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Working alone at a constant rate, machine P produces a widgets in 3...

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Working alone at a constant rate, machine P produces a widgets in 3...  [#permalink]

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New post 08 Jun 2018, 08:38
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Working alone at a constant rate, machine P produces \(a\) widgets in 3 hours. Working alone at a constant rate, machine Q produces \(b\) widgets in 4 hours. If machines P and Q work together for \(c\) hours, then in terms of \(a, b,\) and \(c\) how many widgets will machines P and Q produce?

A) \(\frac{3ac + 4bc}{12}\)

B) \(\frac{4ac + 3bc}{12}\)

C) \(\frac{4ac + 3bc}{6}\)

D) \(4ac + 3bc\)

E) \(\frac{ac + 2bc}{4}\)

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Re: Working alone at a constant rate, machine P produces a widgets in 3...  [#permalink]

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New post 08 Jun 2018, 08:55
1
Drawing a table to document the information provided in the question will be very useful as follows:-

Attachment:
Kp soln1.jpg
Kp soln1.jpg [ 29.83 KiB | Viewed 621 times ]


The rates are additive, hence:-

\(\frac{a}{3} + \frac{b}{4} = \frac{x}{c}\)

Solve for x to get

x = \(\frac{4ac + 3bc}{12}\)
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Re: Working alone at a constant rate, machine P produces a widgets in 3...  [#permalink]

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New post 08 Jun 2018, 10:36
CAMANISHPARMAR wrote:
Working alone at a constant rate, machine P produces \(a\) widgets in 3 hours. Working alone at a constant rate, machine Q produces \(b\) widgets in 4 hours. If machines P and Q work together for \(c\) hours, then in terms of \(a, b,\) and \(c\) how many widgets will machines P and Q produce?

A) \(\frac{3ac + 4bc}{12}\)

B) \(\frac{4ac + 3bc}{12}\)

C) \(\frac{4ac + 3bc}{6}\)

D) \(4ac + 3bc\)

E) \(\frac{ac + 2bc}{4}\)


Let a=3, implying that A produces 3 widgets every 3 hours, the equivalent of 1 widget per hour.
Let b=4, implying that B produces 4 widgets every 4 hours, the equivalent of 1 widget per hour.
Let c=1 hour.
Since A and B each produce 1 widget per hour, the number of widgets produced when they work together for 1 hour = 1+1 = 2.

Now plug a=3, b=4 and c=1 into the answer choices to see while yields a production of 2 widgets.
Only B works:
\(\frac{4ac + 3bc}{12}\) = \(\frac{(4*3*1) + (3*4*1)}{12}\) = 2.


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Re: Working alone at a constant rate, machine P produces a widgets in 3...  [#permalink]

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New post 23 Sep 2018, 07:49
CAMANISHPARMAR wrote:
Working alone at a constant rate, machine P produces \(a\) widgets in 3 hours. Working alone at a constant rate, machine Q produces \(b\) widgets in 4 hours. If machines P and Q work together for \(c\) hours, then in terms of \(a, b,\) and \(c\) how many widgets will machines P and Q produce?

A) \(\frac{3ac + 4bc}{12}\)

B) \(\frac{4ac + 3bc}{12}\)

C) \(\frac{4ac + 3bc}{6}\)

D) \(4ac + 3bc\)

E) \(\frac{ac + 2bc}{4}\)

\(P \cup Q\,\,\, - \,\,\,c\,\,{\text{h}}\,\,\, - \,\,\,\boxed{? = f\left( {a,b,c} \right)}\,\,\,{\text{widgets}}\)


\(P\,\,\, - \,\,\,3\,\,{\text{h}}\,\,\, - \,\,\,a\,\,{\text{widgets}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\,P\,\,\, - \,\,\,12\,\,{\text{h}}\,\,\, - \,\,\,4a\,\,{\text{widgets}}\)

\(Q\,\,\, - \,\,\,4\,\,{\text{h}}\,\,\, - \,\,\,b\,\,{\text{widgets}}\,\,\,\,\, \Rightarrow \,\,\,\,\,Q\,\,\, - \,\,\,12\,\,{\text{h}}\,\,\, - \,\,\,3b\,\,{\text{widgets}}\)


Now let´s apply the UNITS CONTROL, one of the most powerful tools of our method!

\({\text{?}}\,\,\,\,{\text{ = }}\,\,\,\,{\text{c}}\,\,{\text{h}}\,\,\,\left( {\frac{{4a + 3b\,\,\,{\text{widgets}}}}{{12\,\,\,{\text{h}}}}\,\,\begin{array}{*{20}{c}}
\nearrow \\
\nearrow
\end{array}\,\,} \right)\,\,\,\,\, = \,\,\,\,\,\,\frac{{\left( {4a + 3b} \right)c}}{{12}}\,\,\,\,\left[ {{\text{widgets}}} \right]\)

Obs.: arrows indicate licit converter.


The right answer is therefore (B).


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: Working alone at a constant rate, machine P produces a widgets in 3... &nbs [#permalink] 23 Sep 2018, 07:49
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