have mass $140-60=80 \mathrm{~g}$.

Since the total mass of acid in the three bottles is initially $29 \mathrm{~g}$ and the acid in the new mixture has mass $15 \mathrm{~g}$, then the acid in the remaining contents has mass $29-15=14 \mathrm{~g}$. This remaining mixture is thus $\frac{14 \mathrm{~g}}{80 \mathrm{~g}} \times 100 \%=17.5 \%$ acid.

Final answer: 17.5%