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# Hose A runs at a constant rate and can fill a 11,000 gallon

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Hose A runs at a constant rate and can fill a 11,000 gallon [#permalink]

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17 Jan 2013, 10:25
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Hose A runs at a constant rate and can fill a 11,000 gallon pool in 44 hours. How much less time would it take to fill the pool if Hose A and Hose B ran simultaneously at their respective constant rates?

(1) Both Hose A and Hose B can fill the same fraction of the pool in one hour.
(2) It takes Hose B twice as long to fill the pool as it takes Hose A and Hose B running simultaneously to fill the pool.
[Reveal] Spoiler: OA

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Re: Hose A runs at a constant rate and can fill a 11,000 gallon [#permalink]

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17 Jan 2013, 10:51
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shreerajp99 wrote:
Hose A runs at a constant rate and can fill a 11,000 gallon pool in 44 hours. How much less time would it take to fill the pool if Hose A and Hose B ran simultaneously at their respective constant rates?

1.Both Hose A and Hose B can fill the same fraction of the pool in one hour.
2.It takes Hose B twice as long to fill the pool as it takes Hose A and Hose B running simultaneously to fill the pool.

Work is 11000 gallon pool.
Hose A rate rA can be established as 11,000/44 = some value
Hose B rate rB is unknown?

1. It means that their rate is same for the fraction. We know the rate for rA and so rB can be found. Sufficient.
2. Relation between rB and rA. rA is known and can found rB. Sufficient.
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Re: Hose A runs at a constant rate and can fill a 11,000 gallon [#permalink]

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17 Jan 2013, 20:47
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shreerajp99 wrote:
Hose A runs at a constant rate and can fill a 11,000 gallon pool in 44 hours. How much less time would it take to fill the pool if Hose A and Hose B ran simultaneously at their respective constant rates?

1.Both Hose A and Hose B can fill the same fraction of the pool in one hour.
2.It takes Hose B twice as long to fill the pool as it takes Hose A and Hose B running simultaneously to fill the pool.

The concepts being tested here are:

1. Rates are additive. Rate of work of A and B working together is the sum of rates of A and B. i.e.
Rate of work of A = R_A
Rate of work of B = R_B
Rate of work of both A and B working together = R_A + R_B

2. Work done = Rate * Time taken
So if A and B do the same work, R_A*T_A = R_B*T_B
R_A/R_B = T_B/T_A
Ratio of their rates will be inverse of ratio of their time taken.

We are given the work done and time taken for hose A so we can find the rate of work for Hose A. (which is 11000/44 = 250 gallons/hr just for clarity)
To find the time taken when both hoses work together, we need to find their combined rate of work. Hence we need to know the rate of work of Hose B too.

1.Both Hose A and Hose B can fill the same fraction of the pool in one hour.
This tells us that their rate of work is the same. Rate of work of hose B = 250 gallons/hr too. When they work together, they will take half the usual time so they will take 22 hrs i.e. 22 hrs less. Sufficient.

2.It takes Hose B twice as long to fill the pool as it takes Hose A and Hose B running simultaneously to fill the pool.
Time taken by B = 2*Time taken together
Rate of B = 1/2 * Rate of working together (since ratio of rates is inverse)
So Rate of B = Rate of A
This boils down to statement 1 and hence is sufficient too.

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Re: Hose A runs at a constant rate and can fill a 11,000 gallon [#permalink]

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04 Mar 2013, 02:28
Hose A runs at a constant rate and can fill a 11,000 gallon pool in 44 hours. How much less time would it take to fill the pool if Hose A and Hose B ran simultaneously at their respective constant rates?

(1) Both Hose A and Hose B can fill the same fraction of the pool in one hour. This means they both have same rates
(2) It takes Hose B twice as long to fill the pool as it takes Hose A and Hose B running simultaneously to fill the pool. Higher efficiency means lower no. of hrs i.e. twice efficiency means 1/2 the time required to fill i.e. 22 hrs. Now, we can find time taken by both A &B then can subtract
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Re: Hose A runs at a constant rate and can fill a 11,000 gallon [#permalink]

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02 Apr 2013, 21:22
shreerajp99 wrote:
Hose A runs at a constant rate and can fill a 11,000 gallon pool in 44 hours. How much less time would it take to fill the pool if Hose A and Hose B ran simultaneously at their respective constant rates?

(1) Both Hose A and Hose B can fill the same fraction of the pool in one hour.
(2) It takes Hose B twice as long to fill the pool as it takes Hose A and Hose B running simultaneously to fill the pool.

"(1) Both Hose A and Hose B can fill the same fraction of the pool in one hour."

This seems poorly worded to me. Why does the word "can" imply equivalent work rate? To me this means Hose B will the fill the 11,000 gallon pool IN AT LEAST 44 hours, thus Hose B rate is greater than or equal to Hose A rate. Just because Hose B CAN fill the pool at the same rate, doesn't mean that it WILL.

I chose B.

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Re: Hose A runs at a constant rate and can fill a 11,000 gallon [#permalink]

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02 Apr 2013, 21:40
WpackAlumDukeBound wrote:
shreerajp99 wrote:
Hose A runs at a constant rate and can fill a 11,000 gallon pool in 44 hours. How much less time would it take to fill the pool if Hose A and Hose B ran simultaneously at their respective constant rates?

(1) Both Hose A and Hose B can fill the same fraction of the pool in one hour.
(2) It takes Hose B twice as long to fill the pool as it takes Hose A and Hose B running simultaneously to fill the pool.

"(1) Both Hose A and Hose B can fill the same fraction of the pool in one hour."

This seems poorly worded to me. Why does the word "can" imply equivalent work rate? To me this means Hose B will the fill the 11,000 gallon pool IN AT LEAST 44 hours, thus Hose B rate is greater than or equal to Hose A rate. Just because Hose B CAN fill the pool at the same rate, doesn't mean that it WILL.

I chose B.

The question clearly states " ... Hose A and Hose B ran simultaneously at their respective constant rates?"

which means that hose A and hose B have their own constant rates at which they work. 'can' only implies that 'hose is able to complete this much work in this much time running at its constant rate'

On the same lines, you could also argue that 'Hose A runs at a constant rate and can fill a 11,000 gallon pool in 44 hours.' in the question talks about one possible rate of hose A but you are given that it runs at a constant rate.

I think you are reading too much in the word 'can'.
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Re: Hose A runs at a constant rate and can fill a 11,000 gallon [#permalink]

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22 May 2014, 21:11
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Re: Hose A runs at a constant rate and can fill a 11,000 gallon [#permalink]

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29 Mar 2016, 04:45
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Hose A runs at a constant rate and can fill a 11,000 gallon [#permalink]

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20 May 2017, 11:28
VeritasPrepKarishma wrote:
shreerajp99 wrote:
Hose A runs at a constant rate and can fill a 11,000 gallon pool in 44 hours. How much less time would it take to fill the pool if Hose A and Hose B ran simultaneously at their respective constant rates?

1.Both Hose A and Hose B can fill the same fraction of the pool in one hour.
2.It takes Hose B twice as long to fill the pool as it takes Hose A and Hose B running simultaneously to fill the pool.

The concepts being tested here are:

1. Rates are additive. Rate of work of A and B working together is the sum of rates of A and B. i.e.
Rate of work of A = R_A
Rate of work of B = R_B
Rate of work of both A and B working together = R_A + R_B

2. Work done = Rate * Time taken
So if A and B do the same work, R_A*T_A = R_B*T_B
R_A/R_B = T_B/T_A
Ratio of their rates will be inverse of ratio of their time taken.

We are given the work done and time taken for hose A so we can find the rate of work for Hose A. (which is 11000/44 = 250 gallons/hr just for clarity)
To find the time taken when both hoses work together, we need to find their combined rate of work. Hence we need to know the rate of work of Hose B too.

1.Both Hose A and Hose B can fill the same fraction of the pool in one hour.
This tells us that their rate of work is the same. Rate of work of hose B = 250 gallons/hr too. When they work together, they will take half the usual time so they will take 22 hrs i.e. 22 hrs less. Sufficient.

2.It takes Hose B twice as long to fill the pool as it takes Hose A and Hose B running simultaneously to fill the pool.
Time taken by B = 2*Time taken together
Rate of B = 1/2 * Rate of working together (since ratio of rates is inverse)
So Rate of B = Rate of A
This boils down to statement 1 and hence is sufficient too.

Hi Karishma ,

I have 1 doubt how Rb = 1/2 (Ra + Rb) converts to Rb = Ra in next step. (I dont think any where Rb+Ra= 2 Ra is given anywhere in prompt).

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Re: Hose A runs at a constant rate and can fill a 11,000 gallon   [#permalink] 20 May 2017, 11:28
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