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Two pumps working together at their respective constant [#permalink]
18 Dec 2007, 12:35

Two pumps working together at their respective constant rates took 4 hrs to fill a pool. If pump 1 was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool alone?

i think it would have helped if you put the answer choices, but regardless, i got 1.48 hrs.
is that correct?

mbsnyc wrote:

Two pumps working together at their respective constant rates took 4 hrs to fill a pool. If pump 1 was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool alone?

Two pumps working together at their respective constant rates took 4 hrs to fill a pool. If pump 1 was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool alone?

x and 1.5x are rates

1/x + 1/1.5x = 1/4

2.5/1.5x = 1/4 => 1.5x = 10 ( We need to calculate 1.5x here)
So 10 hrs.

Two pumps working together at their respective constant rates took 4 hrs to fill a pool. If pump 1 was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool alone?

x + 1.5x = 4 hrs
x = 8/5 & 1.5x = 24/10

Let Pool = 16
Time = 4 hrs
Speed = 4 pool/hour

Speed of slower = x = 8/5 pool/hr
Speed of faster = 1.5x = 24/10 pool/hr

A = 16
S = 24/10 pool/hr
T = 16/24 * 10 = 20/3 = 6 hrs 40 mins

Two pumps working together at their respective constant rates took 4 hrs to fill a pool. If pump 1 was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool alone?

we have x and y. x= faster pump.

1.5y+y=2.5y total rate--> 1pool/2.5y ---> 1pool/2.5y=4hrs

y=1/10 3/2*1/10= 3/20 --> 20/3 =6.666 or 6hrs and 40min.

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