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| − | {{duplicate|[[2009 AMC 10B Problems|2009 AMC 10B #19]] and [[2009 AMC 12B Problems|2009 AMC 12B #10]]}}
| + | #redirect [[2009 AMC 12B Problems/Problem 10]] |
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| − | == Problem ==
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| − | A particular <math>12</math>-hour digital clock displays the hour and minute of a day. Unfortunately, whenever it is supposed to display a <math>1</math>, it mistakenly displays a <math>9</math>. 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?
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| − | <math>\mathrm{(A)}\ \frac 12\qquad
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| − | \mathrm{(B)}\ \frac 58\qquad
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| − | \mathrm{(C)}\ \frac 34\qquad
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| − | \mathrm{(D)}\ \frac 56\qquad
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| − | \mathrm{(E)}\ \frac {9}{10}</math>
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| − | == Solution ==
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| − | The clock will display the incorrect time for the entire hours of <math>1, 10, 11</math> and <math>12</math>. So the correct hour is displayed <math>\frac 23</math> of the time. The minutes will not display correctly whenever either the tens digit or the ones digit is a <math>1</math>, so the minutes that will not display correctly are <math>10, 11, 12, \dots, 19</math> and <math>01, 21, 31, 41,</math> and <math>51</math>. 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 <math>\frac 23 \cdot \left(1 - \frac {15}{60}\right) = \frac 23 \cdot \frac 34 = \boxed{\frac 12}</math>. The answer is <math>\mathrm{(A)}</math>.
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| − | == See also ==
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| − | {{AMC10 box|year=2009|ab=B|num-b=18|num-a=20}}
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| − | {{AMC12 box|year=2009|ab=B|num-b=9|num-a=11}}
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