John’s clock is broken. The minute hand rotates around the clock correctly, but the hour hand is stuck in the three o’clock position. Suppose John first looks at the clock when the hands are aligned and it shows 3:15. He looks at the clock again and sees that the hour and minute hands of the clock form the arms of an angle measuring –135°. How many degrees could the minute hand have rotated to reach its current position?

Question

A. –45°

B. –315°

C. –405°

D. –495°

Answer 1

Since the minute hand rotates normally, we can ignore it for now and just focus on the hour hand. If the hour hand is stuck in the three o'clock position, then it is pointing directly at the 3 on the clock face. Since the hands were aligned at 3:15, that means the minute hand has moved $\frac{1}{4}$ of the way around the clock face, which is 90 degrees.

Now, to find the angle between the hour hand and the minute hand, we can use the formula $|30h-11m|$, where $h$ is the hour and $m$ is the minute. At the time when the hands are aligned and it shows 3:15, we have $h=3$ and $m=15$, so the angle between the hands is $|30(3)-11(15)|=|90-165|=75$ degrees.

Since we know that the minute hand has moved 90 degrees from its position at 3:00, and the angle between the hands is now -135 degrees (meaning the hour hand is "behind" the minute hand), we know that the minute hand has rotated an additional $90-(-135)=225$ degrees.

Therefore, the minute hand could have rotated a total of $90+225=315$ degrees. However, the question asks for the number of degrees the minute hand could have rotated from its position at 3:15, which is $315-90=\boxed{\textbf{(B)}\,-315}$ degrees.