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One pipe can fill a pool 1.25 times faster than a second

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One pipe can fill a pool 1.25 times faster than a second [#permalink] New post 26 Jan 2014, 14:48
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One pipe can fill a pool 1.25 times faster than a second pipe. When both pipes are opened, they fill the pool in five hours. How long would it take to fill the pool if only the slower pipe is used?

A. 11.25
B. 11.52
C. 1.25
D. 9
E. 7.2

I would appreciate if someone can explain me this using the Rate Time Work chart (as used by Manhattan). That will help me understand better and easier.

Thank you!
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Re: One pipe can fill a pool 1.25 times faster than a second [#permalink] New post 26 Jan 2014, 19:54
Expert's post
Not sure about the chart, but this is how I would approach it...I'm sure there are shortcuts/more efficient methods.

Slower Pipe: 1/x (1 Pool in x hours)
Faster Pipe: 1.25/x (1.25 Pools in x hours)

Combined Rate: 2.25x/x^2
Combined Rate to Fill 1 Pool: x^2/2.25x = 5hours

x=0 or 11.25

Slower Pipe Rate: 1/11.25 (1 pool in 11.25hours)
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Re: One pipe can fill a pool 1.25 times faster than a second [#permalink] New post 27 Jan 2014, 00:43
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flower07 wrote:
One pipe can fill a pool 1.25 times faster than a second pipe. When both pipes are opened, they fill the pool in five hours. How long would it take to fill the pool if only the slower pipe is used?

A. 11.25
B. 11.52
C. 1.25
D. 9
E. 7.2

I would appreciate if someone can explain me this using the Rate Time Work chart (as used by Manhattan). That will help me understand better and easier.

Thank you!


Say the rate of the slower pipe is R pool/hour, then the rate of the faster pipe would be 1.25R=5R/4. Since when both pipes are opened, they fill the pool in five hours, then their combined rate is 1/5 pool/hour.

Thus we have that R + 5R/4 = 1/5 --> R = 4/45 pool/hour --> time is reciprocal of rate thus it's 45/4 =11.25 hours.

Answer: A.

Hope it's clear.
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Re: One pipe can fill a pool 1.25 times faster than a second [#permalink] New post 12 Feb 2014, 08:34
flower07 wrote:
One pipe can fill a pool 1.25 times faster than a second pipe. When both pipes are opened, they fill the pool in five hours. How long would it take to fill the pool if only the slower pipe is used?

A. 11.25
B. 11.52
C. 1.25
D. 9
E. 7.2

I would appreciate if someone can explain me this using the Rate Time Work chart (as used by Manhattan). That will help me understand better and easier.

Thank you!


Hi
Trying to solve the way you want
From the table you can find the rate,R, of smaller pipe. Then time is inverse of rate so answer would be A...


Rgds
Prasannajeet
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Re: One pipe can fill a pool 1.25 times faster than a second [#permalink] New post 07 May 2014, 01:59
Let's say slower pipes speed = v lph (litter per hr)
then faster = 1.25v lph

Together = 2.25v lph
Time = 5 hrs

Volume of Pool = 2.25v * 5
Time taken by slower pipe alone = 2.25v*5/v = 11.25 hrs
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Re: One pipe can fill a pool 1.25 times faster than a second [#permalink] New post 08 May 2014, 12:33
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Rate of pipe 1 - 1/a
pipe2 - 1/b
given rate pipe1=1.125pipe2
1/a=1.25(1/b)
also given 1/a+1/b=1/5
hence b=2.25*5=11.25
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Re: One pipe can fill a pool 1.25 times faster than a second [#permalink] New post 26 Jul 2014, 23:39
entire job is done in 5 hours, hence 20% of the job is done per hour. slower one does 8.88% and faster one does 11.12% of it. so the slower pipe completes the task in 100/8.88 = 11.25 hours
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Re: One pipe can fill a pool 1.25 times faster than a second [#permalink] New post 15 Sep 2014, 06:13
flower07 wrote:
One pipe can fill a pool 1.25 times faster than a second pipe. When both pipes are opened, they fill the pool in five hours. How long would it take to fill the pool if only the slower pipe is used?

A. 11.25
B. 11.52
C. 1.25
D. 9
E. 7.2

I would appreciate if someone can explain me this using the Rate Time Work chart (as used by Manhattan). That will help me understand better and easier.

Thank you!


Let me give you an elaborate explanation on how you can use this (I have also used MGMAT's way of approaching the problems, but adopted according to the time needs)

Let's assume the rate of the slower pipe is: x
So, the rate of the faster pipe would be: 1.25x
Working together, their rate will be x + 1.25x = 2.25x

As the question says, they take 5 hours to fill up the tank working together, so:

Rate x Time = Work
2.25x x 5 = 1
2.25x = 1/5
x = 1/5 / 2.25 = 4/45

So, the rate of the slower pipe, x = 4/45

Now, let's see how long it takes for the slower pipe to complete the task:

Rate x Time = Work
4/45 x Time = 1
Time = 1 / 4/45 = 45/4 = 11.25

Hope it's clear :)
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Re: One pipe can fill a pool 1.25 times faster than a second [#permalink] New post 15 Sep 2014, 06:14
flower07 wrote:
One pipe can fill a pool 1.25 times faster than a second pipe. When both pipes are opened, they fill the pool in five hours. How long would it take to fill the pool if only the slower pipe is used?

A. 11.25
B. 11.52
C. 1.25
D. 9
E. 7.2

I would appreciate if someone can explain me this using the Rate Time Work chart (as used by Manhattan). That will help me understand better and easier.

Thank you!


Let me give you an elaborate explanation on how you can use this (I have also used MGMAT's way of approaching the problems, but adopted according to the time needs)

Let's assume the rate of the slower pipe is: x
So, the rate of the faster pipe would be: 1.25x
Working together, their rate will be x + 1.25x = 2.25x

As the question says, they take 5 hours to fill up the tank working together, so:

Rate x Time = Work
2.25x x 5 = 1
2.25x = 1/5
x = 1/5 / 2.25 = 4/45

So, the rate of the slower pipe, x = 4/45

Now, let's see how long it takes for the slower pipe to complete the task:

Rate x Time = Work
4/45 x Time = 1
Time = 1 / 4/45 = 45/4 = 11.25

Hope it's clear :)
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Re: One pipe can fill a pool 1.25 times faster than a second [#permalink] New post 15 Sep 2014, 16:34
I set it up as 1/r + 1/(5r/4) and derived 1/r + 4/5r and got r=9. What's wrong with that?
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Re: One pipe can fill a pool 1.25 times faster than a second [#permalink] New post 15 Sep 2014, 20:11
Expert's post
bankerboy30 wrote:
I set it up as 1/r + 1/(5r/4) and derived 1/r + 4/5r and got r=9. What's wrong with that?


If r is the rate, then when you reciprocate it you'll get the time, so 1/r + 1/(5r/4) is the sum of the times and cannot equal to the rate (1/5).

Does this make sense?
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Re: One pipe can fill a pool 1.25 times faster than a second   [#permalink] 15 Sep 2014, 20:11
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