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If all machines in a factory are equally efficient and 10 machines

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If all machines in a factory are equally efficient and 10 machines  [#permalink]

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New post 22 Aug 2018, 10:21
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If all machines in a factory are equally efficient and 10 machines take 8 seconds to produce 40 cans, how long will it take 12 machines to produce 600 cans?

A- 25 seconds
B- 50 seconds
C- 1 minute and 40 seconds
D- 2 minutes
E- 3 minutes and 20 seconds

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Re: If all machines in a factory are equally efficient and 10 machines  [#permalink]

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New post 22 Aug 2018, 13:10
rohan2345 wrote:
If all machines in a factory are equally efficient and 10 machines take 8 seconds to produce 40 cans, how long will it take 12 machines to produce 600 cans?

A- 25 seconds
B- 50 seconds
C- 1 minute and 40 seconds
D- 2 minutes
E- 3 minutes and 20 seconds



\(\frac{M1*D1*H1}{W1}\) = \(\frac{M2*D2*H2}{W2}\)

substituting the values

\(\frac{10*8}{40}\)= \(\frac{12*t}{600}\)
t=100 seconds=1 minute 40 seconds
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Re: If all machines in a factory are equally efficient and 10 machines  [#permalink]

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New post 31 Aug 2018, 07:25
rohan2345 wrote:
If all machines in a factory are equally efficient and 10 machines take 8 seconds to produce 40 cans, how long will it take 12 machines to produce 600 cans?

A- 25 seconds
B- 50 seconds
C- 1 minute and 40 seconds
D- 2 minutes
E- 3 minutes and 20 seconds


Excellent opportunity to use UNITS CONTROL, one of our most powerful tools!

(The solution may seem excessively "sophisticated", but it works in much hard/high-level question scenarios!)

\(\boxed{10\,\,mach}\,\,\,\,\, \to \,\,\,\,\frac{{40\,\,{\text{cans}}}}{{8\,\,{\text{s}}}} = \boxed{\frac{{5\,\,{\text{cans}}}}{{1\,\,{\text{s}}}}\,\,\,\begin{array}{*{20}{c}}
\nearrow \\
\nearrow
\end{array}}\)

\(12\,\,mach\,\,,\,\,\,600\,\,cans\,\,\,\,\, \to \,\,\,\,?\,\,\,:\,\,\,{\text{time}}\,\)

\(12\,\,mach\,\,\,\, = \,\,\,\boxed{10\,\,mach}\,\,\left( {\frac{{12}}{{10}}} \right)\,\,\,\, \to \,\,\,\,\,\boxed{\frac{{5\,\,{\text{cans}}}}{{1\,\,{\text{s}}}}}\,\,\,\frac{{12}}{{10}}\,\,\,\,\,\, \to \,\,\,\,\,\frac{{6\,\,\,{\text{cans}}}}{{1\,\,\,{\text{s}}}}\)

\(? = 600\,\,{\text{cans}}\,\,\,\left( {\frac{{1\,\,s}}{{6\,\,{\text{cans}}}}\,\,\begin{array}{*{20}{c}}
\nearrow \\
\nearrow
\end{array}} \right)\,\,\, = \,\,\,100\,\,{\text{s}}\)

Obs.: arrows indicate licit converters.

The above follows the notations and rationale taught in the GMATH method.
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Re: If all machines in a factory are equally efficient and 10 machines  [#permalink]

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New post 31 Aug 2018, 10:00
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rohan2345 wrote:
If all machines in a factory are equally efficient and 10 machines take 8 seconds to produce 40 cans, how long will it take 12 machines to produce 600 cans?

A- 25 seconds
B- 50 seconds
C- 1 minute and 40 seconds
D- 2 minutes
E- 3 minutes and 20 seconds


We can also use some number sense to get to the correct answer...

GIVEN: 10 machines take 8 seconds to produce 40 cans
So, 10 machines take 1 second to produce 5 cans (if we divide the work time by 8, then the output is also divided by 8)
So, 1 machines take 1 second to produce 0.5 cans (if we divide the number of machines by 10, then the output is also divided by 10)
So, 12 machines take 1 second to produce 6 cans (if we multiply the number of machines by 12, then the output is also multiplied by 12)

We now know that 12 machines take 1 second to produce 6 cans
We want the 12 machines to make 600 cans.
600/6 = 100
So, 12 machines take 100 seconds to produce 600 cans

100 seconds = 1 minute and 40 seconds.

Answer: C
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Re: If all machines in a factory are equally efficient and 10 machines  [#permalink]

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New post 27 Sep 2019, 08:49
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rohan2345 wrote:
If all machines in a factory are equally efficient and 10 machines take 8 seconds to produce 40 cans, how long will it take 12 machines to produce 600 cans?

A- 25 seconds
B- 50 seconds
C- 1 minute and 40 seconds
D- 2 minutes
E- 3 minutes and 20 seconds


Since the rate of 10 machines is 40/8 = 5, the rate for 1 machine is 5/10 = 1/2 and the rate for 12 machines is 1/2 x 12 = 6.

Thus, it takes 12 machines 600/6 = 100 seconds = 1 minute 40 seconds to produce 600 cans.

Answer: C
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Re: If all machines in a factory are equally efficient and 10 machines  [#permalink]

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New post 30 Sep 2019, 07:20

Solution



Given
    • All machines of a factory are equally efficient.
    • 10 machines take 8 seconds to produce 40 cans.

To find
    • Time taken by 12 machines to fill 600 cans.

Approach and Working out

10 Machines take 8 seconds to fill 40 cans.
    • So, each machine is working 8 seconds. And total 80 seconds.
      o So, if 1 machine had been working then it would take 80 seconds to fill 40 cans.
         80 seconds to fill 40 cans.
         So, 1 machine will take 2 seconds to fill 1 can.
      • So, 12 machines in 2 seconds can fill 12 cans.
      o 12 cans -> 2 seconds
      o 12 * 50 cans = 600 cans -> 2 * 50 =100 seconds= 1 min 40 sec

Thus, option C is the correct answer.

Correct Answer: Option C
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Re: If all machines in a factory are equally efficient and 10 machines   [#permalink] 30 Sep 2019, 07:20
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