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

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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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

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

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.

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 \(h = \frac{5}{(k+m)}\)

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

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.

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.

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.

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

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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Good One , loved solving it !!

I am attaching 2 Pics ( hope it helps)





Hence IMHO (A)


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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

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

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.