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