1962 AHSME Problems/Problem 35


A man on his way to dinner short after $6: 00$ p.m. observes that the hands of his watch form an angle of $110^{\circ}$. Returning before $7: 00$ p.m. he notices that again the hands of his watch form an angle of $110^{\circ}$. The number of minutes that he has been away is:

$\textbf{(A)}\ 36\frac{2}3\qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 42.4\qquad\textbf{(E)}\ 45$


Let $n$ be the number of minutes after 6:00. Let $h(n)=180+\frac{n}2$ be the angle, in degrees, of the hour hand (with $0^{\circ}$ at the top and increasing in the clockwise direction); similarly, let $m(n)=6n$ be the angle of the minute hand. We want $|h(n)-m(n)|=110$. This is equivalent to \[180-\frac{11n}2=\pm110\] \[-\frac{11n}2\in\{-70,-290\}\] \[\frac{11n}2\in\{70,290\}\] \[11n\in\{140,580\}\] \[n\in\{\frac{140}{11},\frac{580}{11}\}\] The difference between the two values of $n$ is $\frac{440}{11}=\boxed{40\textbf{ (B)}}$.

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