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At the neginning of a certain job, the counter of a...

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by BTGmoderatorLU » Tue Feb 20, 2018 5:45 am
At the beginning of a certain job, the counter of a photocopy machine read 1254. At the end of the job, the counter read 2334. If the running time for the job was 30 minutes, then what was the average operating speed of the machine in copies per second?

A) .6
B) 1.1
C) 6
D) 36
E) 2160

The OA is A.

Experts, any suggestion?

The total of copies during the 30 minutes will be, 2334 - 1254 = 1080. Now I just need to convert the 30 minutes to seconds and it will be 1800 seconds.

Finally, the rate of the machine in those 30 minutes will be, 1080 / 1800 = 0.6 copies per second, right? Thanks in advance.
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by DavidG@VeritasPrep » Tue Feb 20, 2018 9:04 am
LUANDATO wrote:At the beginning of a certain job, the counter of a photocopy machine read 1254. At the end of the job, the counter read 2334. If the running time for the job was 30 minutes, then what was the average operating speed of the machine in copies per second?

A) .6
B) 1.1
C) 6
D) 36
E) 2160

The OA is A.

Experts, any suggestion?

The total of copies during the 30 minutes will be, 2334 - 1254 = 1080. Now I just need to convert the 30 minutes to seconds and it will be 1800 seconds.

Finally, the rate of the machine in those 30 minutes will be, 1080 / 1800 = 0.6 copies per second, right? Thanks in advance.
Sure. Just note that once you have 1080/1800, you know the answer is less than 1. Only A will work. No need to do any more arithmetic than necessary.
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by Jeff@TargetTestPrep » Thu Feb 22, 2018 8:33 am
LUANDATO wrote:At the beginning of a certain job, the counter of a photocopy machine read 1254. At the end of the job, the counter read 2334. If the running time for the job was 30 minutes, then what was the average operating speed of the machine in copies per second?

A) .6
B) 1.1
C) 6
D) 36
E) 2160
Since 30 minutes is 30 x 60 = 1800 seconds, the rate in seconds is:

(2334 - 1254)/1800 = 1,080/1800 = 108/180 = 12/20 = 3/5 = 0.6

Answer: A

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