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

Latest revision as of 16:22, 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 $\frac{440}{11}=\boxed{40\textbf{ (B)}}$ .

@@ Line 5: / Line 5: @@
 ==Solution==
-{{solution}}
+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
+<cmath>180-\frac{11n}2=\pm110</cmath>
+<cmath>-\frac{11n}2\in\{-70,-290\}</cmath>
+<cmath>\frac{11n}2\in\{70,290\}</cmath>
+<cmath>11n\in\{140,580\}</cmath>
+<cmath>n\in\{\frac{140}{11},\frac{580}{11}\}</cmath>
+The difference between the two values of <math>n</math> is <math>\frac{440}{11}=\boxed{40\textbf{ (B)}}</math>.

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

Latest revision as of 16:22, 17 April 2014

Problem

Solution