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Geometry Difficulty 3.1 AMC 10/12 Find the answer

The smaller angle between the hands of a clock at 12:2512:25 p.m. is:
(A) 132∘30′\textbf{(A)}\ 132^\circ 30'(B) 137∘30′\textbf{(B)}\ 137^\circ 30'(C) 150∘\textbf{(C)}\ 150^\circ(D) 137∘32′\textbf{(D)}\ 137^\circ 32'(E) 137∘\textbf{(E)}\ 137^\circ

Multiple choice: answer with the letter of the option you want.

Solution

At 12:2512:25, the minute hand is at 2560⋅360∘\frac{25}{60}\cdot 360^\circ, or 150∘150^\circ.
The hour hand moves 55 'minutes' every hour, or 560⋅360=30∘\frac{5}{60}\cdot 360 = 30^\circ every hour. At 30∘30^\circ every hour, the hour hand moves 12\frac{1}{2} minutes on the clock every minute. At 12:2512:25, the hour hand is at 252∘\frac{25}{2}^\circ. Therefore, the angle between the hands is 150∘−252∘150^\circ - \frac{25}{2}^\circ, 137.5∘137.5^\circ, or 137∘30′  ⟹  137^{\circ} 30^{'} \implies (B)137∘30′\boxed{\mathrm{(B) 137^\circ 30'}}.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.