Machine work

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Machine work

by mysseo » Mon Jan 02, 2012 10:46 pm
Working alone at its own constant rate, a machine seals k cartons in 8 hours, and working alone at its own constant rate, a second machine seals k cartons in 4 hours. If the two machines, each working at its own constant rate and for the same period of time, together sealed a certain number of cartons, what percent of the cartons were sealed by the machine working at the faster rate?

A. 25%
B. 33 and 1/3 %
C. 50%
D. 66 and 2/3 %
E. 75%

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by Anurag@Gurome » Mon Jan 02, 2012 11:04 pm
mysseo wrote:Working alone at its own constant rate, a machine seals k cartons in 8 hours, and working alone at its own constant rate, a second machine seals k cartons in 4 hours. If the two machines, each working at its own constant rate and for the same period of time, together sealed a certain number of cartons, what percent of the cartons were sealed by the machine working at the faster rate?

A. 25%
B. 33 and 1/3 %
C. 50%
D. 66 and 2/3 %
E. 75%
1st machine seals k cartons in 8 hours.
2nd machine seals k cartons in 4 hours OR 2k cartons in 8 hours.
If both the machines are working for the same time, say, 8 hours, then together they seal 3k cartons
2nd machine is the faster machine of the two machines.
So, 2nd machine seals 2k out of 3k cartons, that is it seals 2/3 = 66.66% of the total cartons

The correct answer is D.
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by LalaB » Tue Jan 03, 2012 9:07 am
(1/4) *(8*4/(8+4))=2/3

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by ArunangsuSahu » Tue Jan 03, 2012 7:38 pm
See my Calculation

1/4/ 1/4+1/8=2/3=66 and 2/3%

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by GMATGuruNY » Tue Jan 03, 2012 8:23 pm
mysseo wrote:Working alone at its own constant rate, a machine seals k cartons in 8 hours, and working alone at its own constant rate, a second machine seals k cartons in 4 hours. If the two machines, each working at its own constant rate and for the same period of time, together sealed a certain number of cartons, what percent of the cartons were sealed by the machine working at the faster rate?

A. 25%
B. 33 and 1/3 %
C. 50%
D. 66 and 2/3 %
E. 75%
Let k = 8 cartons.
Rate for the faster machine = w/t = 8/4 = 2 cartons per hour.
Rate for the slower machine = w/t = 8/8 = 1 carton per hour.
Each hour:
(cartons produced by the faster machine)/(total cartons produced) = 2/(2+1) = 2/3 = 66 2/3%.

The correct answer is D.
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by LeoBen » Wed Jan 04, 2012 6:22 am
faster m/c rate = k/4

slower m/c rate = k/8

slower on the critical path, with total time, faster would be 8/12 x 100 % = 66.66 %