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S and T are two water pumps that run at constant rates. If pump T pump

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S and T are two water pumps that run at constant rates. If pump T pump  [#permalink]

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New post 26 Jun 2015, 01:51
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Question Stats:

64% (02:05) correct 36% (02:02) wrong based on 411 sessions

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S and T are two water pumps that run at constant rates. If pump T pumped alone, how many more hours would it take for it to finish pumping a large container than it would take for pump S to accomplish the same task by itself?

(1) When both the water pumps work together, they finish pumping a large container in 2/3 rd the time it takes for pump S to finish the task alone.

(2) Pump T is capable of pumping a large container in twice the time that it takes pump S to accomplish the same task alone.


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Re: S and T are two water pumps that run at constant rates. If pump T pump  [#permalink]

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New post 26 Jun 2015, 02:26
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Bunuel wrote:
S and T are two water pumps that run at constant rates. If pump T pumped alone, how many more hours would it take for it to finish pumping a large container than it would take for pump S to accomplish the same task by itself?

(1) When both the water pumps work together, they finish pumping a large container in 2/3 rd the time it takes for pump S to finish the task alone.

(2) Pump T is capable of pumping a large container in twice the time that it takes pump S to accomplish the same task alone.


Kudos for a correct solution.


immediate thinking on seeing this Q is ...
the answer asked is in terms of hour...
1) stat 1 gives us the ratio between S and T... insuff
2) stat 2 gives us the ratio between S and T again... insuff

combined nothing new.. E
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S and T are two water pumps that run at constant rates. If pump T pump  [#permalink]

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New post 26 Jun 2015, 03:31
Bunuel wrote:
S and T are two water pumps that run at constant rates. If pump T pumped alone, how many more hours would it take for it to finish pumping a large container than it would take for pump S to accomplish the same task by itself?

(1) When both the water pumps work together, they finish pumping a large container in 2/3 rd the time it takes for pump S to finish the task alone.

(2) Pump T is capable of pumping a large container in twice the time that it takes pump S to accomplish the same task alone.


Kudos for a correct solution.


Not sure, but here what you can read out of the Statements:

1: Combined Time is 2/3x where x is the time of Pump S. > IS because there is no information about Pump T to S itself.
2: Pump T needs twice the time of Pump S. Alone it is insufficient.

Statement 1 + 2:
Combined Time 2/3X Hours
Pump S Time: X Hours
Pump T Time: 2X Hours

To solve, we needed to have at least X. Like this, there is no way to figure out the hours.

Answer E
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Re: S and T are two water pumps that run at constant rates. If pump T pump  [#permalink]

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New post 26 Jun 2015, 03:58
Bunuel wrote:
S and T are two water pumps that run at constant rates. If pump T pumped alone, how many more hours would it take for it to finish pumping a large container than it would take for pump S to accomplish the same task by itself?

(1) When both the water pumps work together, they finish pumping a large container in 2/3 rd the time it takes for pump S to finish the task alone.

(2) Pump T is capable of pumping a large container in twice the time that it takes pump S to accomplish the same task alone.


Kudos for a correct solution.


To answer this question, we need to know the relationship between the efficiencies of S and T and the actual amount of time taken by either of them or both of them together to fill the tank

Statement 1:When both the water pumps work together, they finish pumping a large container in 2/3 rd the time it takes for pump S to finish the task alone.

It gives us the relationship between the efficiencies of S and T but the value of time taken by any one of them or together (in hours) is unknown to arrive at any value of the number of hours

NOT SUFFICIENT

Statement 2:Pump T is capable of pumping a large container in twice the time that it takes pump S to accomplish the same task alone.

It gives us the relationship between the efficiencies of S and T but the value of time taken by any one of them or together (in hours) is unknown to arrive at any value of the number of hours

NOT SUFFICIENT

Combining the two statements

It gives us the relationship between the efficiencies of S and T but the value of time taken by any one of them or together (in hours) is unknown to arrive at any value of the number of hours

NOT SUFFICIENT

Answer: option E
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Re: S and T are two water pumps that run at constant rates. If pump T pump  [#permalink]

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New post 26 Jun 2015, 04:05
2
It's a GMATPrep question with very slight changes to the wording:

machines-x-and-y-work-at-their-respective-constant-rates-128011.html
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Re: S and T are two water pumps that run at constant rates. If pump T pump  [#permalink]

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New post 26 Jun 2015, 12:21
1. Time taken by s=S; T+S = 2/3 * S; but we do not know the value of time so insufficient
2. Time taken by t=2S,; we do not know how much time did S take to answer the question, so insufficient

C ) 1+2 still insufficient, as value of time is not mentioned. So Insufficient
Hence answer is E
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Re: S and T are two water pumps that run at constant rates. If pump T pump  [#permalink]

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New post 29 Jun 2015, 06:12
1
(1) When both the water pumps work together, they finish pumping a large container in 2/3 rd the time it takes for pump S to finish the task alone.

1/T+1/S=3/2S => 1/T=1/2S => 2S=T
Actual number of hours??
NS

(2) Pump T is capable of pumping a large container in twice the time that it takes pump S to accomplish the same task alone.
2S=T
Actual number of hours??
NS

Combining
Actual number of hours??
NS

Answer E
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Re: S and T are two water pumps that run at constant rates. If pump T pump  [#permalink]

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New post 03 Jun 2016, 20:49
..my take on this-
The first statement -- when both pumps work together the time it takes to complete the entire task is 2/3 rd of what pump S will take.
let T be the time it takes pump T to finish the task alone and S be for pump S
stat#1 --> 1/T+1/S = 1/(2/3)S
1/T + 1/S = 3/(2S).
1/T = 3/(2S) - 1/S = 1/S*(1/2) = 1/2S => S/T = 1/2. Ratio between S and T cant be solved
stat #2--->Time it takes for T to do the job alone is twice the time it takes for S to do alone.
--> T/S = 2/1 --> this is identical to stat #1.
So we cant generate new info even after combining the statements. So E
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Re: S and T are two water pumps that run at constant rates. If pump T pump  [#permalink]

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New post 13 Aug 2017, 14:26
My 2 cents:
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Re: S and T are two water pumps that run at constant rates. If pump T pump  [#permalink]

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New post 07 Sep 2018, 22:11
Both the statement doesn't cover the main point of the question the no. Of hours more than need by S alone... So option E.

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Re: S and T are two water pumps that run at constant rates. If pump T pump &nbs [#permalink] 07 Sep 2018, 22:11
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