1962 AHSME Problems/Problem 35

Problem

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:

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

Solution

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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1962 AHSME Problems/Problem 35

Problem

Solution