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Difference between revisions of "1962 AHSME Problems/Problem 35"

(Created page with "==Problem== A man on his way to dinner short after <math>6: 00</math> p.m. observes that the hands of his watch form an angle of <math>110^{\circ}</math>. Returning before <math>...")
 
 
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==Solution==
 
==Solution==
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"Unsolved"
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Let <math>n</math> be the number of minutes after 6:00. Let <math>h(n)=180+\frac{n}2</math> be the angle, in degrees, of the hour hand (with <math>0^{\circ}</math> at the top and increasing in the clockwise direction); similarly, let <math>m(n)=6n</math> be the angle of the minute hand. We want <math>|h(n)-m(n)|=110</math>. This is equivalent to
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<cmath>180-\frac{11n}2=\pm110</cmath>
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<cmath>-\frac{11n}2\in\{-70,-290\}</cmath>
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<cmath>\frac{11n}2\in\{70,290\}</cmath>
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<cmath>11n\in\{140,580\}</cmath>
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<cmath>n\in\{\frac{140}{11},\frac{580}{11}\}</cmath>
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The difference between the two values of <math>n</math> is <math>\frac{440}{11}=\boxed{40\textbf{ (B)}}</math>.

Latest revision as of 17:22, 17 April 2014

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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