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Working alone at its constant rate, machine K took 3 hours [#permalink]
29 Nov 2012, 21:44

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A

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D

E

Difficulty:

25% (low)

Question Stats:

76% (02:51) correct
24% (01:56) wrong based on 79 sessions

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?

Re: No hours machine M takes to do last friday's work alone [#permalink]
29 Nov 2012, 22:09

1

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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.

Answer should be D _________________

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Re: No hours machine M takes to do last friday's work alone [#permalink]
29 Nov 2012, 23:21

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?

Re: No hours machine M takes to do last friday's work alone [#permalink]
29 Nov 2012, 23:32

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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

Re: No hours machine M takes to do last friday's work alone [#permalink]
30 Nov 2012, 00:31

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: Working alone at its constant rate, machine K took 3 hours [#permalink]
30 Nov 2012, 02:18

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Expert's post

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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.