Difference between revisions of "1954 AHSME Problems/Problem 50"
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The minute hand moves clockwise at a rate of <math>360^{\circ}/\text{hr}</math>, while the hour hand moves at <math>(1/12)\cdot 360^{\circ}/\text{ hr}=30^{\circ}/\text{hr}</math>. Therefore the minute hand catches up to the hour hand at a rate of <math>360^{\circ}-30^{\circ}=330^{\circ}</math> per hour. Therefore it will take <math>(210^{\circ}-84^{\circ})/(330^{\circ}/\text{hr})=\tfrac{126}{330}\text{hr}</math> for the two hands of the clock to make an <math>84^{\circ}</math> angle. It will also take <math>(210^{\circ}+84^{\circ})/(330^{\circ}/\text{hr})=\tfrac{294}{330}\text{hr}</math> after <math>7</math> o'clock for the hands to make an <math>84^{\circ}</math> angle for the second time. Converting these values to minutes, we see that it will take <math>22\tfrac{10}{11}</math> minutes, and <math>53\tfrac{5}{11}</math> minutes, for the two hands to make <math>84^{\circ}</math> angles. Thus the times, correct to the nearest minute, at which the hands of the clock will form an <math>84^{\circ}</math> angle are <math>\boxed{\textbf{(A)}\ \text{7: 23 and 7: 53}}</math>. | The minute hand moves clockwise at a rate of <math>360^{\circ}/\text{hr}</math>, while the hour hand moves at <math>(1/12)\cdot 360^{\circ}/\text{ hr}=30^{\circ}/\text{hr}</math>. Therefore the minute hand catches up to the hour hand at a rate of <math>360^{\circ}-30^{\circ}=330^{\circ}</math> per hour. Therefore it will take <math>(210^{\circ}-84^{\circ})/(330^{\circ}/\text{hr})=\tfrac{126}{330}\text{hr}</math> for the two hands of the clock to make an <math>84^{\circ}</math> angle. It will also take <math>(210^{\circ}+84^{\circ})/(330^{\circ}/\text{hr})=\tfrac{294}{330}\text{hr}</math> after <math>7</math> o'clock for the hands to make an <math>84^{\circ}</math> angle for the second time. Converting these values to minutes, we see that it will take <math>22\tfrac{10}{11}</math> minutes, and <math>53\tfrac{5}{11}</math> minutes, for the two hands to make <math>84^{\circ}</math> angles. Thus the times, correct to the nearest minute, at which the hands of the clock will form an <math>84^{\circ}</math> angle are <math>\boxed{\textbf{(A)}\ \text{7: 23 and 7: 53}}</math>. | ||
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+ | ==Solution 2== | ||
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+ | Using the formula $(60h-11m)/2 | ||
== See Also == | == See Also == |
Revision as of 10:55, 21 January 2021
Contents
Problem
The times between and o'clock, correct to the nearest minute, when the hands of a clock will form an angle of are:
Solution
At o'clock, the hour hand is at the position clockwise from the o'clock position, and the minute hand is exactly at the o'clock position. Thus the minute hand is ahead of the hour hand, while it is also behind the hour hand. So, when the minute hand first makes an angle with the hour hand, the minute hand will be behind the hour hand, and the second time they make and angle, the minute hand will be ahead the hour hand.
The minute hand moves clockwise at a rate of , while the hour hand moves at . Therefore the minute hand catches up to the hour hand at a rate of per hour. Therefore it will take for the two hands of the clock to make an angle. It will also take after o'clock for the hands to make an angle for the second time. Converting these values to minutes, we see that it will take minutes, and minutes, for the two hands to make angles. Thus the times, correct to the nearest minute, at which the hands of the clock will form an angle are .
Solution 2
Using the formula $(60h-11m)/2
See Also
1954 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 49 |
Followed by Last Question | |
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All AHSME Problems and Solutions |
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