SOLUTION: A 5-horsepower (hp) pump can empty a pool in 5 hours. A smaller, 2-hp pump can empty a pool in 8 hours. The pumps are used together to begin emptying this pool. After two hours, th

Algebra ->  Rate-of-work-word-problems -> SOLUTION: A 5-horsepower (hp) pump can empty a pool in 5 hours. A smaller, 2-hp pump can empty a pool in 8 hours. The pumps are used together to begin emptying this pool. After two hours, th      Log On


   

Question 531209: A 5-horsepower (hp) pump can empty a pool in 5 hours. A smaller, 2-hp pump can empty a pool in 8 hours. The pumps are used together to begin emptying this pool. After two hours, the 2-hp pump breaks down. How long will it take the larger pump to finish emptying the pool?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A 5-horsepower (hp) pump can empty a pool in 5 hours.
A smaller, 2-hp pump can empty a pool in 8 hours.
The pumps are used together to begin emptying this pool.
After two hours, the 2-hp pump breaks down.
How long will it take the larger pump to finish emptying the pool?
:
Let t = additional time to finish emptying the pool
Let the completed job = 1, (an empty pool)
:
%28%28t%2B2%29%29%2F5 + 2%2F8 = 1
multiply by 40 to clear the denominators, results:
8(t+2) + 5(2) = 40
8t + 16 + 10 = 40
8t + 26 = 40
8t = 40 - 26
8t = 14
t = 14/8
t = 1.75 more hrs for the large pump to empty the pool
:
:
Check that (the large pump operates for a total of 3.75 hrs)
3.75%2F5 + 2%2F8 =
.75 + .25 = 1