Tap A fills 10% of a tank in 1.5 minutes. Tap B empties the tank in 20 minutes. How long will it take to fill the tank if both taps are opened?

Tap A fills $10\%$ of the tank in $1\frac{1}{2}$ minutes.
$10\% = \frac{10}{100} = \frac{1}{10}$ of the tank.
$1\frac{1}{2}$ minutes $= \frac{3}{2}$ minutes.
The rate of Tap A is the fraction of the tank filled per minute:

$$\text{Rate}_A = \frac{\frac{1}{10} \text{ tank}}{\frac{3}{2} \text{ minutes}} = \frac{1}{10} \times \frac{2}{3} \frac{\text{tank}}{\text{minute}} = \frac{2}{30} \frac{\text{tank}}{\text{minute}} = \frac{1}{15} \frac{\text{tank}}{\text{minute}}$$

Tap B empties the entire tank (1 whole tank) in 20 minutes.
The rate of Tap B is the fraction of the tank emptied per minute:

$$\text{Rate}_B = \frac{1 \text{ tank}}{20 \text{ minutes}} = \frac{1}{20} \frac{\text{tank}}{\text{minute}}$$

Since Tap B empties the tank, its contribution to filling is negative.

The combined rate is the filling rate minus the emptying rate:

$$\text{Combined Rate} = \text{Rate}_A - \text{Rate}_B$$ $$\text{Combined Rate} = \frac{1}{15} - \frac{1}{20}$$

To subtract these fractions, find a common denominator, which is 60.

$$\text{Combined Rate} = \frac{1 \times 4}{15 \times 4} - \frac{1 \times 3}{20 \times 3}$$ $$\text{Combined Rate} = \frac{4}{60} - \frac{3}{60}$$ $$\text{Combined Rate} = \frac{4 - 3}{60} = \frac{1}{60} \frac{\text{tank}}{\text{minute}}$$

If the combined rate is $\frac{1}{60}$ of the tank per minute, then the time to fill 1 whole tank is:

$$\text{Time} = \frac{\text{Total Tank Capacity}}{\text{Combined Rate}}$$ $$\text{Time} = \frac{1 \text{ tank}}{\frac{1}{60} \frac{\text{tank}}{\text{minute}}} = 1 \times 60 \text{ minutes} = 60 \text{ minutes}$$

The time it will take to fill the tank is 60 minutes.