Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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?

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?

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.

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?

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

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, 10:03, edited 1 time in total.

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
_________________

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

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?

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?

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;

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.
_________________

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?

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

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

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?