SOLUTION: a 5 horsepower(hp) pump can empty a pool in 5 gours. A smaller, 2hp pump empties the same pool in 8 hours. The pumps are used together to being emptying this pool. After two hours,

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: a 5 horsepower(hp) pump can empty a pool in 5 gours. A smaller, 2hp pump empties the same pool in 8 hours. The pumps are used together to being emptying this pool. After two hours,      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 417447: a 5 horsepower(hp) pump can empty a pool in 5 gours. A smaller, 2hp pump empties the same pool in 8 hours. The pumps are used together to being emptying this pool. After two hours, the hp pump breaks down. How long will it take the larger pump top empty the pool?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A 5 hp pump can empty a pool in 5 hrs.
A smaller, 2 hp pump empties the same 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 empty the pool?
:
Let t = time for the larger pump to complete emptying the pool
:
Operating time of the larger pump = (t+2)
Operating time of the smaller pump = 2
:
Let the completed job = 1; (an empty pool)
:
%28%28t%2B2%29%29%2F5 + 2%2F8 = 1
multiply by 40
40*%28%28t%2B2%29%29%2F5 + 40*2%2F8 = 40(1)
cancel the denominators and you have:
8(t+2) + 5(2) = 40
:
8t + 16 + 10 = 40
8t + 26 = 40
8t = 40 - 26
8t = 14
t = 14/8
t = 1.75 hrs for the large pump to complete the job (Total time of large pump = 3.75)
:
:
Check this:
3.75/5 + 2/8 =
.75 + .25 = 1; confirms our solution