2009 AMC 10B Problems/Problem 19
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- The following problem is from both the 2009 AMC 10B #19 and 2009 AMC 12B #10, so both problems redirect to this page.
Contents |
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 
.
See also
| 2009 AMC 10B (Problems • 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 | ||
| 2009 AMC 12B (Problems • 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 | ||
