Mo2men wrote:Hoses A and B spout water at different constant rates, and hose A can fill a certain pool in 6 hours. Hose A filled the pool alone for the first 2 hours and the two hoses, working together, then finished filling the pool in another 3 hours. How many hours would it have taken hose B, working alone, to fill the entire pool?
A 18
B 15
C 12
D 6
E 3
We are given that hose A can fill the pool in 6 hours alone. Thus, the rate of hose A is 1/6.
Since we are not given the rate of hose B, we can let B = the number of hours it takes hose B to fill the pool alone. Thus, the rate of hose B is 1/B.
We are given that hose A worked alone for the first 2 hours and then the two hoses worked together for another 3 hours to complete the job. Thus, the time worked by hose A is 5 hours and the time worked by hose B is 3 hours. Since work = rate x time, we can calculate the work done by hose A and hose B.
Work done by hose A = (1/6) x 5 = 5/6
Work done by hose B = (1/B) x 3 = 3/B
Finally, since the completed job = 1, we can sum the work values of hoses A and B and set the sum to 1.
5/6 + 3/B = 1
Multiplying the entire equation by 6B gives us:
5B + 18 = 6B
18 = B
Thus, hose B, when working alone, can complete the job in 18 hours.
Answer:
A
