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Tanks X and Y contain 500 and 200 gallons of water respectiv

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

m22 q17
[Reveal] Spoiler: OA

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Last edited by Bunuel on 28 Sep 2013, 02:55, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: Tanks X and Y contain 500 and 200 gallons [#permalink] New post 26 Sep 2010, 21:29
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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}

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.

Answer: A.
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Re: Tanks X and Y contain 500 and 200 gallons [#permalink] New post 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 [#permalink] New post 21 Dec 2013, 17:53
TGC wrote:
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)

You forgot to multiply both k and m by 60 since we are looking for hours. :wink:

500-1*60x = 200 + 2*60x
300 = 60x*3
x = 5/3

x = 5/(1 + 2)
which is A
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Re: Tanks X and Y contain 500 and 200 gallons of water respectiv [#permalink] New post 18 Mar 2014, 05:42
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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Re: Tanks X and Y contain 500 and 200 gallons of water respectiv [#permalink] New post 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.

(A) it is.
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Re: Tanks X and Y contain 500 and 200 gallons of water respectiv [#permalink] New post 18 Mar 2014, 20:31
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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


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)

so h = \frac{5}{(k+m)}
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Re: Tanks X and Y contain 500 and 200 gallons of water respectiv [#permalink] New post 18 Mar 2014, 20:57
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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



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] New post 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] New post 19 Mar 2014, 21:02
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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.
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Re: Tanks X and Y contain 500 and 200 gallons of water respectiv [#permalink] New post 19 Mar 2014, 21:34
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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.
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Re: Tanks X and Y contain 500 and 200 gallons of water respectiv [#permalink] New post 19 Mar 2014, 21:49
Karishma / Gyanone

Thank you so much for your reply :)

It was rather wrong to be presumptuous to set 150 as a benchmark for further calculations.
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Re: Tanks X and Y contain 500 and 200 gallons of water respectiv [#permalink] New post 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
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Re: Tanks X and Y contain 500 and 200 gallons of water respectiv [#permalink] New post 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!
Re: Tanks X and Y contain 500 and 200 gallons of water respectiv   [#permalink] 14 Oct 2014, 20:10
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