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M23-16

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Bunuel
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M23-16 [#permalink] New post   16 Sep 2014, 01:19
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When working together, printers A and B can finish printing the entire document in 24 minutes. If printer A by itself can finish printing the document in 60 minutes, how many pages does the document have if printer B prints 5 pages more per minute than printer A?

A. 600
B. 800
C. 1000
D. 1200
E. 1500
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Bunuel
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Re M23-16 [#permalink] New post   16 Sep 2014, 01:19
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Official Solution:

When working together, printers A and B can finish printing the entire document in 24 minutes. If printer A by itself can finish printing the document in 60 minutes, how many pages does the document have if printer B prints 5 pages more per minute than printer A?

A. 600
B. 800
C. 1000
D. 1200
E. 1500


Let \(a\) denote the number of pages Printer A prints in one minute.

Let \(b\) denote the number of pages Printer B prints in one minute.

Given that printers A and B together can print the entire document in 24 minutes, the total number of pages in the document is \(24a + 24b\).

However, when Printer A works alone, it completes the document in 60 minutes, implying the document consists of \(60a\) pages.

Thus, \(24a + 24b = 60a\). Given that \(b = a + 5\), we would get \(24a + 24(a + 5) = 60a\), which results in \(a=10\).

Therefore, the document has \(60a=600\) pages.


Answer: A
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Re: M23-16 [#permalink] New post   07 Feb 2016, 02:44
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shasadou wrote:
Hi Bunuel

Can we solve this using weighted average approach? I am actually stuck at some point: so we derive that the faster machine B takes 40 minutes to complete the job. That is 20 minutes faster (can we come to solution from here somehow given that in 20 minutes B will print 100 pages more) that A alone.

Then I draw the weighted average line and get that: a/b = 16/36 = 4/9. I am trying to plug in further a/(a+5) = 9/4. But I dont get the answer and get stuck. Can you help please?


hi,
I am not sure if it can be done by weighted average, as it is more to do with allegations and mixture and the averages..
It is not the case here..

I'll just tell you two methods here..

1) both finish the work in 24 minutes..
so A does 24/60=2/5 work ..
therefore B does 1-2/5=3/5 work..
Also in 24 minutes B can print 5*24=120 pages..
so 120=(3/5-2/5) of total work..
so 1/5 of total work = 120..
or total work = 600 pages..

2) 1/a + 1/b = 1/24..
1/60 + 1/b = 1/24..
b=40 minutes..
total work = 24(A+B)=24(A+A+5)=48A+120..
A's work individually= 60A..
so 60A=48A+120..
A=10..
so total pages = 60A=600..
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Re: M23-16 [#permalink] New post   06 Feb 2016, 23:51
Hi Bunuel

Can we solve this using weighted average approach? I am actually stuck at some point: so we derive that the faster machine B takes 40 minutes to complete the job. That is 20 minutes faster (can we come to solution from here somehow given that in 20 minutes B will print 100 pages more) that A alone.

Then I draw the weighted average line and get that: a/b = 16/36 = 4/9. I am trying to plug in further a/(a+5) = 9/4. But I dont get the answer and get stuck. Can you help please?
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Re: M23-16 [#permalink] New post   07 Feb 2016, 03:21
chetan2u wrote:
shasadou wrote:
Hi Bunuel

Can we solve this using weighted average approach? I am actually stuck at some point: so we derive that the faster machine B takes 40 minutes to complete the job. That is 20 minutes faster (can we come to solution from here somehow given that in 20 minutes B will print 100 pages more) that A alone.

Then I draw the weighted average line and get that: a/b = 16/36 = 4/9. I am trying to plug in further a/(a+5) = 9/4. But I dont get the answer and get stuck. Can you help please?


hi,
I am not sure if it can be done by weighted average, as it is more to do with allegations and mixture and the averages..
It is not the case here..

I'll just tell you two methods here..

1) both finish the work in 24 minutes..
so A does 24/60=2/5 work ..
therefore B does 1-2/5=3/5 work..
Also in 24 minutes B can print 5*24=120 pages..
so 120=(3/5-2/5) of total work..
so 1/5 of total work = 120..
or total work = 600 pages..

2) 1/a + 1/b = 1/24..
1/60 + 1/b = 1/24..
b=40 minutes..
total work = 24(A+B)=24(A+A+5)=48A+120..
A's work individually= 60A..
so 60A=48A+120..
A=10..
so total pages = 60A=600..


thanks! but we do not need this calculation in the 2nd method, do we?

2) 1/a + 1/b = 1/24..
1/60 + 1/b = 1/24..
b=40 minutes..
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Re: M23-16 [#permalink] New post   07 Feb 2016, 03:32
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shasadou wrote:
chetan2u wrote:
shasadou wrote:
Hi Bunuel

Can we solve this using weighted average approach? I am actually stuck at some point: so we derive that the faster machine B takes 40 minutes to complete the job. That is 20 minutes faster (can we come to solution from here somehow given that in 20 minutes B will print 100 pages more) that A alone.

Then I draw the weighted average line and get that: a/b = 16/36 = 4/9. I am trying to plug in further a/(a+5) = 9/4. But I dont get the answer and get stuck. Can you help please?


hi,
I am not sure if it can be done by weighted average, as it is more to do with allegations and mixture and the averages..
It is not the case here..

I'll just tell you two methods here..

1) both finish the work in 24 minutes..
so A does 24/60=2/5 work ..
therefore B does 1-2/5=3/5 work..
Also in 24 minutes B can print 5*24=120 pages..
so 120=(3/5-2/5) of total work..
so 1/5 of total work = 120..
or total work = 600 pages..

2) 1/a + 1/b = 1/24..
1/60 + 1/b = 1/24..
b=40 minutes..
total work = 24(A+B)=24(A+A+5)=48A+120..
A's work individually= 60A..
so 60A=48A+120..
A=10..
so total pages = 60A=600..


thanks! but we do not need this calculation in the 2nd method, do we?

2) 1/a + 1/b = 1/24..
1/60 + 1/b = 1/24..
b=40 minutes..


yeah, i intuitively wanted to solve this way! thanks
1) both finish the work in 24 minutes..
so A does 24/60=2/5 work ..
therefore B does 1-2/5=3/5 work..
Also in 24 minutes B can print 5*24=120 pages..
so 120=(3/5-2/5) of total work..
so 1/5 of total work = 120..
or total work = 600 pages..


please confirm whether my reasoning is right: A takes 60 min alone and B - 40 mins. Hence we can see how their productivity relates to each other: 40/60 = 2/3. This means taken together A performs 2/5 of the job and B - 3/5. Therefore the difference in the work done between A and B is simply 1/5.

1/5 w = 120 pages. Hence full work = 600!
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Re: M23-16 [#permalink] New post   07 Feb 2016, 03:48
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shasadou wrote:
chetan2u wrote:
shasadou wrote:
Hi Bunuel

Can we solve this using weighted average approach? I am actually stuck at some point: so we derive that the faster machine B takes 40 minutes to complete the job. That is 20 minutes faster (can we come to solution from here somehow given that in 20 minutes B will print 100 pages more) that A alone.

Then I draw the weighted average line and get that: a/b = 16/36 = 4/9. I am trying to plug in further a/(a+5) = 9/4. But I dont get the answer and get stuck. Can you help please?


hi,
I am not sure if it can be done by weighted average, as it is more to do with allegations and mixture and the averages..
It is not the case here..

I'll just tell you two methods here..

1) both finish the work in 24 minutes..
so A does 24/60=2/5 work ..
therefore B does 1-2/5=3/5 work..
Also in 24 minutes B can print 5*24=120 pages..
so 120=(3/5-2/5) of total work..
so 1/5 of total work = 120..
or total work = 600 pages..

2) 1/a + 1/b = 1/24..
1/60 + 1/b = 1/24..
b=40 minutes..
total work = 24(A+B)=24(A+A+5)=48A+120..
A's work individually= 60A..
so 60A=48A+120..
A=10..
so total pages = 60A=600..


thanks! but we do not need this calculation in the 2nd method, do we?

2) 1/a + 1/b = 1/24..
1/60 + 1/b = 1/24..
b=40 minutes..


Hi,
yes you are right .
it is not required here. but can be used in another method..
B does in 40 min and A in 60 min..
In these 40 minutes B does 5*40=200 more than A..
A completes these 200 in next 20 minutes,
so speed =200 in 20= 10 in 1 min...
in 60 minutes = 10*60=600..
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Re: M23-16 [#permalink] New post   25 Jul 2016, 22:34
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It requires two steps.
1. Calculate the number of hours B involves in finishing the task
2. Equate the rates of both A and B w.r.t total job or total pages in this case

Step 1 :
We all know

1/A + 1/B = 1/24 min
=> 1/60 + 1/B = 1/24 with this we get B = 40 mins

Step 2:

It is said that B prints 5 pages more than A. And we know rate = jobs /time

So let "j" be the total number of pages to be printed.
Then the (rate of ) B = ( rate of ) A + 5
=> j /40 = j /60 + 5
calculating for j we get 600.
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Re: M23-16 [#permalink] New post   11 Nov 2019, 12:02
Good question
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Bunuel
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Re: M23-16 [#permalink] New post   21 Aug 2023, 09:55
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M23-16 [#permalink]
21 Aug 2023, 09:55
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