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# M22-17

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Math Expert
Joined: 02 Sep 2009
Posts: 47101

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16 Sep 2014, 01:16
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85% (hard)

Question Stats:

44% (01:23) correct 56% (01:31) wrong based on 90 sessions

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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. $$\frac{60}{M - K}\text{ hours}$$

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16 Sep 2014, 01:16
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Official Solution:

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.

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23 Jan 2017, 20:44
Is a viable method to sub in gallon amounts for M and K, determine the time, t, when the volumes are equal, and then plug in to see which answer gives the correct value? E.g.

If M=100 and K=50, after 1h X has 400 gallons and Y has 250 gallons. After 2h, X has 300 gallons and Y has 300 gallons. So t=2h.

Answer A gives the correct time of 2h, when plugging in (5/3 for M and 5/6 for K). Note that M and K are converted to gallons/min.
Manager
Joined: 13 Dec 2013
Posts: 161
Location: United States (NY)
GMAT 1: 710 Q46 V41
GMAT 2: 720 Q48 V40
GPA: 4
WE: Consulting (Consulting)

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23 Jan 2017, 20:45
Bunuel wrote:
Official Solution:

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.

Hi. Is a viable method to sub in gallon amounts for M and K, determine the time, t, when the volumes are equal, and then plug in to see which answer gives the correct value? E.g.

If M=100 and K=50, after 1h X has 400 gallons and Y has 250 gallons. After 2h, X has 300 gallons and Y has 300 gallons. So t=2h.

Answer A gives the correct time of 2h, when plugging in (5/3 for M and 5/6 for K). Note that M and K are converted to gallons/min.
Intern
Joined: 28 Jan 2018
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12 Mar 2018, 23:32
Hi,

Why do you assume that both tanks need the same amount of time? Should this be not stated in the question?
Because for example if the first tank needs 30 minutes because he has a much higher rate and the second one needs 60 minutes because of a lower rate, the time needed to reach the same level is 60 minutes.

BR
Math Expert
Joined: 02 Sep 2009
Posts: 47101

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12 Mar 2018, 23:59
MarFel wrote:
Hi,

Why do you assume that both tanks need the same amount of time? Should this be not stated in the question?
Because for example if the first tank needs 30 minutes because he has a much higher rate and the second one needs 60 minutes because of a lower rate, the time needed to reach the same level is 60 minutes.

BR

I think you misunderstood the question.

We have two tanks, X and Y, containing 500 and 200 gallons of water. Water is being pumped out of tank X at some rate and water is being added to tank Y at some rate. Now, obviously, after some time both tanks will have equal amount of water. The question asks to find that amount of time.

Hope it's clear.
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13 Mar 2018, 00:14
Bunuel wrote:
MarFel wrote:
Hi,

Why do you assume that both tanks need the same amount of time? Should this be not stated in the question?
Because for example if the first tank needs 30 minutes because he has a much higher rate and the second one needs 60 minutes because of a lower rate, the time needed to reach the same level is 60 minutes.

BR

I think you misunderstood the question.

We have two tanks, X and Y, containing 500 and 200 gallons of water. Water is being pumped out of tank X at some rate and water is being added to tank Y at some rate. Now, obviously, after some time both tanks will have equal amount of water. The question asks to find that amount of time.

Hope it's clear.

Still dont get it.

500−kt=200+mt

My point is that t must not be equal on both sides because they can have different rates at which water is pumped out or in and need thus different amounts of time.
So my formula would be: 500-k*t1=200+m*t2
Which can not be solved for a valid solution.
To be more precise if water is pumped out of Tank X at a rate of 30 gallons per minute and water is pumped into Tank Y at 50 gallons per minute.
Tank X needs to loose 150 gallons and needs thus 150/30= 5 minutes and Tank Y needs to gain 150 gallons and thus 150/50=3 minutes. Thus the time is not equal.

Ahh okay got it now will leave it for others though.
The amount of water is not fixed we will just pump water out of tank X until it reaches the same level of tank Y. And that level is somewhere between 500 and 200.
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05 May 2018, 14:26
The explanation is very good and awesome.
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Kindly press the +1Kudos if you like the explanation. Thanks a lot!!!

Re: M22-17   [#permalink] 05 May 2018, 14:26
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# M22-17

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