Question

A tank can be filled by two taps A and B in 12 hours and 16 hours respectively. The ful tank can be emptied by a third tap in 8 hhours if all the taps be turned on at the same time in hoe how much time will the empty tank be filled up completely

Answer 1

Tap A can fill tank in 12 hrs
Tap B can fill tank in 16 hrs
Tap C can empty the tank in -8hrs

Tap A's 1 hr work = 1/12
Tap B's 1 hr work = 1/16
Tap C's 1 hr work = -1/8
Tap (A+B+C)'s 1 hr work = 1/12 + 1/16 - 1/8
= 7-6/48 = 1/48 = 48 hours

Hence, all the taps will take 48 hrs to fill the tank.

hope it help you!

Answer 2

Answer:

48 hours

Step-by-step explanation:

Tap A can fill tank in 12 hrs

Tap A can fill tank in 1 hour = $\frac{1}{12}$

Tap B can fill tank in 16 hrs

Tap B can fill tank in 1 hour = $\frac{1}{16}$

Tap C can empty the tank in -8hrs

Tap C can empty tank in 1 hour = $\frac{1}{8}$

We are given that The ful tank can be emptied by a third tap

Tap (A+B+C)'s 1 hr work = $\frac{1}{12}+\frac{1}{16}-\frac{1}{8}$

                                       =$\frac{1}{48}$

So, Total time taken by A and B and C = $\frac{48}{1}$

So, if all the taps are opened the tank will be filled in 48 hours