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Tanks X and Y contain 500 and 200 gallons of water respectively. If

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

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New post 27 Jul 2015, 21:42
2
dhawalbp wrote:
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)


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-respectiv-101762.html#p789043
tanks-x-and-y-contain-500-and-200-gallons-of-water-respectiv-101762.html#p1346155
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Re: Tanks X and Y contain 500 and 200 gallons of water respectively. If  [#permalink]

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New post 18 Jan 2016, 00:18
Plug-in SMART numbers and test options.

M=20 and K=5, hence \(\frac{(500-200)}{(20+5)}\) = 12 minutes to equalize the both which is 1/5 h. Answer A works
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Re: Tanks X and Y contain 500 and 200 gallons of water respectively. If  [#permalink]

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New post 18 Jan 2016, 01:22
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



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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New post 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.

M = 5, N = 1. Answer A = 5/6.
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Re: Tanks X and Y contain 500 and 200 gallons of water respectively. If  [#permalink]

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New post 05 Feb 2016, 12:00
Here is how I solve it:

500 gallons - 60K gallons per hour x HOUR = 200 gallons + 60M gallons per hour x HOUR

Translating:
500-60kH=200+60mH
500-200=60mH+60kH
300=60H(m+k)
300/60(m+k)=H - simplify
5/(m+k)=h

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

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New post 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.

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

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New post 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
Capacity.PNG [ 3.4 KiB | Viewed 1265 times ]


Attachment:
Calculations.PNG
Calculations.PNG [ 3.01 KiB | Viewed 1266 times ]


Hence IMHO (A)


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

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

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

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New post 14 Oct 2016, 01:19
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alesia257 wrote:
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


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

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New post 14 Aug 2017, 14:59
work rate in hours for both 60(k+m)
total amount of work to be performed 500-200 = 300
then:
60(k+m) *x hours = 300
X hours = 300/60(k+m) = 5/(k+m)
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Re: Tanks X and Y contain 500 and 200 gallons of water respectively. If  [#permalink]

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New post 13 Jun 2019, 16:31
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


TL;DR



500 - Kx = 200 + Mx
300 = x (K + M)
x = 300/(K + M) minutes
x = 300/(K + M) minutes * (1 hr/60 min)
x = 5/(K + M) hours

ANSWER: A

Veritas Prep Official Solution



There are two tanks with different water levels. Note that the rate of pumping is given as K gallons per min and M gallons per min i.e. they are different. So we cannot say that they both will have equal amount of water when they have 350 gallons. They could very well have equal amount of water at 300 gallons or 400 gallons etc. So when one expects that water in both tanks will be at 350 gallon level, one is making a mistake. The two tanks are working for the same time to get their level equal but their rates are different. So the work done is different. Note here that equal level does not imply equal work done. The equal level could be achieved at 300 gallons when work done would be different – 200 gallons removed from tank X and 100 gallons added to tank Y. The equal level could be achieved at 400 gallons when work done would be different again – 100 gallons removed from tank X and 200 gallons added to tank Y.

To achieve the ‘equal level,’ tank Y needs to gain water and tank X needs to lose water. Total 300 gallons (500 gallons – 200 gallons) of work needs to be done. Which tank will do how much depends on their respective rates.

Work to be done together = 300 gallons

Relative rate of work = (K + M) gallons/minute

The rates get added because they are working in opposite directions – one is removing water and the other is adding water. So we get relative rate (which is same as relative speed) by adding the individual rates.

Note here that rate is given in gallons per minute. But the options have hours so we must convert the rate to gallons per hour.

Relative rate of work = (K + M) gallons/minute = (K + M) gallons/(1/60) hour = 60*(K + M) gallons/hour

Time taken to complete the work = 300/(60(K + M)) hours = 5/(K + M) hours
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Tanks X and Y contain 500 and 200 gallons of water respectively. If  [#permalink]

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New post 10 Dec 2019, 19:08
VeritasKarishma 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


Hi karishma

In relative speed questions -- why do we add the rates [M+K] that are working in opposite directions ? Could you perhaps explain why we are adding in this case and not subtracting

Also, you mentioned above that relative speeds are similar to relative rates...

But i thought relative rates are such that

-- if two people are working together on the project [rates in the same direction] - we can add the rates for a combined rates [ R1 + R2]

-- if one person is constructing and the other person is de-constructing a project [rates in the opposite direction] - the combined rate is obviously R1 - R2

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

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New post 10 Dec 2019, 22:15
jabhatta@umail.iu.edu wrote:
VeritasKarishma 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


Hi karishma

In relative speed questions -- why do we add the rates [M+K] that are working in opposite directions ? Could you perhaps explain why we are adding in this case and not subtracting

Also, you mentioned above that relative speeds are similar to relative rates...

But i thought relative rates are such that

-- if two people are working together on the project [rates in the same direction] - we can add the rates for a combined rates [ R1 + R2]

-- if one person is constructing and the other person is de-constructing a project [rates in the opposite direction] - the combined rate is obviously R1 - R2

Thank you !


Say A and B are standing at two ends of a track of 300 m


A ->----------------- (300 m)---------------------<- B

Now they have to meet so A will start walking towards B (due east direction) and B will start walking towards A (due west direction). They are walking in opposite directions towards each other.

--------------AB---------------------------------

Say they meet here. Together they have covered 300 m. So if A's speed were 100 m/hr and B's speed were 200 m/hr, in 1 hr they would together cover 300 m.

Now say they start walking in opposite directions again away from each other. A starts walking due west and B starts walking due east.

-----------<-A--B->--------------------------------

In one hr, they will again cover 300 m together.

A ------------------ (300 m)---------------------- B

So whenever two objects are moving in opposite directions, their speeds get added.
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Re: Tanks X and Y contain 500 and 200 gallons of water respectively. If  [#permalink]

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New post 10 Dec 2019, 22:28
jabhatta@umail.iu.edu wrote:
VeritasKarishma 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


Hi karishma

In relative speed questions -- why do we add the rates [M+K] that are working in opposite directions ? Could you perhaps explain why we are adding in this case and not subtracting

Also, you mentioned above that relative speeds are similar to relative rates...

But i thought relative rates are such that

-- if two people are working together on the project [rates in the same direction] - we can add the rates for a combined rates [ R1 + R2]

-- if one person is constructing and the other person is de-constructing a project [rates in the opposite direction] - the combined rate is obviously R1 - R2

Thank you !


On the same lines, think about it - one tank is at 500 gallons and another is at 200 gallons. They both need to reach the same level, one by reducing water and the other by gaining water. So they have to together bridge this gap of 300 gallons. Imagine the two tanks sitting side by side at different levels. They are both working in opposite directions to reach the same level such that both are contributing TOWARD removing the 300 gallons difference. Hence their rates will get added.
Always look at the logic of the question.

When one worker is making a wall while the other is destroying, their rates get subtracted because one is contributing towards work that has to be done while the other is taking away from it.
Here, both tanks are contributing toward work that has to be done.

Check out this post on Veritas blog: https://www.veritasprep.com/blog/2015/0 ... -the-gmat/
You will find it helpful.
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Re: Tanks X and Y contain 500 and 200 gallons of water respectively. If  [#permalink]

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New post 12 Dec 2019, 06:47
Thank you karishma

How do one then explain if two people are working on project (say building a house or constructing a fence)

We add the individual rates to form the combined rate ?

Can one not think that both individuals are working in the same direction (i.e towards building a house)

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Re: Tanks X and Y contain 500 and 200 gallons of water respectively. If   [#permalink] 12 Dec 2019, 06:47

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