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Machine M and Machine N working alone at their constant rate

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Machine M and Machine N working alone at their constant rate [#permalink] New post 26 Jul 2014, 09:07
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Machine M and Machine N working alone at their constant rates, non stop, produced 6000 and 8000 nails respectively. Did machine M work longer than machine N?

(1) Machine N produces 2000 more nails than machine M in one hour when each machine work at its constant rate

(2) Machine N produces twice as much as machine M in one hour when each machine work at its constant rate
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Re: Machine M and Machine N working alone at their constant rate [#permalink] New post 26 Jul 2014, 09:37
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Machine M and Machine N working alone at their constant rates, non stop, produced 6000 and 8000 nails respectively. Did machine M work longer than machine N?

(1) Machine N produces 2000 more nails than machine M in one hour when each machine work at its constant rate.

If the rate of M is 2,000 nails per hour and the rate of N is 4,000 nails per hour, then to produce 6,000 nails, M needs 3 hours and to produce 8,000 nails N, needs 2 hours. In this case M worked longer than N.

If the rate of M is 6,000 nails per hour and the rate of N is 8,000 nails per hour, then to produce 6,000 nails, M needs 1 hours and to produce 8,000 nails N, needs 1 hour. In this case M did not work longer than N.

Not sufficient.

(2) Machine N produces twice as much as machine M in one hour when each machine work at its constant rate. In the time M needs to produce 6,000 nails, N can produce 12,000 nails, thus it can produce 8,000 nails in less time than M can produce 6,000 nails. Sufficient.

Answer: B.
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Re: Machine M and Machine N working alone at their constant rate [#permalink] New post 05 Sep 2014, 06:22
Machine M and Machine N working alone at their constant rates, non stop, produced 6000 and 8000 nails respectively. Did machine M work longer than machine N?

Let N is number of nails machine N produced.
Let M is number of nails machine M produced.
And x and y are respective time each machine took to produce 6000 and 8000 nails.

(1) Machine N produces 2000 more nails than machine M in one hour when each machine work at its constant rate

Given N= M + 2000
Mx = 6000 ------(1)
Ny = 8000 ==> (M+2000)y=8000 ---------(2)
Solve 1 and 2 to find ratio of x/y.

(\frac{6000}{x}+2000)y=8000
We can not reduce this equation in x/y form. Thus, (1) insufficient.

(2) Machine N produces twice as much as machine M in one hour when each machine work at its constant rate.
N=2M

Mx = 6000 --------(1)
Ny = 8000 ==> 2My=8000 -------(2)

2y\frac{6000}{x}=8000
Solve for x/y; we can calculate x:y. Thus, statement (2) is sufficient.
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Re: Machine M and Machine N working alone at their constant rate [#permalink] New post 07 Sep 2014, 08:33
Hi Bunuel,

Why are we rounding off the number of hours here?
Am i missing something in the question, why arent we considering minutes, for which im getting the first statement as being sufficient.
Thanks
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Re: Machine M and Machine N working alone at their constant rate [#permalink] New post 07 Sep 2014, 08:43
Expert's post
shishir16 wrote:
Hi Bunuel,

Why are we rounding off the number of hours here?
Am i missing something in the question, why arent we considering minutes, for which im getting the first statement as being sufficient.
Thanks


Cannot follow what you mean...

Where do we round the number of hours? Also, in my solution for (1) there are an example given which gives two different answers to the question, which means that this statement is not sufficient.
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DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Machine M and Machine N working alone at their constant rate   [#permalink] 07 Sep 2014, 08:43
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