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

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

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26 Jul 2008, 08:32
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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

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

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18 Jul 2011, 21:28
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sset009 wrote:
8. Working alone at its constant rate, machine K took 3 hours to produce ¼ 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 unites produced last Friday. How many hours would it have taken machine M, working along at its constant rate, to produce all of the units produced last Friday?
(A) 8
(B) 12
(C) 16
(D) 24
(E) 30

Try using Ratios. It needs some effort in the beginning but is very rewarding. I solve all these questions orally using ratios. I will tell you how.

Working alone at its constant rate, machine K took 3 hours to produce ¼ of the units produced last Friday.

Ok. K takes 3 hrs for 1/4 of units so it takes 12 hrs for all units.

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 unites produced last Friday.

In 6 hrs, they both together produced 3/4 of units. K alone would have taken 9 hrs to produce 3/4 of units (since it takes 3 hrs for 1/4 of units)
Ratio of time taken together : time taken by K = 6:9 = 2:3
Ratio of speed together : speed of K = 3:2
Since speed together is 3 and speed of K is 2, speed of M alone is 1 i.e. speed of K is twice the speed of M. So M will take twice the time taken by K. Since K takes 12 hrs to produce all the units alone, M will take 24 hrs to produce all the units alone.

For more on Ratios, check

http://www.veritasprep.com/blog/2011/03 ... of-ratios/
http://www.veritasprep.com/blog/2011/03 ... os-in-tsd/
http://www.veritasprep.com/blog/2011/03 ... -problems/
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Re: Working alone at its constant rate, machine K took 3 hours to produce [#permalink]

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29 Nov 2012, 21:44
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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

Last edited by Bunuel on 30 Nov 2012, 02:07, edited 1 time in total.
Renamed the topic, edited the question and added the OA.

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

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29 Nov 2012, 22:09
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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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Re: Working alone at its constant rate, machine K took 3 hours to produce [#permalink]

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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?

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

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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

Finally

R * t = W

1/24 * t = 1

there t = 24 hrs

HTH.

Cheers

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

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29 Nov 2012, 23:41
thanks for the explanation jp27.seems ,i was committing a careless mistake before. Hail guys.

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

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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 to produce [#permalink]

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30 Nov 2012, 01:28
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 to produce [#permalink]

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30 Nov 2012, 02:18
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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 to produce [#permalink]

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21 Dec 2015, 15:44
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

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

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09 Sep 2017, 18:03
sset009 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

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