SOLUTION: This is a practice GRE question from a Kaplan helpsite. I believe the answer is wrong, but maybe I'm not understanding it. The question is: One hose can fill a pool 2.25 times

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Question 886388: This is a practice GRE question from a Kaplan helpsite. I believe the answer is wrong, but maybe I'm not understanding it. The question is:
One hose can fill a pool 2.25 times faster than another hose can. When both hoses are used, they fill the pool in 10 hours. How long would it take to fill the pool if only the faster hose is used?
Their answer is 90/13 hours. Logically this doesn't make sense, how can one hose (7 hours) be faster than two hoses (10 hours)? I came up with 14.4 hours (13/9) but can't seem to get the math to look right. Thanks for your help!

Found 2 solutions by stanbon, Theo:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
One hose can fill a pool 2.25 (9/4) times faster than another hose can. When both hoses are used, they fill the pool in 10 hours. How long would it take to fill the pool if only the faster hose is used?
------
Faster hose DATA:
time = x hrs/job ; rate = 1/x job/hr
--------
Slower hose DATA:
time = (9/4)x hrs/job ; rate = (4/9x) job/hr
--------
Together DATA:
time = 10 hrs/job; rate = 1/10 job/hr
------------------------------------------
Equation:
rate + rate = together rate
1/x + (4/(9x)) = 1/10
-----
Multiply thru by 90x to get:
90 + 40 = 9x
9x = 130
x = 130/9 = 14.4 hrs (time for faster hose)
(9/4)x = 32.5 hrs (time for slower hose)
=============================================
You are correct.
=============
Cheers,
Stan H.
-------------------

Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
i believe you are correct.
the correct solutuion appears to be 14.44444444...... which is equivalent to 130/9.

their answer of 90/13 doesn't make much sense if i understand the problem correctly.

here's how i solved it.

rate of the slow hose is x.
rate of the fast hose is 2.25x

when they work together, their rates are additive.

you get the equation of:

(x + 2.25x) * 10 = 1

simplify this to get 3.25x * 10 = 1

solve for x to get x = 1/32.5

that's the rate of the slow hose.

the rate of the fast hose is 2.25/32.5

the fast hose, filling the pool on its own, fills at a rate of 2.25/32.5 of the pool in one hour.

the formula becomes:

2.25/32.5 * T = 1

solve for T to get T = 32.5/2.25 = 14.4444444 which convert to a fraction of 130/9.

the websites aren't perfect and errors do crop up from time to time.
you did everything right.

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