AMC10 2009 10B, Problem 19

A particular $12$-hour digital clock displays the hour and minute of a day. Unfortunately, whenever it is supposed to display a $1$, it mistakenly displays a $9$. 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: The clock will display the incorrect time for the entire hours of $1, 10, 11$ and $12$.  So the correct hour is displayed $\frac 23$ of the time.  The minutes will not display correctly whenever either the tens digit or the ones digit is a $1$, so the minutes that will not display correctly are $10, 11, 12, \dots, 19$ and $01, 21, 31, 41,$ and $51$.  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 $\frac 23 \cdot \left(1 - \frac {15}{60}\right) = \frac 23 \cdot \frac 34  = \fbox{\frac 12}$.  The answer is $\mathrm{(A)}$.

Answer: \frac 12