Question

A tank can be filled by two taps A and B in 8 hours and 6 hours respectively. The full tank can be emptied by a third tap C in 4 hours. If all the taps are turned on at the same time, in how much time will the empty tank be filled up completely?

Answer 1

Answer:

Time taken by tap A to fill the water tank = 8hrs

★ Work done by tap A in one hour = $\dfrac{1}{8}$

Time taken by tap B to fill the water tank= 6 hrs

★ Work done by tap B in one hour = $\dfrac{1}{6}$

Time taken by tap C to empty the water tank = 4 hrs

★ Work done by tap C in one hour = $- \dfrac{1}{4}$

★ Net work done by Taps (A + B + C) in one hour $= \frac{1}{8} + \frac{1}{6} - \frac{1}{4} = \frac{1}{24}$

★ The tank will get completely filled by taps A, B and C in $1 \div \dfrac{1}{24}$

$= 1 \times \dfrac{24}{1} = 24$hrs

Hence, the water tank will get completely filled in 24 hours if all the taps are turned on at the same time.

Answer 2

Answer:

hours, respective The full tank can be emptied by a third tap in 8 hours. If all the taps are turned at the same time will the empty tank ...