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Machine A can process 6000 envelopes in 3 hours. Machines B

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Machine A can process 6000 envelopes in 3 hours. Machines B [#permalink] New post 25 Nov 2010, 07:25
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Machine A can process 6000 envelopes in 3 hours. Machines B and C working together but independently can process the same number of envelopes in 2.5 hours. If Machines A and C working together but independently process 3000 envelopes in 1 hour, then how many hours would it take Machine B to process 12000 envelopes.

A. 2
B. 3
C. 4
D. 6
E. 60/7

[Reveal] Spoiler:
I got this far:

Machine C = 6 hours for 6000 envelopes

Then (1/T) = (1/b) +(1/c)

1/2.5 = (1/b) + (1/6)
b= (30/7) for 6000 envelopes or (60/7) for 12000 envelopes.

Why isnt the answer coming up to exactly 8?


EDITED THE OPTIONS
[Reveal] Spoiler: OA

Last edited by Bunuel on 05 Jun 2013, 23:42, edited 2 times in total.
Edited the options.
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Re: Rate Problem [#permalink] New post 25 Nov 2010, 09:08
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shash wrote:
Machine A can process 6000 envelopes in 3 hours. MAchines B and C working together but independently can process the same number of envelopes in 2.5 hours. If Machines A and C working together but independently process 3000 envelopes in 1 hour, then how many hours would it take Machine B to process 12000 envelopes.

2
3
4
6
8 - Correct Answer

I got this far:

Machine C = 6 hours for 6000 envelopes

Then (1/T) = (1/b) +(1/c)

1/2.5 = (1/b) + (1/6)
b= (30/7) for 6000 envelopes or (60/7) for 12000 envelopes.

Why isnt the answer coming up to exactly 8?


I think you did everything right.

Let the time needed for A, B and C working individually to process 6,000 envelopes be a, b and c respectively.

Now, as "A can process 6,000 envelopes in 3 hours" then a=3;

As "B and C working together but independently can process the same number (6,000) of envelopes in 2.5 hours" then \frac{1}{b}+\frac{1}{c}=\frac{1}{2.5}=\frac{2}{5};

Also, as "A and C working together but independently process 3000 envelopes in 1 hour", then A and C working together but independently process 2*3,000=6,000 envelopes in 2*1=2 hours: \frac{1}{a}+\frac{1}{c}=\frac{1}{2} --> as a=3 then c=6;

So, \frac{1}{b}+\frac{1}{6}=\frac{2}{5} --> b=\frac{30}{7}, which means that B produces 6,000 envelopes in 30/7 hours, thus it produces 12,000 envelopes in 60/7 hours.

Answer: E.
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Re: Rate Problem [#permalink] New post 25 Nov 2010, 20:11
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shash wrote:
Machine A can process 6000 envelopes in 3 hours. MAchines B and C working together but independently can process the same number of envelopes in 2.5 hours. If Machines A and C working together but independently process 3000 envelopes in 1 hour, then how many hours would it take Machine B to process 12000 envelopes.

2
3
4
6
8 - Correct Answer

I got this far:

Machine C = 6 hours for 6000 envelopes

Then (1/T) = (1/b) +(1/c)

1/2.5 = (1/b) + (1/6)
b= (30/7) for 6000 envelopes or (60/7) for 12000 envelopes.

Why isnt the answer coming up to exactly 8?


In work rate questions, generally different people have to complete the same amount of work. In this question, to make it a little tricky, they have given varying amount of work done by the machines.
To make the question straight forward, first thing you can do is make the work the same for all:
Machine A processes 12000 envelopes in - 6 hrs
Machines B and C process 12000 in - 5 hrs
Machines A and C process 12000 in - 4 hrs
I chose to get them all to 12000 since my question has 12000 in it. Also, I easily get rid of all decimals.
Now, I just find 1/6 + 1/c = 1/4 and get c = 12
and 1/b + 1/12 = 1/5 so b = 60/7 hrs
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Re: Rate Problem [#permalink] New post 26 Nov 2010, 03:20
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forget abt the variables a,b,c,d.....in this simple problem

A - 6000 in 3 hrs ==> 2000 in 1 hr
BC- 6000 - in 2.5 hrs == 2400 in 1 hr
AC - 3000 in 1 hr

take AC -3000 in 1 hr , in which A's contribution is 2000 in 1 hr, hence C's contribution is 1000 in 1 hr
take BC -2400 , in which C's contribution is 1000 in 1 hr, hence B's contribution is 1400 in 1 hr.

B - 1400 - 1 hr
==> 12000 in 12000/1400 hrs = 60/7

Regards,
Murali.
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Re: Rate Problem [#permalink] New post 27 Nov 2010, 10:07
muralimba wrote:
forget abt the variables a,b,c,d.....in this simple problem

A - 6000 in 3 hrs ==> 2000 in 1 hr
BC- 6000 - in 2.5 hrs == 2400 in 1 hr
AC - 3000 in 1 hr

take AC -3000 in 1 hr , in which A's contribution is 2000 in 1 hr, hence C's contribution is 1000 in 1 hr
take BC -2400 , in which C's contribution is 1000 in 1 hr, hence B's contribution is 1400 in 1 hr.

B - 1400 - 1 hr
==> 12000 in 12000/1400 hrs = 60/7

Regards,
Murali.



Adding to Murali's approach....

A = 2000
B+C = 2400
A+C = 3000 => C = 3000-1 = 3000-2000 = 1000

=> B = 2400 -C =2400-1000 =1400

hence B can process 1400 Envelopes in 1hour...how much time wud it take B to process 12000 Envelopes = 12000/1400 = 60/7
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Re: Rate Problem [#permalink] New post 27 Nov 2010, 17:58
For 1 hour-

Machine A rate- 2000 envelopes
Machine B+C rate- 2400 envelopes
Since A + C = 3000 envelopes A's rate is 2000 envelopes as above, C has a rate of 1000 envelopes per hour.
Which makes machine B's rate as 1400 envelopes per hour.

Thus, it will take 8 hours to manufacture 12000 envelopes.

Answer:-E
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Re: Machine A can process 6000 envelopes in 3 hours. Machines B [#permalink] New post 30 May 2013, 04:09
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Re: Machine A can process 6000 envelopes in 3 hours. Machines B [#permalink] New post 30 May 2013, 09:50
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shash wrote:
Machine A can process 6000 envelopes in 3 hours. Machines B and C working together but independently can process the same number of envelopes in 2.5 hours. If Machines A and C working together but independently process 3000 envelopes in 1 hour, then how many hours would it take Machine B to process 12000 envelopes.

A. 2
B. 3
C. 4
D. 6
E. 8


Machine A takes 3 hours for 6000 envelopes. Thus, Machine A would take exactly 6 hours for 12000 envelopes. Also, we know that machines B and C, working together, can produce the same no of envelopes in 2.5 hours. Thus, ifr_B andr_C are the rates respectively , we know that(r_B+r_C)*\frac{5}{2} = 6000 --> (r_B+r_C) = 2400. Thus, even if we assume that r_B = 2000 (which is the same rate as that of Machine A), Machine B would again need 6 hours. However, asr_C= 1000, we know for sure thatr_B <2000. Thus, the only option more than 6 hours is E(Assuming that the correct OA is provided with the question).
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Re: Rate Problem [#permalink] New post 05 Jun 2013, 14:36
Sarang wrote:
For 1 hour-

Machine A rate- 2000 envelopes
Machine B+C rate- 2400 envelopes
Since A + C = 3000 envelopes A's rate is 2000 envelopes as above, C has a rate of 1000 envelopes per hour.
Which makes machine B's rate as 1400 envelopes per hour.

Thus, it will take 8 hours to manufacture 12000 envelopes.

Answer:-E


I did this but shouldn't the work take 9 hours instead?
In 8 hours machine B would have made 1400 * 8 = 11200 envelopes.
In order to make 12000 it would require a fraction of an hour to create 200 more envelopes.
Am I mistaken?
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Re: Rate Problem [#permalink] New post 05 Jun 2013, 23:14
Expert's post
PKPKay wrote:
Sarang wrote:
For 1 hour-

Machine A rate- 2000 envelopes
Machine B+C rate- 2400 envelopes
Since A + C = 3000 envelopes A's rate is 2000 envelopes as above, C has a rate of 1000 envelopes per hour.
Which makes machine B's rate as 1400 envelopes per hour.

Thus, it will take 8 hours to manufacture 12000 envelopes.

Answer:-E


I did this but shouldn't the work take 9 hours instead?
In 8 hours machine B would have made 1400 * 8 = 11200 envelopes.
In order to make 12000 it would require a fraction of an hour to create 200 more envelopes.
Am I mistaken?


As mentioned above, the OA is incorrect. In fact, the options are incorrect since none of them is 60/7 hrs (which is the answer).
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Re: Rate Problem [#permalink] New post 05 Jun 2013, 23:44
Expert's post
PKPKay wrote:
Sarang wrote:
For 1 hour-

Machine A rate- 2000 envelopes
Machine B+C rate- 2400 envelopes
Since A + C = 3000 envelopes A's rate is 2000 envelopes as above, C has a rate of 1000 envelopes per hour.
Which makes machine B's rate as 1400 envelopes per hour.

Thus, it will take 8 hours to manufacture 12000 envelopes.

Answer:-E


I did this but shouldn't the work take 9 hours instead?
In 8 hours machine B would have made 1400 * 8 = 11200 envelopes.
In order to make 12000 it would require a fraction of an hour to create 200 more envelopes.
Am I mistaken?


Edited the options.

Check for a solution here: machine-a-can-process-6000-envelopes-in-3-hours-machines-b-105362.html#p823509 or here: machine-a-can-process-6000-envelopes-in-3-hours-machines-b-105362.html#p823655
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Re: Machine A can process 6000 envelopes in 3 hours. Machines B [#permalink] New post 07 Jun 2013, 04:35
How much time should one take in solving these kind of questions which involves though simple yet a lot of calculations?
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Re: Machine A can process 6000 envelopes in 3 hours. Machines B [#permalink] New post 09 Jun 2013, 19:52
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samheeta wrote:
How much time should one take in solving these kind of questions which involves though simple yet a lot of calculations?


This can be easily done in under 2 mins. If you look at the explanation provided above:


To make the question straight forward, first thing you can do is make the work the same for all:
Machine A processes 12000 envelopes in - 6 hrs
Machines B and C process 12000 in - 5 hrs
Machines A and C process 12000 in - 4 hrs
I chose to get them all to 12000 since my question has 12000 in it. Also, I easily get rid of all decimals.

Almost no calculations till here

Now, I just find 1/6 + 1/c = 1/4 and get c = 12
and 1/b + 1/12 = 1/5 so b = 60/7 hrs

You should be comfortable with manipulating fractions.
1/c = 1/4 - 1/6 = 2/24 = 1/12
So c = 12 (Finding c should take just a few seconds)

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Re: Machine A can process 6000 envelopes in 3 hours. Machines B [#permalink] New post 09 Jun 2013, 20:56
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shash wrote:
Machine A can process 6000 envelopes in 3 hours. Machines B and C working together but independently can process the same number of envelopes in 2.5 hours. If Machines A and C working together but independently process 3000 envelopes in 1 hour, then how many hours would it take Machine B to process 12000 envelopes.

A. 2
B. 3
C. 4
D. 6
E. 60/7


You can either take the amount of work done as the same as Karishma has done or take the work done by each in the same time. I will do the latter

1. Work done in 1 hr by A is 2000 envelopes
2. Work done in 1 hr by A and C is 3000 envelopes
3. So work done in 1 hr by C is 1000 envelopes
4. Work done in 1 hr by B and C is 2400 envelopes
5. So work done in 1 hr by B is 1400 envelopes
6. So to process 12000 envelopes B will take 12000/1400 hrs = 60/7 hrs

So the answer is choice E
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Re: Machine A can process 6000 envelopes in 3 hours. Machines B [#permalink] New post 17 Jun 2014, 08:27
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Re: Machine A can process 6000 envelopes in 3 hours. Machines B   [#permalink] 17 Jun 2014, 08:27
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