1955 AHSME Problems/Problem 2

Revision as of 12:33, 4 May 2020 by Fortytwok (talk | contribs)

Problem

The smaller angle between the hands of a clock at $12:25$ p.m. is:

$\textbf{(A)}\ 132^\circ 30'\qquad\textbf{(B)}\ 137^\circ 30'\qquad\textbf{(C)}\ 150^\circ\qquad\textbf{(D)}\ 137^\circ 32'\qquad\textbf{(E)}\ 137^\circ$

Solution

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

Solution 2

Using the formula $\frac{|(60h-11m)|}{2}$, and plugging in 12 for h and 25 for m, we get the angle between to be 222.5°. Since the problem wants the smaller angle, we do $360-222.5 = 137.5 \Rightarrow 137°30'$.

See Also

1955 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions


The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS