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At the Scholarly Text Printing Company, each of n printing presses ca

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At the Scholarly Text Printing Company, each of n printing presses ca  [#permalink]

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New post 04 Feb 2018, 20:34
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Question Stats:

54% (02:08) correct 46% (02:12) wrong based on 36 sessions

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At the Scholarly Text Printing Company, each of n printing presses can produce on the average t books every m minutes. If all presses work without interruption, how many hours will be required to produce a run of 10,000 books?

(A) 10,000*60mn/t
(B) 10,000*60tm/n
(C) 10,000mn/(60t)
(D) 10,000m/(60nt)
(E) 10,000/(60mnt)

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At the Scholarly Text Printing Company, each of n printing presses ca  [#permalink]

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New post 05 Feb 2018, 02:49
Bunuel wrote:
At the Scholarly Text Printing Company, each of n printing presses can produce on the average t books every m minutes. If all presses work without interruption, how many hours will be required to produce a run of 10,000 books?

(A) 10,000*60mn/t
(B) 10,000*60tm/n
(C) 10,000mn/(60t)
(D) 10,000m/(60nt)
(E) 10,000/(60mnt)


Algebraic Method

Since 1 printing press produces t books in m minutes, 1 printing press produces \(\frac{t}{m}\) books in a minute.

In an hour, the printing press will produce \(\frac{60t}{m}\) books. 'n' printing presses would produce \(\frac{60nt}{m}\) books per hour.
In order to product 10000 books, the presses will need \(\frac{10000m}{60nt}\) hours (Option D)


Assuming numbers

1 printing press will produce (t = 10) books in (m = 6) minutes.
Therefore, 1 printing press produces \(\frac{5}{3}\) books in a minute.
In an hour, the printing presses will produce \(60(\frac{5}{3}) = 100\) books.
If there are (n=10) machines to do the work, every hour 1000 books are printed

In order to print 10000 books, it took the n printing presses \(\frac{10000}{1000} = 10\) hours.

Substituting the values of t,m,and n in the answer options,

(A) 10,000*60mn/t = 10000*60*6*10/10 = 3600000 hours
(B) 10,000*60tm/n = 10000*60*10*6/10 = 3600000 hours
(C) 10,000mn/(60t) = 10000*10*10/60*6 = 100000/36 hours

(D) 10,000m/(60nt) = 10000*6/60*10*10 = 10 hours
(E) 10,000/(60mnt) = 10000/60*10*6*10 = 10000/36000 hours (Option D)
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Re: At the Scholarly Text Printing Company, each of n printing presses ca  [#permalink]

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New post 05 Feb 2018, 20:06
Bunuel wrote:
At the Scholarly Text Printing Company, each of n printing presses can produce on the average t books every m minutes. If all presses work without interruption, how many hours will be required to produce a run of 10,000 books?

(A) 10,000*60mn/t
(B) 10,000*60tm/n
(C) 10,000mn/(60t)
(D) 10,000m/(60nt)
(E) 10,000/(60mnt)


It should be D
Each press produces \(t\) books every \(m\) minutes --> \(\frac{t}{m}\) = \(\frac{t}{m}*60 \frac{minutes}{hour}\) = \(\frac{60t}{m}\) books per hour

\(n\) presses would make \(\frac{n*60t}{m}\) books per hour.

\(Rate * Time = Work\)
\(Time = Work/Rate\)
\(Time = \frac{10000}{60nt/m}\)
\(Time = 10000m/60nt\)
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Re: At the Scholarly Text Printing Company, each of n printing presses ca   [#permalink] 05 Feb 2018, 20:06
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