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Two pumps working together at their respective constant

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mbsnyc
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Two pumps working together at their respective constant [#permalink] New post   18 Dec 2007, 13:35
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?



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Re: Two pumps working together at their respective constant [#permalink] New post   18 Dec 2007, 13:46
Suppose slower pump rate is x gallons/Hr
Faster will be 1.5x gallons/Hr

In 4 Hr both combined will pump: 4x + 1.5x*4 = 10x

Time taken by faster pump to fill: 1.5x*T = 10x

Hence Time Taken, T = 6.667 Hrs.
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Re: Two pumps working together at their respective constant [#permalink] New post   18 Dec 2007, 13:46
hi mbsnyc,

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?
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Re: Two pumps working together at their respective constant [#permalink] New post   18 Dec 2007, 13:49
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michaelny2001 wrote:
i got 1.48 hrs. is that correct?

No.

Together 4hrs. Therefore your answer should be more than 4 hrs.
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Re: Two pumps working together at their respective constant [#permalink] New post   18 Dec 2007, 14:14
the answer was (16/3) - sorry I didn't bring all the answer choices with me to work - just had written down the question.

Was anyone able to come up with the solution 16/3?
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Re: Two pumps working together at their respective constant [#permalink] New post   18 Dec 2007, 18:19
dkverma wrote:
Suppose slower pump rate is x gallons/Hr
Faster will be 1.5x gallons/Hr

In 4 Hr both combined will pump: 4x + 1.5x*4 = 10x

Time taken by faster pump to fill: 1.5x*T = 10x

Hence Time Taken, T = 6.667 Hrs.


This is what I got too:

Let x be the number of gallons the slower pipe pumps per hour.

so in 4 hours it pumps = 4x gallons

The faster pipe pumps = 4 (1.5X) = 6x gallons

So the capacity of the tank is = 10x gallons.

The faster pipe pumped 6x in 4 hours
so to fill 10x it would take

= 10x * 4 / 6x = 20/3 = 6.66 hours
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Re: Two pumps working together at their respective constant [#permalink] New post   18 Dec 2007, 19:48
Here is my way:

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

2.5 / 1.5x = 1/4
1.5x = 10
x = 100/15

Hence,
1.5x = 100 * 3 / 15 * 2 = 10

Answer is 10.

Checking myself:

1/10 + 1/ (100/15) = 1/T
T = 4
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Re: Two pumps working together at their respective constant [#permalink] New post   18 Dec 2007, 20:05
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?




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.
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Re: Two pumps working together at their respective constant [#permalink] New post   18 Dec 2007, 20:45
Guys

What is the OA ?

I am getting 20/3 similar to dkverma.

Total Size of tank = 10N gallons. (assuming pump rates to be N & 1.5N/hr)

So to fill the tank with the larger pump it would take

10/1.5 = 20/3 Hrs = 6.667

Pls advise.
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Re: Two pumps working together at their respective constant [#permalink] New post   18 Dec 2007, 23:23
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4*V+4*1.5V=1
V=1/10

t*1.5/10=1
t=20/6=6h40m
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Re: Two pumps working together at their respective constant [#permalink] New post   19 Dec 2007, 00:05
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?


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
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Re: Two pumps working together at their respective constant [#permalink] New post   19 Dec 2007, 06:48
Whatever wrote:
Here is my way:

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

2.5 / 1.5x = 1/4
1.5x = 10
x = 100/15

Hence,
1.5x = 100 * 3 / 15 * 2 = 10

Answer is 10.

Checking myself:

1/10 + 1/ (100/15) = 1/T
T = 4


I had the same approach - I don't understand why you can't apply it in this case?!
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Re: Two pumps working together at their respective constant [#permalink] New post   19 Dec 2007, 07:10
Whatever wrote:
Here is my way:

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

2.5 / 1.5x = 1/4
1.5x = 10
x = 100/15

Hence,
1.5x = 100 * 3 / 15 * 2 = 10

Answer is 10.

Checking myself:

1/10 + 1/ (100/15) = 1/T
T = 4


This seems to be the correct method.
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Re: Two pumps working together at their respective constant [#permalink] New post   19 Dec 2007, 14:43
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?



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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Re: Two pumps working together at their respective constant [#permalink]
19 Dec 2007, 14:43
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