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M12-10

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M12-10  [#permalink]

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New post 16 Sep 2014, 00:46
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With both valves open, the pool will be filled with water in 48 minutes. The first valve alone would fill the pool in 2 hours. If the second valve emits 50 cubic meters of water more than the first every minute, then what is the capacity of the pool?

A. 9000 cubic meters
B. 10500 cubic meters
C. 11750 cubic meters
D. 12000 cubic meters
E. 12500 cubic meters

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Re M12-10  [#permalink]

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New post 16 Sep 2014, 00:46
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Official Solution:

With both valves open, the pool will be filled with water in 48 minutes. The first valve alone would fill the pool in 2 hours. If the second valve emits 50 cubic meters of water more than the first every minute, then what is the capacity of the pool?

A. 9000 cubic meters
B. 10500 cubic meters
C. 11750 cubic meters
D. 12000 cubic meters
E. 12500 cubic meters


Let the rate of the first valve be \(x\) cubic meters per minute, then the rate of the second valve will be \(x+50\) cubic meters per minute.

As both valves open fill the pool in 48 minutes then the capacity of the pool equals to \(C=\text{time * combined rate}=48(x+x+50)=48(2x+50)\).

But as the first valve alone fills the pool in 2 hours (120 minutes) then the capacity of the pool also equals to \(C=\text{time * rate}=120x\);

So, \(120x=48(2x+50)\), which gives \(x=100\), therefore \(C=120x=12,000\).


Answer: D
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Re: M12-10  [#permalink]

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New post 20 Dec 2014, 09:51
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I had a different approach, which is more lengthy, since I could not even consider the method used above :)
rate of pump A = 1/120 (2h = 120 min)
rate A+B = 1/48
Rate B = 1/48 - 1/120 = 1/80
1/80 - 1/120 is 50m^3
1/240 = 50m^3
1 pool = 50m^3 * 240 = 12,000
i am not sure whether this is a correct method :)
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Re: M12-10  [#permalink]

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New post 10 Feb 2016, 04:37
Bunuel wrote:
With both valves open, the pool will be filled with water in 48 minutes. The first valve alone would fill the pool in 2 hours. If the second valve emits 50 cubic meters of water more than the first every minute, then what is the capacity of the pool?

A. 9000 cubic meters
B. 10500 cubic meters
C. 11750 cubic meters
D. 12000 cubic meters
E. 12500 cubic meters



I have a little different approach.
We have to find number of minutes in which the second valve alone would fill the pool (let's denote this number x):
1/120 + 1/x = 1/48
So x = 80.
The capacity of pool we can find solving the following equation:
120y = 80 (y+50), where y is a volume of water that emit the first valve.
y = 100.
The capacity of pool = 100 * 120 = 12 000 cub.m.
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Re: M12-10  [#permalink]

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New post 18 May 2016, 23:40
Hi chetan2u / @Experts

I m quiet confused by the language of this question. First statement says that if we open both valves then the pool will be filled with water in 48 mins.

so, we can have 2 scenarios.
Lets name the valves to A and B.

Scenario 1- A and B, both are inlet valves, which fills the tank and tank will take 48 mins to get filled

scenario 2- A is inlet valve and B is outlet valve.
2a) if rate of valve B is LESS then that of A. Tank will get filled in 48 mins
2b) if rate of valve B is MORE then that of A. In this case the tank can not be filled--> Therefore this case will not exist.

reading further the question says- The first valve alone would fill the pool in 2 hours.
If case 1- Then we can take the first valve as A and say, Valve A take 120 mins to fill the tank..Till here its ok

Reading further the question says- If the second valve EMITS 50 cubic meters of water more than the first every minute,
After reading this statement, its clear that there is 1 valve which emits the water.
Therefore, our existing possibilty MUST BE from scenario2.

But this part of the sentence also says that the second valve emits 50 cubic meter MORE water.
If rate of emission is more than the inlet rate, how would the tank be filled..??

Can you please assist.



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Re: M12-10  [#permalink]

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New post 19 May 2016, 00:25
PrakharGMAT wrote:
Hi chetan2u / @Experts

I m quiet confused by the language of this question. First statement says that if we open both valves then the pool will be filled with water in 48 mins.

so, we can have 2 scenarios.
Lets name the valves to A and B.

Scenario 1- A and B, both are inlet valves, which fills the tank and tank will take 48 mins to get filled

scenario 2- A is inlet valve and B is outlet valve.
2a) if rate of valve B is LESS then that of A. Tank will get filled in 48 mins
2b) if rate of valve B is MORE then that of A. In this case the tank can not be filled--> Therefore this case will not exist.

reading further the question says- The first valve alone would fill the pool in 2 hours.
If case 1- Then we can take the first valve as A and say, Valve A take 120 mins to fill the tank..Till here its ok

Reading further the question says- If the second valve EMITS 50 cubic meters of water more than the first every minute,
After reading this statement, its clear that there is 1 valve which emits the water.
Therefore, our existing possibilty MUST BE from scenario2.

But this part of the sentence also says that the second valve emits 50 cubic meter MORE water.
If rate of emission is more than the inlet rate, how would the tank be filled..??

Can you please assist.



Thanks


Hi,
You are wrogly considering that one valve is filling and the other is emptying it...
both are filling it..
so, A takes longer time to fill up that is 2hours..
and both combined fill up in 48 minutes..
therefore B is filling up at faster rate and that is given as 50 cum more than A
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Re: M12-10  [#permalink]

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New post 19 May 2016, 00:29
Hi chetan2u,

I am thinking so because the question explicitly says-

If the second valve EMITS 50 cubic meters of water more than the first every minute

i hope EMIT = discharge, release.

Please assist.
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Re: M12-10  [#permalink]

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New post 19 May 2016, 00:35
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PrakharGMAT wrote:
Hi chetan2u,

I am thinking so because the question explicitly says-

If the second valve EMITS 50 cubic meters of water more than the first every minute

i hope EMIT = discharge, release.

Please assist.


Yes you are correct emit can be taken as discharge, BUT it can be discharge in the tank NOT necessarily discharge from the tank....

But if you read the following, the Q should be clear..
1) If the second valve EMITS 50 cubic meters of water more than the first every minute..
This should tell you that BOTH are doing the same JOB -- MORE THAN----
2) If two fill up in less time and ONE of them takes more time, BOTH have to be filling
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3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: M12-10  [#permalink]

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New post 19 May 2016, 00:42
Hi chetan2u,

I got your point..Thanks for assistance
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M12-10  [#permalink]

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New post 25 Sep 2016, 09:04
I did it a little differently.

I set up this equation:

1/48 = 1/120 + 1/x

then find common denominator

5/240 = 2/240 + 1/x = 2/240 + 3/240

Since second pump flows 50 cu in more, that would equal to be the difference of rate between first and second pump, which is:

3/240 - 2/240 = 1/240 is 50 cu in / min

therefore, since 1/240 is 50 cu in / min, 2/240 would equal 100 cu in / min, making the rate of the first pump 100 cu in / min.

Since it takes 2 hrs (120 min) for first pump to completely fill the tank,

100 X 120 = 12,000

Therefore D.
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Re: M12-10  [#permalink]

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New post 15 Mar 2018, 01:55
Hi Experts,

I followed the rate table method but I'm confused as to how you ended up combining 120x = 48(2x + 50) into an equation to find x, which is then used to multiply with 120 to get the capacity.

Please explain the logic/working behind 120x=48(2x+50)

I couldn't make that leap to turn that into an equation.

Thanks.
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Re: M12-10  [#permalink]

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New post 15 Mar 2018, 02:59
ttaiwo wrote:
Hi Experts,

I followed the rate table method but I'm confused as to how you ended up combining 120x = 48(2x + 50) into an equation to find x, which is then used to multiply with 120 to get the capacity.

Please explain the logic/working behind 120x=48(2x+50)

I couldn't make that leap to turn that into an equation.

Thanks.


The first valve alone would fill the pool in 2 hours, so in 120 minutes. If the rate of the first valve is \(x\) cubic meters per minute, then the capacity of the pool = time*rate = 120x.


With both valves open, the pool will be filled with water in 48 minutes. Again, the rate of the first valve is \(x\) cubic meters per minute, and the rate of the second valve is \(x+50\) cubic meters per minute. The capacity of the pool = time*combined rate = 48(x + x + 50) = 48(2x + 50).

Since we have the same pool, then 120x=48(2x+50).

Hope it's clear.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M12-10 &nbs [#permalink] 15 Mar 2018, 02:59
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