One tap can fill a cistern in 3 hours and the waste pipe can empty the full cistern in 5 hours. In what time will the empty cistern be full, if the tap and the waste pipe are kept open together?

Tap's 1 hour work = $\dfrac{1}{3}$

Waste tap's 1 hour work = $-\dfrac{1}{5}$

Both tap's 1 hour work = $\dfrac{1}{3} - \dfrac{1}{5}$

= $\dfrac{5 - 3}{15}$

= $\dfrac{2}{15}$

Total time taken by both pipe = $\dfrac{15}{2}$ hours

Hence, with both taps kept open, the empty cistern will be full in $7\dfrac{1}{2}$ hours.