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M23-16

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M23-16 [#permalink]

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Working together, printer A and printer B could finish a task in 24 minutes. If printer A alone could finish the task in 60 minutes, how many pages does the task contain if printer B prints 5 pages a minute more than printer A ?

A. 600
B. 800
C. 1000
D. 1200
E. 1500
[Reveal] Spoiler: OA

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Official Solution:

Working together, printer A and printer B could finish a task in 24 minutes. If printer A alone could finish the task in 60 minutes, how many pages does the task contain if printer B prints 5 pages a minute more than printer A ?

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


Let \(A\) be the number of pages printed by Printer A in one minute. In 60 minutes, printer A will print \(60A\) pages.

Let \(B\) be the number of pages printed by Printer B in one minute. \(B = A + 5\) and we can compose equation:

\(24A + 24(A + 5) = 60A\) from where \(A = 10\). The number of pages in the task \(= 60A = 600\).


Answer: A
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Re: M23-16 [#permalink]

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New post 06 Feb 2016, 22:51
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]

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New post 07 Feb 2016, 01:44
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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]

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New post 07 Feb 2016, 02:21
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]

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New post 07 Feb 2016, 02:32
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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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New post 07 Feb 2016, 02:48
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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]

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New post 25 Jul 2016, 21:34
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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]

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New post 26 Sep 2017, 10:38
Bunuel,

I got this question correct and I am trying to understand your solution. Can you please help me to understand that how did you arrive at this equation 24A+24(A+5)=60A ?

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Re: M23-16 [#permalink]

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New post 26 Sep 2017, 20:22
msk0657 wrote:
Bunuel,

I got this question correct and I am trying to understand your solution. Can you please help me to understand that how did you arrive at this equation 24A+24(A+5)=60A ?


A is the rate of printer A;
B is the rate of printer B, which equals to A + 5;

Working together, printer A and printer B could finish a task in 24 minutes: (job done) = (combined rate)(time) = 24(A + (A + 5));
A alone could finish the task in 60 minutes: (job done) = (rate)(time) = 60A.

Equate these two to get 24(A + (A + 5)) = 60A.

Hope it's clear.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M23-16 [#permalink]

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New post 26 Oct 2017, 13:55
Same as above to get rate:

\(\frac{1}{60} + \frac{1}{40} = \frac{1}{24}\)

From 60:40 relationship we can intuitively see what the possible answer would be: 10 pages per minute and 15 pages per minute, 240+(15*24)=600

Not sure about the validity of the process but working back from the answer made sense and afterwards I checked to make sure the math worked out.

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Re: M23-16   [#permalink] 26 Oct 2017, 13:55
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