Difference between revisions of "2009 AMC 12B Problems/Problem 10"
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==Video Solution== | ==Video Solution== | ||
https://www.youtube.com/watch?v=VNJrZ-ABtS4 | https://www.youtube.com/watch?v=VNJrZ-ABtS4 | ||
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+ | https://youtu.be/rmRQK_flGOM | ||
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+ | ~savannahsolver | ||
== See also == | == See also == |
Latest revision as of 21:18, 20 January 2021
- The following problem is from both the 2009 AMC 10B #19 and 2009 AMC 12B #10, so both problems redirect to this page.
Problem
A particular -hour digital clock displays the hour and minute of a day. Unfortunately, whenever it is supposed to display a , it mistakenly displays a . For example, when it is 1:16 PM the clock incorrectly shows 9:96 PM. What fraction of the day will the clock show the correct time?
Solution
Solution 1
The clock will display the incorrect time for the entire hours of and . So the correct hour is displayed of the time. The minutes will not display correctly whenever either the tens digit or the ones digit is a , so the minutes that will not display correctly are and and . This amounts to fifteen of the sixty possible minutes for any given hour. Hence the fraction of the day that the clock shows the correct time is . The answer is .
Solution 2
The required fraction is the number of correct times divided by the total times. There are 60 minutes in an hour and 12 hours on a clock, so there are 720 total times.
We count the correct times directly; let a correct time be , where is a number from 1 to 12 and and are digits, where . There are 8 values of that will display the correct time: 2, 3, 4, 5, 6, 7, 8, and 9. There are five values of that will display the correct time: 0, 2, 3, 4, and 5. There are nine values of that will display the correct time: 0, 2, 3, 4, 5, 6, 7, 8, and 9. Therefore there are correct times.
Therefore the required fraction is .
Video Solution
https://www.youtube.com/watch?v=VNJrZ-ABtS4
~savannahsolver
See also
2009 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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 | ||
All AMC 10 Problems and Solutions |
2009 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 9 |
Followed by Problem 11 |
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 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.