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GMAT PREP PS Problem

Expert replies
by alex.gellatly » Sat Apr 21, 2012 4:05 am
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 1/3%
(C) 50%
(D) 66 2/3%
(E) 75%

Thanks
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Source: — Problem Solving |

by Birottam Dutta » Sat Apr 21, 2012 5:45 am
In 8 hours, the faster machine will seal 2k cartons and the slower machine will seal k cartons

Total sealed = 3k.

Percentage sealed by the faster machine = 2k/3k = 66 2/3%.

Hence, D!
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by aneesh.kg » Sat Apr 21, 2012 6:33 am
In any period of time, the second machine seals double the number cartons than the first machine.
If, in a certain period of time, the first machine seals 'x' number of machines, the second machine will seal '2x' number of machines.

Percent of cartons sealed by the second machine = (Number of cartons sealed by the second machine)/ (Total number of cartons sealed)
= (2x)/(2x + x)]*100
= 66 2/3 %

[D] is the answer
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by Brent@GMATPrepNow » Sat Apr 21, 2012 10:02 am
alex.gellatly 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 1/3%
(C) 50%
(D) 66 2/3%
(E) 75%

Thanks
We can also solve this question by plugging in a convenient value for k.

Let's say that k=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 2 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 (66 2/3%) of the cartons.

So, the answer is D

Cheers,
Brent
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by Anurag@Gurome » Sun Apr 22, 2012 5:50 pm
alex.gellatly 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 1/3%
(C) 50%
(D) 66 2/3%
(E) 75%

Thanks

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 = [spoiler]66.66%[/spoiler] of the total cartons

The correct answer is D.
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hihi

by Jeff@TargetTestPrep » Tue Apr 10, 2018 10:18 am
alex.gellatly 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 1/3%
(C) 50%
(D) 66 2/3%
(E) 75%
Since the second machine can seal k cartons in 4 hours, it can seal 2k cartons in 8 hours, whereas the first machine can seal only k cartons in 8 hours. So the second machine is twice as fast as the first machine and thus it will seal twice as many cartons as the the first machine. Therefore, when working together, the second machine (the faster machine) will finish 2/3 or 66 2/3% of the work whereas the first machine (the slower machine) will finish 1/3 or 33 1/3% of the work.

Alternate Solution:

In 8 hours, the slower machine will seal k cartons, and the faster machine will seal 2k cartons. In total, k + 2k = 3k cartons will be sealed, and, of these cartons, the faster machine will seal 2k/3k = 2/3 = 66 2/3% of them.

Answer: D

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