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by phoenix9801 » Thu May 31, 2012 4:29 pm
Hose 1 takes twice as long as hose 2 to fill an empty pool. Working together at their respective constant rates, the hoses can fill the pool in 6 hours. How many hours would it take the slower hose to fill the tank working alone?

A) 18
B) 15
C) 12
D) 9
E) 8
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by GMATGuruNY » Thu May 31, 2012 4:56 pm
phoenix9801 wrote:Hose 1 takes twice as long as hose 2 to fill an empty pool. Working together at their respective constant rates, the hoses can fill the pool in 6 hours. How many hours would it take the slower hose to fill the tank working alone?

A) 18
B) 15
C) 12
D) 9
E) 8
Time and rate are RECIPROCALS.
If H1 takes TWICE AS LONG as H2, then H1 works HALF AS FAST as H2.

Let the rate for H1 = 1 unit per hour and the rate for H2 = 2 units per hour.
Their combined rate = 1+2 = 3 units per hour.
When they work together for 6 hours, the amount of work produced = r*t = 3*6 = 18 units. This is the value of the pool.

Time for H1 alone to fill the pool = w/r = 18/1 = 18 hours.

The correct answer is A.
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by GMAT Kolaveri » Thu May 31, 2012 6:50 pm
Backsolving:

Assume slower one to be =18
Faster one = 9 (since slower take twice the time)

1/18 + 1/9
= 1/18 +2/18
=3/18

Total time taken(combined) = 18/3 = 6 (given in the question) Hence slower one =18
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