Taps A, B and C can fill a tank in 6 hours, 8 hours and 12 hours, respectively. Tap D
can empty out the tank in 4 hours. The four taps, A, B, C and D, are allowed to run
for 4 hours, after which Tap D is closed. How long will it take for the 3 taps to fill the
remaining tank?
Answers
Time taken by tap A to fill the tank = 6 hrs.
Work done in 1 hour by tap A = 1/6
Time taken by tap B to fill the tank = 8 hrs.
Work done in 1 hour by tap B = 1/8
Time taken by tap C to empty the tank = 4 hrs.
Work done in 1 hour by tap C = 1/4
∴ Work done to fill the tank if all of them are opened together =
1/6 + 1/8 - 1/4= (4+3-6)/24 =1/24
∴ Time taken to fill the tank if all are opened together is = 24 hours= 1 day
Answer:
The time taken by tap A to fill the tank = 6 hrs
Work done in 1 hr by tap A = 1/6
Time taken by tap B to fill the tank = 8 hrs
Work done in 1 hr by tap B = 1/8
Time taken by tap C to fill the tank = 12 hrs
Work done by Tap C in 1 hr = 1/12 hrs
Time taken by tap D to empty the tank = 4 hrs
Work done by tap D in 1 hr = - 1/4
Work done by all the four taps in 4 hr =4( 1/6 + 1/12 + 1/8 - 1/4)
= 2/3 +1/3 + 1/2 -1)
= (4+2+3-6 )/ 6 = 3/6 = 1/2
The remaining part of tank = 1 - 1/2= 1/2
Work done by 3 taps in 1 hr = 1/8 + 1/12 + 1/6 = (3+2+4)/24 = 9/24
Let the time taken to fill the remaining part be x.
Therefore, 9/24 × x = 1/2
x = 1/2× 24/9 = 4/3 hrs or 4/3 × 60(mins) = 1 hr 20 mins