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Tanks X and Y contain 500 and 200 gallons of water respectiv [#permalink]

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26 Sep 2010, 19:04

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Tanks X and Y contain 500 and 200 gallons of water respectively. If water is being pumped out of tank X at a rate of K gallons per minute and water is being added to tank Y at a rate of M gallons per minute, how many hours will elapse before the two tanks contain equal amounts of water?

A. \(\frac{5}{M + K}\text{ hours}\) B. \(6(M + K)\text{ hours}\) C. \(\frac{300}{M + K}\text{ hours}\) D. \(\frac{300}{M - K}\text{ hours}\) E. \(\frac{60}{M - K}\text{ hours}\)

Tanks X and Y contain 500 and 200 gallons of water respectively. If water is being pumped out of tank X at a rate of K gallons per minute and water is being added to tank Y at a rate of M gallons per minute, how many hours will elapse before the two tanks contain equal amounts of water?

A. \(\frac{5}{M + K}\text{ hours}\) B. \(6(M + K)\text{ hours}\) C. \(\frac{300}{M + K}\text{ hours}\) D. \(\frac{300}{M - K}\text{ hours}\) E. \(\frac{60}{M - K}\text{ hours}\)

Say \(t\) minutes are needed the two tanks to contain equal amounts of water, then we would have that \(500-kt=200+mt\). Find \(t\): \(t=\frac{300}{m+k}\) minutes or \(\frac{1}{60}*\frac{300}{m+k}=\frac{5}{m+k}\) hours.

Re: Tanks X and Y contain 500 and 200 gallons [#permalink]

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28 Sep 2013, 01:21

Consider k=1,m=2

Then per the question,

500 - 1x = 200 + 2x

x=100

Check answer options for No. of hours (x)=100 for k=1 and m=2

Hence (C)
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Re: Tanks X and Y contain 500 and 200 gallons of water respectiv [#permalink]

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18 Mar 2014, 05:42

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Although as many of them did I also forgot to convert minutes into hours in last step, here is my approach to correct answer.

Capacity X=500 Rate of drain = K/min

Capacity Y=200 Rate of filling = M/min

After t minutes the volume taken out from X and filled in Y should be same. Hence,

500 - Kt = 200 + Mt

t=300/(M+K) minutes

t= 5/(M+K) hours
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Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

Re: Tanks X and Y contain 500 and 200 gallons of water respectiv [#permalink]

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18 Mar 2014, 11:12

The analytical approaches above are correct. Here is a way to think about this and get to the right answer with minimal calculations.

If M or K increases, the tanks should reach equivalence faster. So M and K both need to be in the denominator. Eliminate (B). If M or K increases, the tanks reach equivalence faster. Therefore K cannot be negative in the denominator. Eliminate (D) and (E). To choose between (A) and (C), put in M=K=1. Then (C) says equivalence will take 150 hours, which is absurd given that at this rate, the bigger tank will drop to 300L in a little over 3 hours, and equivalence needs to be obviously be reached at a level between 300L and 500L for the two tanks. So (A) has to be right.

Tanks X and Y contain 500 and 200 gallons of water respectively. If water is being pumped out of tank X at a rate of K gallons per minute and water is being added to tank Y at a rate of M gallons per minute, how many hours will elapse before the two tanks contain equal amounts of water?

A. \(\frac{5}{M + K}\text{ hours}\) B. \(6(M + K)\text{ hours}\) C. \(\frac{300}{M + K}\text{ hours}\) D. \(\frac{300}{M - K}\text{ hours}\) E. \(\frac{60}{M - K}\text{ hours}\)

m22 q17

Let time taken to reach water to same level in both tanks be h hours Water is being pumped out of tank x at the rate of K gallons per minute OR 60k gallons per hour. Water is being added to tank y at the rate of M gallons per minute OR 60m gallons per hour.

After h hours amount of water in both the tanks will be the same. ------> 500 - 60kh = 200+60mh -------------> 300=60kh+60mh -----> 5 = h(k+m)

Tanks X and Y contain 500 and 200 gallons of water respectively. If water is being pumped out of tank X at a rate of K gallons per minute and water is being added to tank Y at a rate of M gallons per minute, how many hours will elapse before the two tanks contain equal amounts of water?

A. \(\frac{5}{M + K}\text{ hours}\) B. \(6(M + K)\text{ hours}\) C. \(\frac{300}{M + K}\text{ hours}\) D. \(\frac{300}{M - K}\text{ hours}\) E. \(\frac{60}{M - K}\text{ hours}\)

m22 q17

This is a relative speed question.

Distance to be covered together = 300 gallons (= 500 gallons - 200 gallons) Relative speed (rate of work) = (K+M) gallons per minute OR 60*(K+M) gallons per hour (The rates get added because they are working in opposite directions)

Time taken = 300/60(K+M) hours = 5/(K+M) hours
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Re: Tanks X and Y contain 500 and 200 gallons of water respectiv [#permalink]

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19 Mar 2014, 20:49

I started up like this; We require both tanks to have same amount of water i.e 350 gallons means 150 gallons have to be removed from Tank X & 150 gallons have to be added to Tank Y

It will take \(\frac{150}{60K}\) Hrs to remove water from Tank X & \(\frac{150}{60M}\) Hrs to add water to Tank Y

I stuck at this point; Bunuel / Karishma can you please suggest how to continue using this approach
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Re: Tanks X and Y contain 500 and 200 gallons of water respectiv [#permalink]

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19 Mar 2014, 21:02

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Quote:

I started up like this; We require both tanks to have same amount of water i.e 350 gallons means 150 gallons have to be removed from Tank X & 150 gallons have to be added to Tank Y

It will take \frac{150}{60K} Hrs to remove water from Tank X & \frac{150}{60M} Hrs to add water to Tank Y

You are assuming here that water leaves the first tank at the same rate at which it enters the second tank (i.e. M=K). This can be seen from your own calculations. As the time needs to be equal in both cases, according to your calculations, 150/60K = 150/60M => K = M

If this were the case, then indeed the two tanks would have been level at 350L each. However, this is not given to be so. Therefore it is not essential that they draw level at 350L each.
_________________

I started up like this; We require both tanks to have same amount of water i.e 350 gallons means 150 gallons have to be removed from Tank X & 150 gallons have to be added to Tank Y

It will take \(\frac{150}{60K}\) Hrs to remove water from Tank X & \(\frac{150}{60M}\) Hrs to add water to Tank Y

I stuck at this point; Bunuel / Karishma can you please suggest how to continue using this approach

Your starting point is the problem Paresh. Think about it. There are two people standing on a number line, one at 200 and the other at 500. They have a distance of 300 steps between them. They want to meet by walking towards each other and hence be at the same point. Will they necessarily meet at the center point? No. It depends on their speed where they meet. If the person at 200 is very slow and the other very fast, they will meet very close to 200 because the person at 200 would not have covered much distance and most distance will be covered by the person at 500. Hence the assumption that both need to have 350 ml is incorrect. Perhaps the filling up of 200 gallon tank is very slow while the emptying of 500 gallon is very fast. Then they both might have equal volumes of 250 gallons.

Check out the posts above for alternative solutions.
_________________

Re: Tanks X and Y contain 500 and 200 gallons of water respectiv [#permalink]

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18 Jul 2014, 23:42

BestGMATEliza wrote:

an important thing to keep in mind is that the rates are given to you in minutes and the question asks for hours, don't get caught in that GMAT trap!

There are 500 gallons of water in tank X and they are being pumped out of the tank at a rate of K gallons per minute. Thus, X contains 500-K(60t) gallons. We want t to be in hours and K is in minutes so we must multiply t by 60.

There are 200 gallons in tank Y and they are being pumped into the tank at a rate of M gallons per minute. Thus, Y contains 200 +M(60t) gallons.

The question asks for when (ie what time t) the two contain the same amount of water ie when:

500-K(60t) = 200 +M(60t) in order to figure out the answer, we need to isolate t on one side. Let's start by subtracting 200 from and adding K(60t) to both sides:

300=M(60t)+K(60t) now lets pull out the 60t and we get 300=(M+K)60t divide both sides by (M+K) and then 60 and you get t=300/(60(M+K)) which simplifies to A

I think this subtle hrs vs minutes was the only thing which could lead to error. Barring this the question is a sitter to me

Re: Tanks X and Y contain 500 and 200 gallons of water respectiv [#permalink]

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14 Oct 2014, 20:10

I really hate VIC problems but I found a relatively fast way to do this one by plugging in numbers. Let´s see if you find it helpful.

So the difference between tanks X and Y is 300 liters. So if half of that (150 liters) is being pumped out of tank X at a certain rate and into tank Y at the same rate (I mean that K = M) then the time would be the same since both things would happen simultaneously.

What I did was I just went ahead and picked numbers (I know my limits and I know I can screw it up with Algebra under a 2-minute constraint)

K = 150 gallons / 30 minutes = 5 M = 150 gallons / 30 minutes = 5

(Btw, I first chose 150 gallons / 60 minutes, but the result was an ugly fraction)

So the target number I am looking for is 0.5 hours (which is 30 minutes)

A) 5/M+K hours --> 5/5+5 hours --> 0.5 hours. Bingo B) C) D) E)

Hope it helps!
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Consider giving me Kudos if I helped, but don´t take them away if I didn´t!

Re: Tanks X and Y contain 500 and 200 gallons of water respectiv [#permalink]

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21 Oct 2014, 17:52

so one thing to keep in mind when doing this problem is that the rates M and K are in minutes, but the question asks for how many hours has elapsed. The GMAT tries to trick you with things like that. So now let's look at what we know: Tank X has 500 gallons and is secreting water at a rate of K gallons per minute, thus 500-K(60t) this will give us the time in hours Tank Y has 200 gallons and it is gaining water at a rate of M gallons per minute, so 200+M(60t)

so we want the t for when 500-K(60t) = 200+M(60t) To isolate the t to one side, we get: 300=t(60M+60K) 300/(60(M+K))=t simplified it becomes 5/(M+K)=t in hours, thus A

note that answer choice C is what you would get if you forgot to multiply everything by 60, because they are trying to trip you up if you didn't notice the minutes vs hours difference
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Re: Tanks X and Y contain 500 and 200 gallons of water respectiv [#permalink]

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30 Nov 2014, 12:26

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ichha148 wrote:

Tanks X and Y contain 500 and 200 gallons of water respectively. If water is being pumped out of tank X at a rate of K gallons per minute and water is being added to tank Y at a rate of M gallons per minute, how many hours will elapse before the two tanks contain equal amounts of water?

A. \(\frac{5}{M + K}\text{ hours}\) B. \(6(M + K)\text{ hours}\) C. \(\frac{300}{M + K}\text{ hours}\) D. \(\frac{300}{M - K}\text{ hours}\) E. \(\frac{60}{M - K}\text{ hours}\)

m22 q17

Though forming algebraic equation is easy, plug-in numbers for k & m will take lesser time.

Assume k=100 gallons/min & M= 200 g/m

So in one min X becomes 400 and Y also becomes 400. So by substituting k=100 & M=200 in the choices we should get 1/60. A gives us 1/60

Hence answer A.

------------- If this post helps, please give Kudos.

Re: Tanks X and Y contain 500 and 200 gallons of water respectiv [#permalink]

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02 Dec 2014, 00:44

Take k=20 and M =10 then 500 gallons -20 gallons per minute should be equal to 200 gallons +10 gallons per minute check the options 1) 5/K+M hours putting the values 5/30 hrs = 10 minutes so we get that 10 gallons per minuted added in 10 minutes Y gallons will have 300 gallons and 20 gallons per minute removed from X means after 10 minutes 200 gallons removed from X. So we get that after 10 minutes both will have equal amount.

Re: Tanks X and Y contain 500 and 200 gallons of water respectiv [#permalink]

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25 Apr 2015, 10:33

I think that this question contains some small portion of ambiguity. Im saying this because is not very clear what the author ment( maybe because im not native english), does the water goes from the 500G tank to the 200G tank, or they independetly work on their own. That is why PareshGmat and I was also in the same basket assuming that teh tanks are conected and need to equalize and tehre is a limited 700G and of course need to have 350G each, and got stuck. Of course if they are not concetd than the equation is streight forward. Does someone else had the same issue?

I think that this question contains some small portion of ambiguity. Im saying this because is not very clear what the author ment( maybe because im not native english), does the water goes from the 500G tank to the 200G tank, or they independetly work on their own. That is why PareshGmat and I was also in the same basket assuming that teh tanks are conected and need to equalize and tehre is a limited 700G and of course need to have 350G each, and got stuck. Of course if they are not concetd than the equation is streight forward. Does someone else had the same issue?

It doesn't matter whether the water goes from one tank to another or whether it is pumped out and in independently. Note that you are clearly given that "water is being pumped out of tank X at a rate of K gallons per minute and water is being added to tank Y at a rate of M gallons per minute". The pumping out and pumping in take place at different rates. Now whether the water is pumped out of the big tank at a certain rate, stored and then pumped in the smaller tank at another rate or whether they have independent water sources - it doesn't matter. We want the water in the two tanks to be equal.

Say two people A and B are standing 300 steps away and they start walking toward each other. When will they meet? It their speeds are equal, then they will meet in the center i.e. 150 steps away from each end (that would be the case if K = M). But if their speeds are different, they could meet anywhere else. So if speed of A is twice the speed of B, A will cover 200 steps while B will cover 100 steps and they will meet 200 steps away from A's original position. Similarly if K is twice of M, in the time that the big tank loses 200 gallons, the small tank will gain 100 gallons - they will both have 300 gallons of water each.
_________________

Re: Tanks X and Y contain 500 and 200 gallons of water respectiv [#permalink]

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27 Jul 2015, 09:30

VeritasPrepKarishma wrote:

PareshGmat wrote:

I started up like this; We require both tanks to have same amount of water i.e 350 gallons means 150 gallons have to be removed from Tank X & 150 gallons have to be added to Tank Y

It will take \(\frac{150}{60K}\) Hrs to remove water from Tank X & \(\frac{150}{60M}\) Hrs to add water to Tank Y

I stuck at this point; Bunuel / Karishma can you please suggest how to continue using this approach

Your starting point is the problem Paresh. Think about it. There are two people standing on a number line, one at 200 and the other at 500. They have a distance of 300 steps between them. They want to meet by walking towards each other and hence be at the same point. Will they necessarily meet at the center point? No. It depends on their speed where they meet. If the person at 200 is very slow and the other very fast, they will meet very close to 200 because the person at 200 would not have covered much distance and most distance will be covered by the person at 500. Hence the assumption that both need to have 350 ml is incorrect. Perhaps the filling up of 200 gallon tank is very slow while the emptying of 500 gallon is very fast. Then they both might have equal volumes of 250 gallons.

Check out the posts above for alternative solutions.

karishma: What PareshGmat did was pretty right....The catch is if both tank X and y are taking same time say, t hr to equal the level that means rate M=K...so time taken t=150/(K*60)=5/(2K)=5/(K+M)

gmatclubot

Re: Tanks X and Y contain 500 and 200 gallons of water respectiv
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27 Jul 2015, 09:30

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