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# Working alone at its constant rate, machine K took 3 hours

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Working alone at its constant rate, machine K took 3 hours [#permalink]

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29 Nov 2012, 22:44
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Working alone at its constant rate, machine K took 3 hours to produce 1/4 of the units produced last Friday. Then machine M started working and the two machines, working simultaneously at their respective constant rates, took 6 hours to produce the rest of the units produced last Friday. How many hours would it have taken machine M, working alone at its constant rate, to produce all of the units produced last Friday?

A. 8
B. 12
C. 16
D. 24
E. 30
[Reveal] Spoiler: OA

Last edited by Bunuel on 30 Nov 2012, 03:07, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: No hours machine M takes to do last friday's work alone [#permalink]

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29 Nov 2012, 23:09
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Lets assume the work to be something like moving 24 bricks. (Three fourths of 24 gives a multiple of 6 & one fourth of 24 gives a multiple of 3. Hence we are selecting 24 for ease of calculation. This method should work fine for any number)
Machine K works at 2 bricks per hour
K & M together work at 3 bricks per hour
So, M works at 1 brick per hour
So, M working alone would have took 24 hours.

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Intern
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Re: No hours machine M takes to do last friday's work alone [#permalink]

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30 Nov 2012, 00:21
the explanation sounds logical and the correct answer to this is D.Though,that's the (pick an answer an work upon) approach.how can we approach the same question through the conventional method (conceptual), by taking out the respective rates etc.can any one explain?
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Re: No hours machine M takes to do last friday's work alone [#permalink]

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30 Nov 2012, 00:32
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samdarsh88 wrote:
the explanation sounds logical and the correct answer to this is D.Though,that's the (pick an answer an work upon) approach.how can we approach the same question through the conventional method (conceptual), by taking out the respective rates etc.can any one explain?

1st 2nd and 3 columns are Rate, Time and Work done.

R * 3 = 1/4
so rate of K = 1/12

Now 3/4 of the job is done by both K and M in 6 hrs

(1/12 + rate of M ) * 6 = 3/4
solving we get rate of M = 1/24

Finally

R * t = W

1/24 * t = 1

there t = 24 hrs

HTH.

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Re: No hours machine M takes to do last friday's work alone [#permalink]

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30 Nov 2012, 00:41
thanks for the explanation jp27.seems ,i was committing a careless mistake before. Hail guys.
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Re: No hours machine M takes to do last friday's work alone [#permalink]

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30 Nov 2012, 01:31
samdarsh88 wrote:
the explanation sounds logical and the correct answer to this is D.Though,that's the (pick an answer an work upon) approach.how can we approach the same question through the conventional method (conceptual), by taking out the respective rates etc.can any one explain?

As I mentioned in my previous comment, I was not picking an answer. Rather I was just picking a smart number. In this case both happened to be the same. The same method would work just as well with any other number.

Illustration :

Lets assume the work to be making 30 bricks :

$$Rate of Machine K = \frac{30}{4*3} = \frac{30}{12}$$

$$Rate of K & M together = \frac{30*3}{4*6} = \frac{90}{24} = \frac{45}{12}$$

$$Rate of machine M = \frac{45}{12} - \frac{30}{12} = \frac{15}{12} = \frac{5}{4}$$

$$Machine M alone = \frac{30}{5/4} = \frac{120}{5} = 24$$

So, we can see that answer is still 24. I had picked 24 only to avoid fractions.
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Re: No hours machine M takes to do last friday's work alone [#permalink]

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30 Nov 2012, 02:28
rate of machine k = (1/4)/3 (i.e it takes 3 hours to complete 1/4 of the work) ----> 1/12

Now both M and K working together at their respective rate complete 3/4 (the remaining work i.e 1- 1/4) of the work in 6 hours.

1/12 + 1/M = (3/4)/6 ------M = 24

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Re: Working alone at its constant rate, machine K took 3 hours [#permalink]

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30 Nov 2012, 03:18
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samdarsh88 wrote:
Working alone at its constant rate, machine K took 3 hours to produce 1/4 of the units produced last Friday. Then machine M started working and the two machines, working simultaneously at their respective constant rates, took 6 hours to produce the rest of the units produced last Friday. How many hours would it have taken machine M, working alone at its constant rate, to produce all of the units produced last Friday?

A. 8
B. 12
C. 16
D. 24
E. 30

Given that machines M and K, working together, can produce 3/4 of units in 6 hours. Since also given that machine K can produce 1/4 of units in 3 hours, then in 6 hours it can produce 2/4 of units. Which means that when working together, machine M produced 3/4-2/4=1/4 of units in 6 hours, thus it needs 6*4=24 hours to produce all of the units.

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Re: Working alone at its constant rate, machine K took 3 hours [#permalink]

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20 Nov 2014, 17:37
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Re: Working alone at its constant rate, machine K took 3 hours [#permalink]

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18 Dec 2015, 10:06
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Working alone at its constant rate, machine K took 3 hours [#permalink]

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21 Dec 2015, 14:03
Would be curious to know whether veritasprep's joint variation method would apply here. Or am I missing the point of the joint variation method to begin with...
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Re: Working alone at its constant rate, machine K took 3 hours [#permalink]

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21 Dec 2015, 16:44
samdarsh88 wrote:
Working alone at its constant rate, machine K took 3 hours to produce 1/4 of the units produced last Friday. Then machine M started working and the two machines, working simultaneously at their respective constant rates, took 6 hours to produce the rest of the units produced last Friday. How many hours would it have taken machine M, working alone at its constant rate, to produce all of the units produced last Friday?

A. 8
B. 12
C. 16
D. 24
E. 30

Machine K in 6 hours produced 2/4
As Machine K already had produced 1/4, so machine M produced only 1/4 in 6 hours

1/4x = 6
x = 24
Re: Working alone at its constant rate, machine K took 3 hours   [#permalink] 21 Dec 2015, 16:44
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