Difference between revisions of "2009 AMC 10B Problems/Problem 19"
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We count the correct times directly; let a correct time be <math>x:yz</math>, where <math>x</math> is a number from 1 to 12 and <math>y</math> and <math>z</math> are digits, where <math>y<6</math>. There are 8 values of <math>x</math> that will display the correct time: 2, 3, 4, 5, 6, 7, 8, and 9. There are five values of <math>y</math> that will display the correct time: 0, 2, 3, 4, and 5. There are nine values of <math>z</math> that will display the correct time: 0, 2, 3, 4, 5, 6, 7, 8, and 9. Therefore there are <math>8\cdot 5\cdot 9=40\cdot 9=360</math> correct times. | We count the correct times directly; let a correct time be <math>x:yz</math>, where <math>x</math> is a number from 1 to 12 and <math>y</math> and <math>z</math> are digits, where <math>y<6</math>. There are 8 values of <math>x</math> that will display the correct time: 2, 3, 4, 5, 6, 7, 8, and 9. There are five values of <math>y</math> that will display the correct time: 0, 2, 3, 4, and 5. There are nine values of <math>z</math> that will display the correct time: 0, 2, 3, 4, 5, 6, 7, 8, and 9. Therefore there are <math>8\cdot 5\cdot 9=40\cdot 9=360</math> correct times. | ||
| − | Therefore the required fraction is <math>\frac{360}{720}=\frac{1}{2}\Rightarrow \boxed\ | + | Therefore the required fraction is <math>\frac{360}{720}=\frac{1}{2}\Rightarrow \boxed{\mathrm{(A)}}</math>. |
== See also == | == See also == | ||
{{AMC10 box|year=2009|ab=B|num-b=18|num-a=20}} | {{AMC10 box|year=2009|ab=B|num-b=18|num-a=20}} | ||
{{AMC12 box|year=2009|ab=B|num-b=9|num-a=11}} | {{AMC12 box|year=2009|ab=B|num-b=9|num-a=11}} | ||
Revision as of 11:19, 2 March 2009
- 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 
.
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 | |
