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Machine rate problem [#permalink]
 07 Jun 2011, 14:00
Question Stats:
100% (03:32) correct
0% (00:00) wrong based on 0 sessions
Hi All,
Can you please help me solve this problem?
Machine A @ constant rate 9,000/H
Machine B @ Constant rate 7,000/H
If both together, then it can produces 100,000 in 8 hours, then what is the minumum hours is needed for machine B to run?
The answer was at least 4 hours.
This is the equation I think is correct: 9000/A + 7000/B = 10000/8. However, I don't know how to get another equation since i have 2 unknowns.
Thanks!
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Re: Machine rate problem [#permalink]
07 Jun 2011, 16:17
It seems that something is wrong with your question. A&B together have 16,000/h rate and over 8h it will be 128,000 (not 100,000). Or am I missing something?
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Re: Machine rate problem [#permalink]
07 Jun 2011, 18:27
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We have to find out the minumum no. hours that machine B HAS to run. Now we have the output for 8 hrs for both the machines ( nothing given about the no. of hrs each runs) Lets assume Machine A runs for full 8 hrs..it produces 72000 units. Now the remaining 28000 have to be produced by B..which it would do in 28000/7000 = 4 hrs... I would slightly tweak the wording in the problem to say "if both of them operate simultenously or otherwise for 8 hrs..." (but I guess it would make the question far to easy!!)
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Re: Machine rate problem [#permalink]
08 Jun 2011, 01:51
So you're saying B HAS to run for atleast 4 hours? But what if A runs less than 8 hrs.. Is it given that it has to run for atleast 7 hours? I think than the answer would change unless the result has to be an integer multiple of both rates.. I was actually thinking the same thing as walker about the combined rate.. Anyone?
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Re: Machine rate problem [#permalink]
08 Jun 2011, 02:26
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l0rrie wrote: So you're saying B HAS to run for atleast 4 hours? But what if A runs less than 8 hrs.. Is it given that it has to run for atleast 7 hours? I think than the answer would change unless the result has to be an integer multiple of both rates.. I was actually thinking the same thing as walker about the combined rate.. Anyone? The question asks the min. no of hrs B should run..we have a given output and no. Of hrs..A can run for a maximum of 8 hrs..and the shortfall in the output would be covered by B.. I agree the language of the question can be improved but this is what I think given the problem at hand... Posted from my mobile device 
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Re: Machine rate problem [#permalink]
08 Jun 2011, 09:59
I think maverick04 is correct. I see how you solved the problem now. Thanks so much. Sorry for wording the question so badly. I got it off this online problem generating website and I am unable to regenerator the same problem. I wrote down the problem in short-hand style and so I don't remember the exact wording. Again, Thanks so much for all your help.
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Re: Machine rate problem [#permalink]
08 Jun 2011, 18:15
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teebumble wrote: I think maverick04 is correct. I see how you solved the problem now. Thanks so much. Sorry for wording the question so badly. I got it off this online problem generating website and I am unable to regenerator the same problem. I wrote down the problem in short-hand style and so I don't remember the exact wording. Again, Thanks so much for all your help. Anytime mate 
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Re: Machine rate problem [#permalink]
10 Jun 2011, 21:40
16,000*x + 9000(8-x) = 1,00,000 so x will be 4 hrs
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Re: Machine rate problem [#permalink]
11 Aug 2011, 06:20
A+B in 1 hr produce 9+7 = 16k in 8hrs = 128k but we need only 100k in 8hrs. so we have 28k extra. let's give B some rest. 128k/7k = 4hrs so B will rest for 4 hours and, thus, will work for 4 hrs (8-4)
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Re: Machine rate problem [#permalink]
12 Aug 2011, 00:25
100- 9*8 = 28 thus 7*4 hrs = 28 hence 4
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Re: Machine rate problem
[#permalink]
12 Aug 2011, 00:25
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