2009 AMC 10B Problem 19

Question

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?

\dfrac12

\dfrac58

\dfrac34

\dfrac56

\dfrac{9}{10}

The hour is wrong only for

1,

10,

11,

12

A minute is wrong when either of its digits is a

1;

count those out of

60

Among the hours

1

through

12,

exactly

1,

10,

11,

12

contain a

1,

so the hour is correct

\dfrac{8}{12}=\dfrac23

of the time.

A minute is displayed wrong when its tens digit is

1

(minutes

10

–

19

) or its units digit is

1

(

01,11,\dots,51

), which is

15

of the

60

minutes. So the minute is correct

\dfrac{45}{60}=\dfrac34

of the time.

The clock is correct

\dfrac23\cdot\dfrac34=\dfrac12

of the day.

Thus, the correct answer is A.

Problem 19 in other years