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Six machines, each working at the same constant rate, togeth [#permalink]
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27 Nov 2006, 03:04
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Six machines, each working at the same constant rate, together can complete a certain job in 12 days. How many additional machines, each working at the same constant rate, will be needed to complete the job in 8 days? A. 2 B. 3 C. 4 D. 6 E. 8 OPEN DISCUSSION OF THIS QUESTION IS HERE: sixmachineseachworkingatthesameconstantrate127112.html
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Last edited by Bunuel on 25 May 2014, 14:01, edited 1 time in total.
Renamed the topic, edited the question and added the OA.



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1/x+1/x+1/x+1/x+1/x+1/x=1/12
6/x=1/12
x=72
So.
x/72=1/8
8x=72
x=9
96=3



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Re: Six Machines, [#permalink]
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27 Nov 2006, 09:28
each working at the same constant rate, together can complete a certain job in 12 days. How many additional machines each working at the same constant rate, will be needed to complete the job in 8 days?
Assume N machines are needed and each machine does M amount of work
6/M=1/12==> M =72
N*(1/72)=1/8
N*(1/72)=1/8
N=72/8=9
Additional machines =96=3



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6:12
x:8
flip one ratio
6:8
x:12
multiply 6*12 = 72 so 8*x=72 ...x=9
3 more machines needed
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Just had this, here is my take:
I used the work formula (like the RT=D) Time*Rate= Work
T*R=W
12*6=72
8*(6+x) = 72
48+8x=72
8x=24
x=3
Seemed to work out for me.



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Re: PS machines working together [#permalink]
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05 Feb 2009, 07:42
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Let R be the rate of working of each machine 6/R=1/12 hence, R=72 Let X be the total no of machines which can complete work in 8 days X/R=1/8 Put value of R=72 Solving X=9 Hence additional no of machines required =96=3 Hence B jairus wrote: Six machines, each working at the same constant rate, together can complete a certain job in 12 days. How many additional machines, each woring at the same constant rate, will be needed to complete the job in 8 days?
A) 2
B) 3
C) 4
D) 6
E) 8



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Re: PS machines working together [#permalink]
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05 Feb 2009, 20:31
jairus wrote: Six machines, each working at the same constant rate, together can complete a certain job in 12 days. How many additional machines, each woring at the same constant rate, will be needed to complete the job in 8 days?
A) 2
B) 3
C) 4
D) 6
E) 8 Work Rate = 1/Time taken Work Rate for 6 machines = 1/12 Work Rate for 1 machine = 1/12*6 = 1/72 Wor Rate for X machines = x/72 = 1/8 > x=9 Answ = 96=3 B
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Re: machines? [#permalink]
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09 Aug 2009, 06:47
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For work related problems, number of person days or machine days is always constant Let x be the number of machines that were used to complete the work in 8 days. So, 6*12 = 8*x x = 9 Number of machines to be added = 96 = 3



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Re: machines? [#permalink]
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09 Aug 2009, 07:14
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Here number of machines and days are inversely proportional. 6 machine can complete the work in 12 days, so in 1 day all together can complete 1/12 part of work. Now to complete 1/8 part of work, 6  1/12 x  1/8 x =6/8*12 x=9 Number of machines to be added = 96 = 3
/Prabu



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Re: PS machines working together [#permalink]
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16 Aug 2009, 11:55
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Here is how I solved the problem:
6 machines each working at the same constant rate together can complete a job in 12 days. How many additional machines, each working at the same constant rate, will be needed to complete the job in 8 days?
I think I saw this one in the first PowerPrep Test:
This is a simple two equation problem:
let x = the number of additional machines needed to complete the job in 8 days let y = the rate for one machine to complete the job
Equation 1:
Rate * Time = Work 6*y*12 = 1 job
Equation 2:
(6+x)*y*8 = 1 job
so... y = 1/(6*12) (6+x)*(1/(6*12))*8 = 1 x = 3



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Re: PS machines working together [#permalink]
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17 Aug 2009, 10:27
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This solution is much faster and doesn't require you to calculate the rate of each of the machines  6 machines > 12 days ? machines > 8 days. the trick is to consider proportions since all machines are running at constant rate. so the number of machines is inversely proportional to number of days taken to complete the job. More number of machines to finish in lesser number of days  plain logic. Under such situations, the blind approach is to multiply 6 with 12 and divide by 8, gives total number of machines to finish in 8 days, additional machines would be to just subtract with original number of machines. 9  6 = 3. If the proportion was straight and not inverse, we would have multiplied 6 with 8 and divided with 12, but in inverse proportions, same line elements are multiplied.
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Re: Six Machines, [#permalink]
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17 Aug 2009, 12:42
18.Six machines, each working at the same constant rate, together can complete a certain job in 12 days. How many additional machines, each working at the same constant rate, will be needed to complete the job in 8 days?
A) 2 B) 3 C) 4 D) 6 E) 8
Rate of one machine = 1 job / (12*6) days
let X = number of machines needed to complete the job in 8 days
1/(6*12) * 8 * X = 1 job
X = 9
96 = 3
ANSWER: B. 3



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Re: machines? [#permalink]
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17 Aug 2009, 12:43
18.Six machines, each working at the same constant rate, together can complete a certain job in 12 days. How many additional machines, each working at the same constant rate, will be needed to complete the job in 8 days?
A) 2 B) 3 C) 4 D) 6 E) 8
Rate of one machine = 1 job / (12*6) days
let X = number of machines needed to complete the job in 8 days
1/(6*12) * 8 * X = 1 job
X = 9
96 = 3
ANSWER: B. 3



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Re: machines? [#permalink]
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17 Aug 2009, 13:52
let the rate of machines be x
then 6 machines * 12 days * x = n machines * 8 days * x
> 9 machines, thus 3 more are required



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Re: machines? [#permalink]
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17 Aug 2009, 19:10
arvs212 wrote: got this question on gmatprep...i know its an easy question but dunno why today i jus wasnt able to come 2 eqtns>>
Q> 6 machines at the same constant rate can do a work in 12 days,if the work has to be done in 8 days, how many machines which work at the same constant rate, need to be added.??? Follow this formula, it will be always easy.. * Different number of machines (or men) , change the number of days(hours) , but the product of the number of machines and number of days remain the same M1 * D1 = M2 * D2 M= number of machines D= number of days 6 * 12 = M2 * 8 M2 = 9, so 3 more * another formula when amount of work is different from the original, will be useful in some problems. (M1 * D1) / W1 = (M2 * D2)/W2 W = amount of work



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Re: Six Machines Working Together GMAT Prep [#permalink]
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25 May 2014, 13:05
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Re: Six machines, each working at the same constant rate, togeth [#permalink]
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25 May 2014, 14:01
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Re: Six machines, each working at the same constant rate, togeth
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