SOLUTION: A swimming pool can be emptied in 6 hours using a 10-horsepower pump along with a 6-horsepower pump. The 6-horsepower pump requires 5 hours more than the 10horsepower pump to empty

Algebra ->  Rate-of-work-word-problems -> SOLUTION: A swimming pool can be emptied in 6 hours using a 10-horsepower pump along with a 6-horsepower pump. The 6-horsepower pump requires 5 hours more than the 10horsepower pump to empty      Log On


   

Question 674181: A swimming pool can be emptied in 6 hours using a 10-horsepower pump along with a 6-horsepower pump. The 6-horsepower pump requires 5 hours more than the 10horsepower pump to empty the pool when working by itself. How long would it take to empty the pool using just the 10-horsepower pump?
How do I go about this? I have tried just about everything I can think of. My book does not provide an example, so I can't work backwards.
I was thinking this,
let 6hp pump =x
10hp pump =y
10hp + 6hp = c
so the rate of emptying the pool with pump c= -(1/6)
pump x= -(1)/(5+y)
pump y= -(1)/(x-5)
Assuming I applied these rates correctly, where do I go from here?
Thanks.
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

The 6-horsepower pump requires 5 hours more than the 10horsepower pump to empty the pool when working by itself

10horsepower pump highlight%28x%29

6-horsepower pump highlight%28x%2B5%29

PER hr KEY: emptied in 6 hours using both

Using both Algebraically

+1%2Fx+%2B+1%2F%28x%2B5%29+=+1%2F6+ |Multiplying thru by 6x(x+5) so as all denominators = 1

 6(x+5) + 6x = x(x+5)

 6x + 30 + 6x = x^2 + 5x

   x^2 - 7x - 30 = 0

factoring

(x+3)(x-10) = 0   ||x = 10hr, time it would take the 10hp by itself

Note: Negative solution is tossed out for unit measrement