Difference between revisions of "1954 AHSME Problems/Problem 50"
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At <math>7</math> o'clock, the hour hand is at the position <math>\frac{7}{12}\cdot 360^{\circ}=210^{\circ}</math> clockwise from the <math>12</math> o'clock position, and the minute hand is exactly at the <math>12</math> o'clock position. Thus the minute hand is <math>360^{\circ}-210^{\circ}=150^{\circ}</math> ahead of the hour hand, while it is also <math>210^{\circ}</math> behind the hour hand. So, when the minute hand first makes an <math>84^{\circ}</math> angle with the hour hand, the minute hand will be <math>84^{\circ}</math> behind the hour hand, and the second time they make and <math>84^{\circ}</math> angle, the minute hand will be <math>84^{\circ}</math> ahead the hour hand. | At <math>7</math> o'clock, the hour hand is at the position <math>\frac{7}{12}\cdot 360^{\circ}=210^{\circ}</math> clockwise from the <math>12</math> o'clock position, and the minute hand is exactly at the <math>12</math> o'clock position. Thus the minute hand is <math>360^{\circ}-210^{\circ}=150^{\circ}</math> ahead of the hour hand, while it is also <math>210^{\circ}</math> behind the hour hand. So, when the minute hand first makes an <math>84^{\circ}</math> angle with the hour hand, the minute hand will be <math>84^{\circ}</math> behind the hour hand, and the second time they make and <math>84^{\circ}</math> angle, the minute hand will be <math>84^{\circ}</math> ahead the hour hand. | ||
− | 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})=\frac{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})=\frac{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\frac{10}{11}</math> minutes, and <math>53\frac{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>\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})=\frac{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})=\frac{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\frac{10}{11}</math> minutes, and <math>53\frac{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 also == | == See also == |
Revision as of 18:10, 18 July 2015
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 |
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