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Machine rate problem

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Machine rate problem [#permalink] New post 07 Jun 2011, 13:00
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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] New post 07 Jun 2011, 15:17
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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] New post 07 Jun 2011, 17:27
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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] New post 08 Jun 2011, 00:51
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] New post 08 Jun 2011, 01:26
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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...

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Re: Machine rate problem [#permalink] New post 08 Jun 2011, 08:59
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] New post 08 Jun 2011, 17:15
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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] New post 10 Jun 2011, 20:40
16,000*x + 9000(8-x) = 1,00,000

so x will be 4 hrs
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Re: Machine rate problem [#permalink] New post 11 Aug 2011, 05:20
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] New post 11 Aug 2011, 23:25
100- 9*8 = 28

thus 7*4 hrs = 28

hence 4
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Re: Machine rate problem   [#permalink] 11 Aug 2011, 23:25
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