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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

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)

Did you read the explanation I gave to him on why his logic is not sound? Having equal level of water does not mean they did equal work. After some time, they both could have been at 300 gallons. So the tank with 200 gallons would have increased its water level by 100 gallons and tank with 500 gallons would have decreased its water level by 200 gallons. Note that the second tank would have done twice the work as done by first tank in the same amount of time so in this case, its rate would be twice the rate of first tank. They do have equal level at the end but to reach there, they might have done different amount of work. So their rate need not be equal. This is the catch of this question. Kindly go through my explanation given above and then check out the solutions given here:

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

Hi, the Q can be done in two steps... the total change, both filling in one and draining in other, equals 500-200=300.. what are the rates working towards it .. M+K gallons per minute or (M+K)*60 gallons per hour... \(answer =\frac{300}{{(M+K)*60 }}=\frac{5}{{M+K}}\).. A
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Re: Tanks X and Y contain 500 and 200 gallons of water respectively. If [#permalink]

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18 Jan 2016, 02:01

Plug in numbers to find an answer.

Lets take 250 gallons as the common amount that we need to reach with each tank.

If water is pumped out of X at 5 gallons per min, it will take 50 mins to pump out 250 gallons (500 - 250 = 250 gallons left). If water is pumped into Y at 1 gallon per min, then in 50 mins, 50 gallons will be pumped in. Hence Y will also reach 250.

It will take 50 mins or 5/6 hours for both of them to reach an equal level.

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

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05 Feb 2016, 20:10

Bunuel 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. \(\frac{60}{M - K}\text{ hours}\)

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.

This approach is the one I can formulate easiest, however I do have two questions.

1) W = RT, which is why we take k*t and m*t is that correct? 2) Why do we use k and m, rather than 1/k and 1/m? Are we not supposed to use rates? Perhaps I'm missing what kt and mt really mean.

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

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06 Feb 2016, 10:22

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

Good One , loved solving it !!

I am attaching 2 Pics ( hope it helps)

Attachment:

Capacity.PNG [ 3.4 KiB | Viewed 593 times ]

Attachment:

Calculations.PNG [ 3.01 KiB | Viewed 593 times ]

Hence IMHO (A)

_________________

Thanks and Regards

Abhishek....

PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS

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

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12 Oct 2016, 19:11

Bunuel 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. \(\frac{60}{M - K}\text{ hours}\)

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.

BUNUEL, pls explain me why t should be equal for both tanks. Why cant one tank take 10 min more to reach the same amount of water that another tank already has?

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

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12 Oct 2016, 19:21

VeritasPrepKarishma wrote:

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

honestly, I didnt realize it is this type of question. Can you explain pls how to spot relative speed questions , especially in a different context like this one. thank you

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

honestly, I didnt realize it is this type of question. Can you explain pls how to spot relative speed questions , especially in a different context like this one. thank you

It is not necessary to do it using the relative speed concept. You can do it using algebra too. But it is good if you recognise that there is an equivalence in time-speed-distance and work-rate-time. In fact, time-speed-distance is just a special case of work-rate-time. In this case, work done is the 'distance covered' and rate is the 'speed'. So the concepts of TSD can be applied to the generic work-rate case too.

Distance covered is work done (300 gallons) Relative speed is relative rate of work = 60(K+M) So time taken = Distance/Speed = 300/60*(K+M) = 5/(K+M) hrs
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Re: Tanks X and Y contain 500 and 200 gallons of water respectively. If [#permalink]

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14 Oct 2016, 14:00

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 h=number of hours 200+h*60M=500-h*60K h*(60M+60K)=300 h=300/(60M+60K) h=5/(M+K) A.