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Two machines

by dbowden » Tue Dec 04, 2012 12:47 pm
Hi,

Can someone help me solve this question?

Working alone at its own constant rate, a machine seals n cartons in 8 hours, and working alone at its own constant rate, a second machine seals n 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 1/3%
(C) 50%
(D) 66 2/3%
(E) 75%
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by Brent@GMATPrepNow » Tue Dec 04, 2012 12:51 pm
dbowden wrote:Hi,

Can someone help me solve this question?

Working alone at its own constant rate, a machine seals n cartons in 8 hours, and working alone at its own constant rate, a second machine seals n 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 1/3%
(C) 50%
(D) 66 2/3%
(E) 75%
One option is to plug in a convenient value for n.

Let's say that n=8.
So, the slower machine seals 8 cartons in 8 hours, which means it seals 1 carton per hour.
This also means that the faster machine seals 8 cartons in 4 hours, which means it seals 2 carton per hour.

Now the question considers a scenario in which the 2 machines, working together, and for the same period of time, sealed a certain number of cartons.

Well, let's say they worked for one hour.
This means they sealed a total of 3 cartons, and the faster machine sealed 2 of those cartons.
So, the faster machine sealed 2/3 ([spoiler]or 66 2/3%[/spoiler]) of the cartons.

So, the answer is D

Cheers,
Brent
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by mcressy » Tue Dec 04, 2012 12:56 pm
Hi,

The best approach to this problem is to think about is to think about the two machines' relative rates. The first machine can complete 1/8 of the task in an hour (if it takes 8 hours to do the job, then the machine can get 1/8 done in one hour), and the second machine can complete ¼ (or 2/8) of the task in the same hour (similar logic as above). We can then combine 2/8+1/8 and figure out that in a single hour, the two machines can get 3/8 of the job done.

So, if the machines together get 3/8 of the job done in an hour, and the second machine accomplishes 2/8 of the job, and the first machine accomplishes 1/8 of the job, then machine b represents 2/3 of the total work done (2/8 divided by 3/8 = 2/3, or 66 2/3%.)


Alternatively, you could consider it from the aspect of getting the jobs done. If both machines work for 8 hours, how many times would the task get done? The answer is 3, the first machine could do it once in 8 hours, the second machine could do it twice in 8 hours (four hours for the first, four more for the second). So if a task the size n could get done 3 times in eight hours, and the second machine does 2 of those 3 times, then the answer is 2/3 (66 2/3%)

Correct answer is D.
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by GMATGuruNY » Tue Dec 04, 2012 6:02 pm
dbowden wrote:Hi,

Can someone help me solve this question?

Working alone at its own constant rate, a machine seals n cartons in 8 hours, and working alone at its own constant rate, a second machine seals n 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 1/3%
(C) 50%
(D) 66 2/3%
(E) 75%
Time and rate are RECIPROCALS.
The TIME RATIO for the two machines = slower time : faster time = 8:4.
Thus, the RATE RATIO for the two machines = slower rate : faster rate = 4:8.
Implication: for every 4 cartons sealed by the slower machine, the faster machine seals 8 cartons.
Thus, of every 12 cartons, 8 are sealed by the faster machine:
8/12 = 2/3 = 66 2/3 %.

The correct answer is D.
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