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# M12-10

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14 May 2008, 03:17
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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. What is the capacity of the pool if every minute the second valve admits 50 cubic meters of water more than the first?

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

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

I sweated when time gone!
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14 May 2008, 04:00
I get D

let x mcube be the vol

then
60*x/(x+3000) = 48
=> x = 12000
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15 May 2008, 03:45
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sondenso 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. What is the capacity of the pool if every minute the second valve admits 50 cubic meters of water more than the first?

9000 cubic meters
10500 cubic meters
11750 cubic meters
12000 cubic meters
12500 cubic meters

I sweated when time gone!

let first taps speed be x cu/min
then second taps speed be x+50 cu/min
then
48(x+(x+50))= total capacity---------(1)

since first tap can fill in 2 hrs
60*2(x) is total capacity------------(2)

equating
x=100

total capacity=100*2*60 =12000

hope this is clear
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04 Feb 2009, 17:17
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I'm confused here:

Shouldn't we add the quantity of valve1 + quantity of valve2 = total quantity? Please tell me where I'm going wrong.

For example:

This is what I did:

1. Figure out the time Valve 2 took to fill the pool by itself

1/Valve1 + 1/Valve2 = 1/48
1/120 mins (or 2 hours) + 1/Valve2 = 1/48 mins
Valve2 = 80 mins

2. Find the Quantity to fill the pool X

120(X) + 80 (X+50) = 48 (2x + 50)

So this is where it goes down hill...
not sure why we don't include 80 (X+50) in the equation?
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05 Feb 2009, 09:18
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I'm confused here:

Shouldn't we add the quantity of valve1 + quantity of valve2 = total quantity? Please tell me where I'm going wrong.

For example:

This is what I did:

1. Figure out the time Valve 2 took to fill the pool by itself

1/Valve1 + 1/Valve2 = 1/48
1/120 mins (or 2 hours) + 1/Valve2 = 1/48 mins
Valve2 = 80 mins

2. Find the Quantity to fill the pool X

120(X) + 80 (X+50) = 48 (2x + 50)

So this is where it goes down hill...
not sure why we don't include 80 (X+50) in the equation?

Step 1 is correct in that you found that Valve 2 can fill a tank in 80 min.

Step 2 is wrong.

Valve 1 can fill the tank in 120 Min with x being the volume filled per min - So total volume = 120*x ----> Eq1
Valve 2 can fill the tank in 80 min with (x+50) being the volume filled per min. So the total volume = 80(x+50) ---> Eq2

Together can fill the tank in 48 min with (x + (x + 5)) being the volume filled per min. So the total volume = 48(x + (x + 5)) ---> Eq 3.

To solve for x, pick any of the 2 of the 3 eq, ie either equate Eq 1 = Eq 2 or Eq 1 = Eq 3 or Eq2 = Eq 3 and get the

ie 120X = 80(x + 50) ==> x = 100.

We get x = 100.

Now substitute value of x in any of the 3 eq and you should get the total volume.
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05 Feb 2009, 09:57
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sondenso wrote:
rohit929 wrote:
sondenso 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. What is the capacity of the pool if every minute the second valve admits 50 cubic meters of water more than the first?

9000 cubic meters
10500 cubic meters
11750 cubic meters
12000 cubic meters
12500 cubic meters

I sweated when time gone!

let first taps speed be x cu/min
then second taps speed be x+50 cu/min
then
48(x+(x+50))= total capacity---------(1)

since first tap can fill in 2 hrs
60*2(x) is total capacity------------(2)

equating
x=100

total capacity=100*2*60 =12000

hope this is clear

So sooooooo great, rohit! Thanks! I lost many basic, I think. I approach this by using work rate = work/time. And stucked!

Indeed "rate = work/time" approach is same as rohit used.

capacity=work = rate*time = (x +x+50)*48
capacity= rate*time = x*120
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16 Aug 2009, 10:12
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Here is how I solved the problem:

9. 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. What is the capacity of the pool if every minute the second valve admits 50 cubic meters of water more than the first?
* 9000 cubic meters
* 10500 cubic meters
* 11750 cubic meters
* 12000 cubic meters
* 12500 cubic meters

let B = # of mins for the second valve alone to fill the pool

1/(time it takes first valve to fill the pool) + 1/B = 1/(total time its takes to fill the pool)
1/120 + 1/B = 1/48
B = 80

let C = the capacity of the pool
C/80 - C/120 = 50
C = 12000

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17 Aug 2009, 14:27
Here x and x+50 are nothing but the rates of the pipes.
R1 + R2 = R ( R=combined rate )

x+x+50 = Capacity/48, Now capacity = 120x
=> 2x+50 = 120x/48
=> x=100
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21 Jan 2010, 12:57
48 min = 0.8 hrs, then if x is pool's volume
both pipes open
for 0.8 hrs - x cubic m (multiply both sides by 5)
for 4 hrs - 5x cubic m
Within 4 hrs
the first pipe alone could fill 2 pools,
the second 3 pools.
So, the second could fill one pool more than the 1st being open for 4hrs.
x=4*60*50=12000 (D).

Last edited by nvgroshar on 26 Jan 2011, 11:03, edited 1 time in total.
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23 Jan 2010, 23:18
It's actually an easy word question - just take your time to set up the x variable - let x be the amount of water in cube meter the first valve admits per minute, x+50 will be for the second valve

48(x+(x+50))=60*2x

=> x=100 substitute back to the right hand side of the equation =>
total capacity=100*120 =12,000

Usually when I have enough time, I put the x value in the left side to see if I get 12,000 as well - just for sanity check
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25 Jan 2010, 07:18
from the given info we know that

1/x+1/y=48 ( both valve will fill in 48 mins)
1/x=1/120 (fist valve will fill in 2 hrs or 120 mins)

so by solving
1/y=1/80 (second valve will take 80 mins)

first valve =1/120
second valve =1/80

if the tank can be filled by first valve in 120 minutes and the second one by 80 mins

then answer should be exactly divided by 120 and 80 only choice D satisfies this criterion so I picked ans D ( of courseif it contains more than one choice divisble by 120 and 80 then you'll have to solve it precisely)

or you can solve this by solving for x tradionaly by applying formula as everyone stated above

1/120(x)+1/80(x+5)=1/48(2x+5)
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28 Nov 2010, 14:39
 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?(C) 2008 GMAT Club - m12#10 * 9000 cubic meters * 10500 cubic meters * 11750 cubic meters * 12000 cubic meters * 12500 cubic meters

still puzzled how to do it my way....

found out that the rate of V1 is 1/120 and V2 is 1/100....correct? if yes, then how to find volume?

thanks.....
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28 Nov 2010, 15:14
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zisis 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?(C) 2008 GMAT Club - m12#10 * 9000 cubic meters * 10500 cubic meters * 11750 cubic meters * 12000 cubic meters * 12500 cubic meters

still puzzled how to do it my way....

found out that the rate of V1 is 1/120 and V2 is 1/100....correct? if yes, then how to find volume?

thanks.....

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 $$C=time*combined \ rate=48(x+x+50)=48(2x+50)$$;
But as the first valve alone fills the pool in 2 hours (120 minutes) then the capacity of the pool also equals to $$C=time*rate=120x$$;

So, $$120x=48(2x+50)$$ --> $$x=100$$ --> $$C=120x=12,000$$.

Hope it's clear.
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25 Nov 2011, 06:16
Total Volume = x
First valve and second valve take 48 minutes to fill the pool
The both valves filled x/48 in 1 minute
First valve filled x/120 part in 1 minute
So, Second valve filled in 1 minute = x/48 - x/120 = 3x/240 part

Given that 3x/240 - x/120 = 50
x =12000
Ans. D.
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25 Nov 2011, 20:18
good question..........and an amazing explanation
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03 Dec 2011, 07:59
Gurus,

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. What is the capacity of the pool if every minute the second valve admits 50 cubic meters of water more than the first?

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

I tried this question and tried this approach but did not get me the answer? Please could you tell me what I did wrong?
W= RT
R=w/t
c/120 +x =C/48
X( rate at which second valve fills up) .I got stuck here cos i had two many unknowns
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31 Jan 2012, 13:03
Jiggyjee1

X is C/120+50 which is given so only 1 unknown.

+1 D
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30 Jan 2013, 06:28
We can use plugin approach also here.

From the options given, assume that capacity of tank is = 12000

Tap 1 fills 12000 in 2 hrs. In 1 min = 100 cu/m
Tap 2 in 1 min = 100 + 50 = 150.

So in 48 mins = 48(100 (t1) + 150 (t2) = 12000.
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30 Jan 2013, 17:03
Consider the time the second pipe takes to fill the tank as 1/x = 1/48 - 1/120
= 1/80 .

therefore x takes 80 mins to fill the pipe . But since each minute pipe B fills in an extra 50 from A using a smart number 100
B takes = 80 * 150 = 12000 would be the capacity
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16 Jan 2014, 10:07
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. What is the capacity of the pool if every minute the second valve admits 50 cubic meters of water more than the first?

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

Given:
First valve takes 120 mins to fill the pool
Both valve take 48 mins to fill the pool

Let capacity of pool be $$x$$ cubic meters
Inlet rate of First valve per minute $$= x/120$$
Combined inlet rate per minute of both valve $$= x/48$$

Inlet rate of second valve per minute :$$( x/48) - (x/120 ) = x/80$$

Since, every minute second valve admits 50 cubic meters of water more than the first valve,

$$(x/80) - (x/120) = 50$$
Or,$$x = 12000$$ cubic meters

Re: M12-10   [#permalink] 16 Jan 2014, 10:07

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# M12-10

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