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 see the application of 6, 2, and 3 in the question, so we take the LCM of 6, 2, and 3 = 18 lts. as the capacity of the pool.

Now A can fill 18 lts in 6 hrs, thus (18/6)*2 = 6 lts. in the first two hours of the sole operation.

Thus, 18 - 6 = 12 lts is left to be filled by both A and B together in 3 hours.

A will fill (18/6)*3 = 9 lts. alone in the joint operation. This mean that 12 - 9 = 3 lts was filled by B alone in 3 hours.

=> Rate of B = (3/3)*18 = 18 hrs.

OA: A

Hope this helps!

-Jay

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Hi All,

There are a number of different ways to approach this prompt, depending on how you 'see' the question. Here's an approach that's based on simple algebra and some note-taking/logic:

To start, we're told that Hose A can fill a pool in 6 hours; thus, it fills 1/6 of the pool in 1 hour.

Hose A starts filling the pool for 2 hours; thus, it fills 2(1/6) = 2/6 of the pool during that time.

For the next 3 hours, Hose A AND Hose B fill the pool - and finish filling the pool. In those 3 hours, Hose A will fill 3/6 of the pool (and since it already filled 2/6 of the pool before, that Hose is responsible for filling 5/6 of the pool). By extension, during those 3 hours, Hose B filled the remaining 1/6 of the pool.

So if it takes Hose B 3 hours to fill 1/6 of the pool, then it will take (3)(6) = 18 hours to fill the pool by itself.

Final Answer: A

As an aside, it's worth noting that one of the wrong answers (Answer D) is what you would end up with if you FORGOT that Hose B spent 3 hours to fill 1/6 of the pool.

GMAT assassins aren't born, they're made,
Rich

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