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At the beginning of a certain job, the counter on a photocopy machine

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At the beginning of a certain job, the counter on a photocopy machine  [#permalink]

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New post 19 Feb 2018, 21:59
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Question Stats:

77% (01:22) correct 23% (01:25) wrong based on 58 sessions

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At the beginning of a certain job, the counter on 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

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Re: At the beginning of a certain job, the counter on a photocopy machine  [#permalink]

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New post 19 Feb 2018, 22:11
No of copies read by the machine during the operation= 2334-1254= 1080

Running time= 30*60= 1800 secs

Average operating speed= \(\frac{1080}{1800}\)= 0.6

Answer: A.
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At the beginning of a certain job, the counter on a photocopy machine  [#permalink]

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New post 20 Feb 2018, 10:36
Bunuel wrote:
At the beginning of a certain job, the counter on 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

Number of copies made: 2,334 - 1,254 = 1,080
Time: 30 minutes

Copies per second?
\(\frac{1080c}{30mins}=\frac{36c}{1min}\)

\(\frac{36c}{1min}*\frac{1min}{60secs}=\frac{36c}{60secs}\)

\(\frac{36c}{60secs}=\frac{3c}{5secs}\)

\(\frac{3c}{5secs}=\frac{c?}{1sec}\)

Denominator goes from 5 to 1; 5 is divided by 5. Do the same to numerator; divide 3 by 5. Number of copies per second =
\(\frac{\frac{3}{5}c}{1sec}=\frac{.6c}{1sec}\)

Answer A
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At the beginning of a certain job, the counter on a photocopy machine &nbs [#permalink] 20 Feb 2018, 10:36
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At the beginning of a certain job, the counter on a photocopy machine

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