Math, asked by sumandas41cu, 1 year ago

Two taps A and B can fill a tank in 5 hours and 20 hours respectively. If both the taps are open then due to a leakage, it took 30 minutes more to fill the tank. If the tank is full, how long will it take for the leakage alone to empty the tank?

Answers

Answered by knjroopa
43

Answer:

36

Step-by-step explanation:

Given Two taps A and B can fill a tank in 5 hours and 20 hours respectively. If both the taps are open then due to a leakage, it took 30 minutes more to fill the tank. If the tank is full, how long will it take for the leakage alone to empty the tank?

  • In 1 hour, part filled by pipe A will be 1/5  
  • In 1 hour, part filled by pipe B will be 1/20
  • So in 1 hour, A and B will fill the tank in 1/5 + 1/20 = 1/4
  • Now A and B together can fill the tank in 4 hours
  • It is given due to leakage it took 30 min more to fill the tank.
  • So the tank was filled in 4 30/60 hours = 4 ½ hours = 9/2.
  • Now total filled and leakage by pipe A and B will be 2/9 hours.
  • Again part emptied due to leakage in 1 hour will be  

                  1/4 – 2/9

            = 9 – 8 / 36

            = 1/36

So the leakage can empty the tank in 36 hours.

Answered by amitnrw
22

Answer:

leakage can empty tank in 36 Hrs alone

Step-by-step explanation:

Two taps A and B can fill a tank in 5 hours and 20 hours respectively.

Tap A fills in 1 hr  = 1/5

Tap B fills in 1 hr  = 1/20

Tap A & B together fill in 1hr = 1/5 + 1/20 = (4 + 1)/20 = 1/4

Tap A & B together fill tank in = 1/(1/4) = 4 Hr

Let say leakage can empty tank alone in =  L hrs

Leakage in 1 hr = 1/L

Tap A + Tap B + Leakage in 1 hr = 1/4 - 1/L

Tap A + Tap B + Leakage fill tank in = 4 hr + 30 minutes = 9/2 hrs

Tap A + Tap B + Leakage fill tank in  1 hr = 2/9

1/4 - 1/L = 2/9

=> 9 - 36/L = 8

=> 1 = 36/L

=> L = 36

leakage can empty tank in 36 Hrs alone

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