Difference between revisions of "1954 AHSME Problems/Problem 50"
m (→Solution) |
Yjc64002776 (talk | contribs) (→Solution) |
||
| (2 intermediate revisions by 2 users not shown) | |||
| Line 14: | Line 14: | ||
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>. | ||
| − | == See | + | == See Also == |
{{AHSME 50p box|year=1954|num-b=49|after=Last Question}} | {{AHSME 50p box|year=1954|num-b=49|after=Last Question}} | ||
Latest revision as of 10:59, 21 January 2021
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 
.
See Also
| 1954 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 49 |
Followed by Last Question | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
