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Two water pumps, working simlutaneosly at their respective [#permalink]
12 Nov 2006, 04:19
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Two water pumps, working simlutaneosly at their respective rates, took exactly 4 hours to fill a certain swimming pool. If the constant rate of one pump was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool if it had worked alone at its constant rate?
Finding x is easy, but i found difficulty in finding the total work done....
Since the rate of the faster one is 3/2 of the slower machine and the rate is expressed as 1/x and 1/y for the 2 machine.
I have: 1/x = (3/2) (1/y) <=> y = (3x)/2 <1>
another equation is 1/x +1/y = 1/4 <2>
Substitute the value of y into the second equation and solve for x, that would be the number of hrs x will take to work alone (20/3)