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# 6 machines working at the same rate takes 12 hours to finish

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6 machines working at the same rate takes 12 hours to finish [#permalink]  28 Aug 2007, 19:25
6 machines working at the same rate takes 12 hours to finish 1 job.
How many machines, all working at the same rate, does it take to finish 1 job in 8 hours?
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Re: Work [#permalink]  28 Aug 2007, 19:34
ywilfred wrote:
6 machines working at the same rate takes 12 hours to finish 1 job.
How many machines, all working at the same rate, does it take to finish 1 job in 8 hours?
6r = 1/12 --> r = 1/72 (individual rate of each machine)

8r/72 = 1 --> 8r = 72 --> r = 9

I get 9 machines.
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6 machines = 12 h
0.5 machine = 1 h
Difference between 12 and 8 = 4 h
so 4 h = 4*0.5 = 2 machines

6 + 2 = 8 machines.

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Whatever wrote:
6 machines = 12 h
0.5 machine = 1 h

Difference between 12 and 8 = 4 h
so 4 h = 4*0.5 = 2 machines

6 + 2 = 8 machines.

if 6 machines take 12 hrs to complete a job how can 1/2 a machine take 1 hr to complete a job? Am I reading you correctly?
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ggarr wrote:
Whatever wrote:
6 machines = 12 h
0.5 machine = 1 h

Difference between 12 and 8 = 4 h
so 4 h = 4*0.5 = 2 machines

6 + 2 = 8 machines.

if 6 machines take 12 hrs to complete a job how can 1/2 a machine take 1 hr to complete a job? Am I reading you correctly?

I thought in terms of machines
6 machines = 12 h
1 machine work for 2 h in this bundle
if the rate is the same, we need 2 more machines to complete the work in 8 hours (for me it was easier to devide by 12 at that point).

Actually, I'm not sure about this method, 'cause I'm still testing it, but if I'm right - this method is nice and elegant in comparison with usual 1/x + 1/y = 1/T =)
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Re: Work [#permalink]  28 Aug 2007, 21:15
ywilfred wrote:
6 machines working at the same rate takes 12 hours to finish 1 job.
How many machines, all working at the same rate, does it take to finish 1 job in 8 hours?

I got 9
Machine:Time:Job
6:12:1
6*12/8 :12*8/12:1
9:8:1
9 machines
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Re: Work [#permalink]  28 Aug 2007, 21:33
bkk145 wrote:
I got 9
Machine:Time:Job
6:12:1
6*12/8 :12*8/12:1
9:8:1
9 machines

Good job!
6 : 12 : 1
x : 8 : 1

to get 8 from 12 we should multiply 12 * 2/3, but when I multiplied 6*2/3 I got 4. Could you please tell me why you turned up the fraction?
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Re: Work [#permalink]  28 Aug 2007, 21:50
ywilfred wrote:
6 machines working at the same rate takes 12 hours to finish 1 job.
How many machines, all working at the same rate, does it take to finish 1 job in 8 hours?

R 1/12hrs
T 12hrs
D 1 job
lets say x is one machine

so the rate of 1 machine is
6x=1/12 --> x=1/72

R 1/72
T 8hrs
D 1/9 of the job is complete

so from this it looks like its going to take 9 machines

9(1/72)=1/8

Now plug in to see if this works...

R: 1/8
T: 8hrs
D 1job

D=rt D=1job/8hrs*8hrs = 1 job

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Re: Work [#permalink]  28 Aug 2007, 21:56
ywilfred wrote:
6 machines working at the same rate takes 12 hours to finish 1 job.
How many machines, all working at the same rate, does it take to finish 1 job in 8 hours?

12*6/8 = 9 machines
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Re: Work [#permalink]  28 Aug 2007, 22:17
Whatever wrote:
bkk145 wrote:
I got 9
Machine:Time:Job
6:12:1
6*12/8 :12*8/12:1
9:8:1
9 machines

Good job!
6 : 12 : 1
x : 8 : 1

to get 8 from 12 we should multiply 12 * 2/3, but when I multiplied 6*2/3 I got 4. Could you please tell me why you turned up the fraction?

Machine and time have inverse relationship.
Re: Work   [#permalink] 28 Aug 2007, 22:17
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