M27-11

Official Solution:


Bonnie can paint a stolen car in \(x\) hours, and Clyde can paint the same car in \(y\) hours. They start working simultaneously and independently at their respective constant rates at 9:45 am. If both \(x\) and \(y\) are odd integers, does \(x\) equal \(y\)?

When Bonnie and Clyde work together, they complete the painting of the car in \(\frac{xy}{x+y}\) hours (the sum of the rates equals the combined rate, which is the reciprocal of the total time: \(\frac{1}{x}+\frac{1}{y}=\frac{1}{T}\). From this, we can deduce that \(T=\frac{xy}{x+y}\)). Now, if \(x=y\), then the total time would be: \(\frac{x^2}{2x}=\frac{x}{2}\). Since \(x\) is odd, this time would be \(\frac{odd}{2}\): 0.5 hours, 1.5 hours, 2.5 hours, ...

(1) \(x^2+y^2 \lt 12\).

It's possible that \(x\) and \(y\) are odd and equal to each other if \(x=y=1\), but it's also possible that \(x=1\) and \(y=3\) (or vice versa). This information is not sufficient.

(2) Bonnie and Clyde finish painting the car together at 10:30 am.

They complete the job in \(\frac{3}{4}\) of an hour (45 minutes). Since this time is not \(\frac{odd}{2}\), \(x\) and \(y\) cannot be equal. This information is sufficient.


Answer: B