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Twelve workers pack boxes at a constant rate of 60 boxes in

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Twelve workers pack boxes at a constant rate of 60 boxes in [#permalink] New post 18 Jun 2018, 14:50
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Question Stats:

84% (01:52) correct 15% (03:21) wrong based on 44 sessions
Twelve workers pack boxes at a constant rate of 60 boxes in 9 minutes. How many minutes would it take 27 workers to pack 180 boxes, if all workers pack boxes at the same constant rate?

(A) 12
(B) 13
(C) 14
(D) 15
(E) 16
[Reveal] Spoiler: OA

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Re: Twelve workers pack boxes at a constant rate of 60 boxes in [#permalink] New post 01 Jul 2018, 23:15
sandy wrote:
Twelve workers pack boxes at a constant rate of 60 boxes in 9 minutes. How many minutes would it take 27 workers to pack 180 boxes, if all workers pack boxes at the same constant rate?

(A) 12
(B) 13
(C) 14
(D) 15
(E) 16



Here,

We know Work = No. of individual * Rate * Time

so 60 = 12 * Rate * 9

or Rate =\(\frac{60}{(12 * 9)}\) = \(\frac{5}{9}\)

Now for 180 boxes ;

Work = No. of individual * Rate * Time

\(180 = \frac{5}{9} * 27 * Time\)

or Time = \(\frac{(180 * 9)}{(5 * 27)}\) = 12
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Retired Moderator
User avatar
Joined: 07 Jun 2014
Posts: 4803
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
Followers: 171

Kudos [?]: 2912 [0], given: 394

Re: Twelve workers pack boxes at a constant rate of 60 boxes in [#permalink] New post 07 Jul 2018, 04:55
Expert's post
Explanation

To solve a Rates & Work problem with multiple workers, modify the standard formula Work = Rate × Time to this:

Work = Individual Rate × Number of Workers × Time

Use the first sentence to solve for an individual worker’s rate. Plug in the fact that 12 workers pack boxes at a constant rate of 60 boxes in 9 minutes:
Work = Individual Rate × Number of Workers × Time

60 boxes = (R)(12)(9 minutes)

\(R =\frac{5}{9}\) boxes per minute.


In other words, each worker can pack \(\frac{5}{9}\) of a box per minute. Plug that rate back into the formula, but use the details from the second sentence in the problem:

Work = Individual Rate × Number of Workers × Time

\(180 = \frac{5}{9} \times 27 \times T\)

\(180=15T\)

\(T=12\).
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Re: Twelve workers pack boxes at a constant rate of 60 boxes in [#permalink] New post 09 Jul 2018, 09:02
sandy wrote:
Twelve workers pack boxes at a constant rate of 60 boxes in 9 minutes. How many minutes would it take 27 workers to pack 180 boxes, if all workers pack boxes at the same constant rate?

(A) 12
(B) 13
(C) 14
(D) 15
(E) 16


Use the following equation:

\(\frac{workers * time}{output} = \frac{workers*time}{output}\)

In the equation above:
Workers and time are INVERSELY PROPORTIONAL.
As the number of workers increases, the amount of time required to produce the same output decreases.
Workers and output are DIRECTLY PROPORTIONAL.
As the number of workers increases, the amount of output also increases.
Time and output are also DIRECTLY PROPORTIONAL.
As the amount of time increases, the amount of output also increases.

In the problem above:
Since 12 workers take 9 minutes to pack 60 boxes, and we want to know the time required for 27 workers to pack 180 boxes, we get:
\(\frac{12*9}{60} = \frac{27t}{180}\)

\(\frac{9}{5} = \frac{3t}{20}\)

\(15t = 180\)

\(t = 12\)

[Reveal] Spoiler:
A

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Re: Twelve workers pack boxes at a constant rate of 60 boxes in [#permalink] New post 09 Jul 2019, 06:28
sandy wrote:
Twelve workers pack boxes at a constant rate of 60 boxes in 9 minutes. How many minutes would it take 27 workers to pack 180 boxes, if all workers pack boxes at the same constant rate?

(A) 12
(B) 13
(C) 14
(D) 15
(E) 16


hard type question
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Re: Twelve workers pack boxes at a constant rate of 60 boxes in   [#permalink] 09 Jul 2019, 06:28
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