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Machine A can process 6000 envelopes in 3 hours. Machines B [#permalink]
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Updated on: 06 Jun 2013, 00:42
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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 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
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Originally posted by shash on 25 Nov 2010, 08:25.
Last edited by Bunuel on 06 Jun 2013, 00:42, edited 2 times in total.
Edited the options.



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Re: Rate Problem [#permalink]
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25 Nov 2010, 10:08
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]
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25 Nov 2010, 21:11
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]
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26 Nov 2010, 04:20
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]
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27 Nov 2010, 11:07
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 = 30001 = 30002000 = 1000 => B = 2400 C =24001000 =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]
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27 Nov 2010, 18:58
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]
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30 May 2013, 05:09



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Re: Machine A can process 6000 envelopes in 3 hours. Machines B [#permalink]
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30 May 2013, 10:50
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, if\(r_B\) and\(r_C\) 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, as\(r_C\)= 1000, we know for sure that\(r_B\) <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]
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05 Jun 2013, 15:36
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]
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06 Jun 2013, 00:14
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]
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06 Jun 2013, 00:44



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Re: Machine A can process 6000 envelopes in 3 hours. Machines B [#permalink]
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07 Jun 2013, 05:35
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]
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09 Jun 2013, 20:52
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 hereNow, 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]
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09 Jun 2013, 21:56
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]
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05 Dec 2014, 02:30
shash wrote: ... If Machines A and C working together but independently ... Is this a tricky way of saying "working together"? I mean, can I treat that phrase just like how you would when you combine the rates of machine A and machine C? Is there a question where GMAT asks two or more entities working together depending on each other?



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Re: Machine A can process 6000 envelopes in 3 hours. Machines B [#permalink]
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05 Dec 2014, 07:51



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Re: Machine A can process 6000 envelopes in 3 hours. Machines B [#permalink]
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14 Feb 2015, 09:45
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 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 OPTIONS1/A = 2000 1/B+1/C = 6000/2.5=2400 (1) 1/A+ 1/C = 3000 (2) (1)  (2) 1/B1/A =24003000 > 1/B  2000 = 24003000 > 1/B = 1400 Rate = 1400 Work= 12000 Time= 12000/1400 =60/7 Feed me kudos if it is helpful for you :D



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Re: Machine A can process 6000 envelopes in 3 hours. Machines B [#permalink]
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24 Dec 2015, 16:54
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 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 OPTIONSThis is not the original question. In the original question (from Kaplan) the time given for mashines B and C working together is 2\frac{2}{5} and the correct answer choice is E, which equals to 8 and not \(\frac{60}{7}\)
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Machine A can process 6000 envelopes in 3 hours. Machines B [#permalink]
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05 Mar 2017, 15:58
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
let a,b,c=respective rates a=2000 (a+c)(b+c)=ab=600 b=1400 12000/1400=60/7 hours E



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Re: Machine A can process 6000 envelopes in 3 hours. Machines B [#permalink]
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08 Mar 2018, 12:48
Instead of 2.5 hours , 2 hours and (2/5) mins then answer will be 4 hours . Correct? Posted from GMAT ToolKit




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