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

Question

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

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

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

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

D)  4ac + 3bc

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

CAMANISHPARMAR · Answer

Drawing a table to document the information provided in the question will be very useful as follows:-

Attachment:

Kp soln1.jpg [ 29.83 KiB | Viewed 5259 times ]

The rates are additive, hence:-

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

Solve for x to get

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

GMATGuruNY · Answer

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.

B

fskilnik · Answer

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

P\,\,\, - \,\,\,3\,\,{	ext{h}}\,\,\, - \,\,\,a\,\,{	ext{widgets}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\,P\,\,\, - \,\,\,12\,\,{	ext{h}}\,\,\, - \,\,\,4a\,\,{	ext{widgets}}

Q\,\,\, - \,\,\,4\,\,{	ext{h}}\,\,\, - \,\,\,b\,\,{	ext{widgets}}\,\,\,\,\, \Rightarrow \,\,\,\,\,Q\,\,\, - \,\,\,12\,\,{	ext{h}}\,\,\, - \,\,\,3b\,\,{	ext{widgets}}

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

{	ext{?}}\,\,\,\,{	ext{ = }}\,\,\,\,{	ext{c}}\,\,{	ext{h}}\,\,\,\left( {\frac{{4a + 3b\,\,\,{	ext{widgets}}}}{{12\,\,\,{	ext{h}}}}\,\,\begin{array}{*{20}{c}}
 
earrow \ 
 
earrow 
\end{array}\,\,} ight)\,\,\,\,\, = \,\,\,\,\,\,\frac{{\left( {4a + 3b} ight)c}}{{12}}\,\,\,\,\left

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.

bumpbot · Answer

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