Fill the pool with water through one pipe in 10 hours, through the other – in 8. What part of the pool will remain

Fill the pool with water through one pipe in 10 hours, through the other – in 8. What part of the pool will remain to be filled if both pipes are operated simultaneously for 1 hour.

To solve the problem, it is necessary to determine the performance of each pipe.

Imagine the entire volume of water in the pool as 100% or 1.

In this case, in 1 hour of operation, the first pipe will fill:

1/10 = 1/10 of the pool.

The second pipe will fill:

1/8 = 1/8 of the pool.

We find the productivity of the work of two pipes when working together.

To do this, we summarize the productivity of each pipe.

1/10 + 1/8 = (Common denominator 40) = 4/40 + 5/40 = 9/40.

In this case, after 1 hour of joint work, it remains to fill:

1 – 9/40 = 31/40 part of the pool.