2009 AMC 12B Problem 10

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?

\dfrac{1}{2}

\dfrac{5}{8}

\dfrac{3}{4}

\dfrac{5}{6}

\dfrac{9}{10}

The hour is shown correctly unless it contains a

1,

namely hours

1, 10, 11, 12

A minute is correct only when neither of its digits is a

1

; count the bad minutes out of

60

The hours containing a

1

are

1, 10, 11, 12,

so

8

of the

12

hours display correctly, a fraction

\dfrac{2}{3}.

A minute is wrong if either digit is

1

: the tens digit gives

10\text{-}19

(

10

minutes), and the ones digit adds

01, 21, 31, 41, 51

(

5

more),

15

in all. So

\dfrac{45}{60} = \dfrac{3}{4}

of minutes are correct.

The fraction of the day is

\dfrac{2}{3} \cdot \dfrac{3}{4} = \dfrac{1}{2}.

Thus, the correct answer is A.

Problem 10 in other years