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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] New post 26 Jul 2008, 08:32
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




-------------------
to do this, i went by this methodology
K takes 3 hours to produce 1/4 units, therefore, K would take 12 hours to produce all the total number of units.

when M starts working, amount of wokr left = 0.75. denoting x as the time M would take to produce all the units,
therefore,
0.75 ( 1/12 + 1/x) = 1/6
1/12 + 1/x = 2/9
1/x = 5/36

x = 36/5
which is not even one of the answers.
:cry:
help!!!!
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Re: rates question [#permalink] New post 26 Jul 2008, 08:48
Working Alone machine K takes = 4*3 = 12 Hours

3/4 M+K takes 6 hours
Complete time M+K takes = 6*4/3 = 8 hours

Machine M rate per hour = 1/8 - 1/12 (Total minus machine k) = 1/24

Total time taken by machine M = 24 hours

Proof of Answer (In 6 hours M finishes) = 1/24*6 = 1/4 work
In 6 hours k finishes = 1/12*6 = 1/2 work
Total work M+K finish in 6 hours = 1/2+1/4 = 3/4

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Re: rates question [#permalink] New post 18 Jul 2011, 03:24
Let total unit be 24.
For K
In 3 hours K produce 1/4 of total units.
K's rate => 3 hours =6 units
= 2units/hour

For M
Let M's rate be m units/hour.

Together, for K and M rate = 2+m units/hour.

i.e Together K and M finish 2+m units in 1 hour
Together K and M finish 18(6 units already done by K alone) units in = 1/2+m * 18 = 6 hours
solving m=1

Hence M's rate is 1 unit in 1 hour
Therefor it will take M 24 hours to do all the units (24).

OA D.
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Re: rates question [#permalink] New post 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: rates question   [#permalink] 18 Jul 2011, 21:28
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