Machine A and Machine B can produce 1 widget in 3 hours working together at their respective constant rates. If Machine A's speed were doubled, the two machines could produce 1 widget in 2 hours working together at their respective rates. How many hours does it currently take Machine A to produce 1 widget on its own?
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Correct answer: E
Let \(a\) be the number of hours it takes Machine A to produce \(1\) widget on its own. Let \(b\) be the number of hours it takes Machine B to produce \(1\) widget on its own.<br><br>
The question tells us that Machines A and B together can produce \(1\) widget in \(3\) hours. Therefore, in \(1\) hour, the two machines can produce \(\frac{1}{3}\) of a widget.<br><br>
In \(1\) hour, Machine A can produce \(\frac{1}{a}\) widgets and Machine B can produce \(\frac{1}{b}\) widgets. Together in \(1\) hour, they produce \(\frac{1}{a} + \frac{1}{b} = \frac{1}{3}\) widgets.<br><br>
If Machine A's speed were doubled it would take the two machines \(2\) hours to produce \(1\) widget. When one doubles the speed, one cuts the amount of time it takes in half. Therefore, the amount of time it would take Machine A to produce \(1\) widget would be \(\frac{a}{2}\). Under these new conditions, in \(1\) hour Machine A and B could produce \(\frac{1}{\frac{a}{2}} + \frac{1}{b} = \frac{1}{2}\) widgets.<br><br>
We now have two unknowns and two different equations. We can solve for \(a\).<br><br>
The two equations:<br>
\(\frac{2}{a} + \frac{1}{b} = \frac{1}{2}\) (Remember, \(\frac{1}{\frac{a}{2}} = \frac{2}{a}\))<br>
\(\frac{1}{a} + \frac{1}{b} = \frac{1}{3}\)<br><br>
Subtract the bottom equation from the top:<br>
\(\frac{2}{a} - \frac{1}{a} = \frac{1}{2} - \frac{1}{3}\)<br>
\(\frac{1}{a} = \frac{3}{6} - \frac{2}{6}\)<br>
\(\frac{1}{a} = \frac{1}{6}\)<br><br>
Therefore, \(a = 6\).<br><br>
The correct answer is E.