# Tap A can fill a tank in 6 hours and tap B can empty the same tank in 10 hours. If both taps are opened together, then how much time (in hours) will b

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Tap A can fill a tank in 6 hours and tap B can empty the same tank in 10 hours. If both taps are opened together, then how much time (in hours) will be taken to fill the tank?
1. 16
2. 15
3. 18
4. 20

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Correct Answer - Option 2 : 15

Given:

Tap A can fill the tank in 6 hours and tap B can empty the same tank in 10 hours.

Concept Used:

If a pipe can fill or empty a tank in x hours then in 1 hour it can fill or empty 1/x part of the tank

Calculation:

Let, the total tank is 1 part

Tap A can fill the tank in 6 hours

That means in 6 hours tap A can fill 1 part of the tank

⇒ In 1 hour tap A can fill 1/6 part of the tank

Tap B can empty the tank in 10 hours

That means in 10 hours tap B can empty 1 part of the tank

⇒ In 1 hour tap B can empty 1/10 part of the tank

In 1 hour part of the tank will be filled when both taps are opened together is (1/6 – 1/10)

⇒ (5 - 3)/30

⇒ 2/30

⇒ 1/15

To fill the tank time taken is 1/(1/15) hours

⇒ 15 hours

∴ If both taps are opened together the tank will be filled in 15 hours.

Alternative Solution:

Tap A can fill the tank in 6 hours and tap B can empty the same tank in 10 hours

LCM of 6 and 10 is 30

Let, the tank be of 30 units

In 1 hour tap A can fill (30/6) = 5 unit of the tank

In 1 hour tap B can empty (30/10) = 3 unit of the tank

When both the taps are opened together in 1 hour tank will be fill (5 - 3) = 2 unit

∴ To fill the tank total time needs (30/2) = 15 hours