Question

There are three inlet taps A, B and C in a tank. They can fill the tank in 5 h, 30 h and 105 h respectively, if opened individually. The tank is empty. At first, all the 3 taps are opened simultaneously. After one hour, tap C is closed and A and B are kept running. After another one hour, tap B is also closed. The remaining portion of the tank was filled by tap A alone. What percentage of the tank was filled by tap A in this process (answer to the nearest percentage)?

Answer 1

Answer:

92%

Step-by-step explanation:

Tap A can fill the tank in 5 hours

1 hour = 1/5 of the tank

Tap B can fill the tank in 30 hours

1 hour = 1/30 of the tank

Tap C can fill the tank in 105 hours

1 hour = 1/105 of the tank

Find the amount of water filled in the 1st hour

Tap A, B and C are opened

1 hour = 1/5 + 1/30 + 1/105 = 17/70 of the tank

Find the amount of water filled in the 2nd hour

Tap A and B are opened

1 hour = 1/5 + 1/30 = 7/30 of the tank

Find the rest of the water filled in by Tap A:

1 - 17/70 - 7/30 = 11/21 of the tank

Find the amount of water filled by Tap A:

Amount of water = 1/5 + 1/5 + 11/21 = 97/105

Find the percentage of water filled by Tap A:

Percentage of water = 97/105 x 100 = 92%

Answer: 92%