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# Tanks X and Y contain 500 and 200 gallons

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Tanks X and Y contain 500 and 200 gallons [#permalink]  26 Sep 2010, 20:04
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32% (02:11) correct 67% (01:15) wrong based on 1 sessions
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 much time will elapse before the two tanks contain equal amounts of water?

(C) 2008 GMAT Club - m22#17

* \frac{5}{M + K}\text{ hours}
* 6(M + K)\text{ hours}
* \frac{300}{M + K}\text{ hours}
* \frac{300}{M - K}\text{ hours}
* \frac{60}{M - K}\text{ hours}

OA to be shared later. Please provide detailed explaination
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Re: Tanks X and Y contain 500 and 200 gallons [#permalink]  26 Sep 2010, 22:29
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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 much time will elapse before the two tanks contain equal amounts of water?

(C) 2008 GMAT Club - m22#17

* \frac{5}{M + K}\text{ hours}
* 6(M + K)\text{ hours}
* \frac{300}{M + K}\text{ hours}
* \frac{300}{M - K}\text{ hours}
* \frac{60}{M - K}\text{ hours}

OA to be shared later. Please provide detailed explaination

500-kt=200+mt --> mt+kt=300 --> t=\frac{300}{m+k}, where t is the time in minutes need.

I think question should specify unit of time, as if question asks how many minutes are needed then the answers would be \frac{300}{m+k} (C) BUT if question asks how many hours are needed then the answer would be \frac{300}{m+k}*\frac{1}{60}=\frac{5}{m+k} (A).

Question needs revision.

Hope it's clear.
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Re: Tanks X and Y contain 500 and 200 gallons [#permalink]  12 Oct 2010, 02:38
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Chait4R0811 wrote:
Hi .. i am pretty new to this forum..please correct me if my way of answering the question is wrong..

Backsolving
Assume k=10 gallons per minute and M=20 gallons per minute
then say in 10 mins tank x would become 400 gallons and tank y the same 400 gallons..
substituting the values in thh answer options ( A ) is the right answer since it is asking in hours and should be multiplied by 60

(A) [5][/M+K] * 60 mins = [5][/30]*60=10 mins

IMO A

This is my first post and i have been going through a lot of quant and verbal questions in the forum and they are very helpful..

Bunuel u are truely amazing in the way u answer the questions !!

The original question was:

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 much time will elapse before the two tanks contain equal amounts of water?

A. 5/(m+k)

B. 6(m+k)

C.300/(m+k)

D.300/(m-k)

E. 60/(m-k)

As you can see there is no "HOURS" mentioned in answer choices. I see that now in Gmat Club Tests the question is edited and the word "HOURS" is added to answer choices. Also someone edited the question here as well.

See below my solution:
"500-kt=200+mt --> mt+kt=300 --> t=\frac{300}{m+k}, where t is the time in minutes need.

I think question should specified unit of time, as if question asks how many minutes are needed then the answers would be \frac{300}{m+k} (C) BUT if question asks how many hours are needed then the answer would be \frac{300}{m+k}*\frac{1}{60}=\frac{5}{m+k} (A)."

So you can see that without specifying the unit of time the answers given by walker and geturdream could be considered as correct. In its current form when the unit of time is specified (HOURS) then the answer is A.

Hope it's clear.
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Re: Tanks X and Y contain 500 and 200 gallons [#permalink]  26 Sep 2010, 20:14
C

1. the difference that need to be eliminated is 500-200 = 300 gallons
2. rate at which the difference is being eliminated: K+M
3. t = volume/rate = 300/(K+M)

Edit: it was a trap
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Re: Tanks X and Y contain 500 and 200 gallons [#permalink]  26 Sep 2010, 20:20
this is not the OA and i got the same answer as well
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Re: Tanks X and Y contain 500 and 200 gallons [#permalink]  26 Sep 2010, 20:34
The diff is 500-200=300
The relative rate at which this diff is being eliminated = M-(-K)=M+K

so the time required = 300/(M+K)
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Re: Tanks X and Y contain 500 and 200 gallons [#permalink]  11 Oct 2010, 22:36
Hi .. i am pretty new to this forum..please correct me if my way of answering the question is wrong..

Backsolving
Assume k=10 gallons per minute and M=20 gallons per minute
then say in 10 mins tank x would become 400 gallons and tank y the same 400 gallons..
substituting the values in thh answer options ( A ) is the right answer since it is asking in hours and should be multiplied by 60

(A) [5][/M+K] * 60 mins = [5][/30]*60=10 mins

IMO A

This is my first post and i have been going through a lot of quant and verbal questions in the forum and they are very helpful..

Bunuel u are truely amazing in the way u answer the questions !!
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Re: Tanks X and Y contain 500 and 200 gallons [#permalink]  12 Oct 2010, 00:30
You are right, the answer is indeed A. Where most people have gone wrong in the explanations above, is that M & K are rates per minute and the answers are given in hours. So in the end a divide by 60 is needed. So answer is (A).

But having said that, I am not a fan of the way you solved this because, you plugged in numbers to see which expression gives you the right answer. If you use this technique, in some cases you will see more than one choice giving you the right answer in which case you will have to substitute another set of values to differentiate between the two or three expressions that matched out. So be careful with such an approach.

Chait4R0811 wrote:
Hi .. i am pretty new to this forum..please correct me if my way of answering the question is wrong..

Backsolving
Assume k=10 gallons per minute and M=20 gallons per minute
then say in 10 mins tank x would become 400 gallons and tank y the same 400 gallons..
substituting the values in thh answer options ( A ) is the right answer since it is asking in hours and should be multiplied by 60

(A) [5][/M+K] * 60 mins = [5][/30]*60=10 mins

IMO A

This is my first post and i have been going through a lot of quant and verbal questions in the forum and they are very helpful..

Bunuel u are truely amazing in the way u answer the questions !!

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Re: Tanks X and Y contain 500 and 200 gallons [#permalink]  12 Oct 2010, 00:45
shrouded1 wrote:
You are right, the answer is indeed A. Where most people have gone wrong in the explanations above, is that M & K are rates per minute and the answers are given in hours. So in the end a divide by 60 is needed. So answer is (A).

But having said that, I am not a fan of the way you solved this because, you plugged in numbers to see which expression gives you the right answer. If you use this technique, in some cases you will see more than one choice giving you the right answer in which case you will have to substitute another set of values to differentiate between the two or three expressions that matched out. So be careful with such an approach.

Chait4R0811 wrote:
Hi .. i am pretty new to this forum..please correct me if my way of answering the question is wrong..

Backsolving
Assume k=10 gallons per minute and M=20 gallons per minute
then say in 10 mins tank x would become 400 gallons and tank y the same 400 gallons..
substituting the values in thh answer options ( A ) is the right answer since it is asking in hours and should be multiplied by 60

(A) [5][/M+K] * 60 mins = [5][/30]*60=10 mins

IMO A

This is my first post and i have been going through a lot of quant and verbal questions in the forum and they are very helpful..

Bunuel u are truely amazing in the way u answer the questions !!

Thanks shroud !! will keep tht in mind
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Re: Tanks X and Y contain 500 and 200 gallons [#permalink]  26 Feb 2011, 20:20
Let time = t mins
500 -Kt = 200 + Mt
t = 300 / (K+M) mins = 5/(K+M) hr
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Re: Tanks X and Y contain 500 and 200 gallons [#permalink]  04 Apr 2011, 12:53
500-Kt =200+Mt
=> t = 300/(M+K) minutes
= 5/(M+K) hours

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Re: Tanks X and Y contain 500 and 200 gallons [#permalink]  18 Sep 2011, 02:56
Question needs to be edited to include HOUR in it for A to be the answer.
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Re: Tanks X and Y contain 500 and 200 gallons [#permalink]  18 Sep 2011, 04:40
the difference is 300 and the rate at which the difference is getting reduced is M+K
so the time to make the difference 0 is 300/M+K min or 5/M+K hours
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Re: Tanks X and Y contain 500 and 200 gallons [#permalink]  22 Oct 2011, 12:01
Got it wrong for units:(
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Re: Tanks X and Y contain 500 and 200 gallons [#permalink]  16 Nov 2012, 00:05
The trap got me!

My solution:

We need the two tanks to have equal volume. X gallons = Y gallons

500gallons - Kt gallons = 200gallons + Mt gallons, t is in minutes...

t = \frac{300}{K+M} minutes x \frac{1hr}{60minutes}=\frac{5}{M+K}hours
Re: Tanks X and Y contain 500 and 200 gallons   [#permalink] 16 Nov 2012, 00:05
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