tap A and B can fill a tank in 2 and 4 hours respectively and tap B can empty it in 12 hours if all the three taps are opened together when the tank is empty after how many hours will the tank be full ?
Answers
Answer:
2+4+12 = 18
Step-by-step explanation:
18/3= 6
answer is 6 hours
Step-by-step explanation:
One day's work = (1/Number of days to complete the work)
Number of days needed to complete the work = (1/one day's work)
Time required to do a certain work = (work to be done/one day's work)
Calculation:
In one hour,
Tap A fills (1/4) of the tank.
Tap B fills (1/6) of the tank.
Tap C empty (1/12) of the tank.
Now,
In one hour, the portion of the tank filled by the taps A, b and C all together
⇒ (1/4) + (1/6) - (1/12)
Using by L.C.M
⇒ (3 + 2 - 1)/12
⇒ (4/12)
⇒ (1/3)
∴ All the three tap A, B and C together will fill the tank in 3 hours.
Let the total work be 12 units(LCM of 4, 6 and 12).
When all three pipes works together,
⇒ Total efficiency = 3 + 2 - 1
⇒ Total efficiency = 4
Time required to complete to work together = (Total work/Total time)
⇒ Time required to complete to work together = 12/4
⇒ Time required to complete to work together = 3 hours
∴ All the three tap A, B and C together will fill the tank in 3 hours.