Pipe A can fill a swimming pool in 10 hours.Pipe B can fill the same pool in 8 hours. If pipe A and pibe B work together for the first 2 hours and then pipe B stops working, how long will it take pipe A to finish the job?
in 2 hours, they can fill
2(1/10 + 1/8) = 9/20 of the pool. So, at 1/10 pool/hr, A takes an additional
(11/20) / (1/10) = 11/2 hours
To solve this problem, we need to determine how much of the job can be completed in the first 2 hours when both Pipe A and Pipe B are working together. Then we can calculate how much of the job still needs to be done and how long it would take Pipe A to complete it.
First, let's find the rate at which each pipe can fill the pool per hour. Pipe A takes 10 hours to fill the pool, so its rate is 1/10 of the pool per hour. Similarly, Pipe B takes 8 hours to fill the pool, so its rate is 1/8 of the pool per hour.
When both pipes A and B work together, their combined fill rate is the sum of their individual rates, which means they can fill 1/10 + 1/8 = 9/40 of the pool per hour.
Now, let's calculate how much of the job they can complete together in the first 2 hours. Multiplying their combined fill rate (9/40) by the time (2 hours), we get (9/40) * 2 = 9/20 of the pool.
Therefore, in the first 2 hours, they can fill 9/20 of the pool, leaving (20/20 - 9/20) = 11/20 of the pool still needing to be filled.
Since Pipe B stops working after the initial two hours, Pipe A will complete the remaining job on its own. Pipe A's rate is 1/10 of the pool per hour, so it will take Pipe A (11/20) / (1/10) = 11/2 = 5.5 hours to finish filling the pool.
Therefore, it will take Pipe A 5.5 hours to finish the job after Pipe B stops working.