Two taps can fill a tank in 20 min and 25 min respectively. Both are opened for 5 min. and then second tap is turned off. In how many minutes more is the tank completely filled.

Given two taps,

tap A fills in 20 min $\to$ in 1 min $it fills{\frac{1}{20}}^{ }part$

tap B fills in 25 min $\to$ in 1 min $it fills {\frac{1}{25}}^{ }part$

Both are on for 5 min

Then in 1 min both can fill $\to 5(\frac{1}{20}+\frac{1}{25})=5\left(\frac{9}{100}\right)=\frac{9}{20}$

remaining part of the work=$\frac{11}{20}$

let A completes the remaining work in x days

$\frac{1}{20}x=\frac{11}{20}\to x=11 minutes$