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**