Difference between revisions of "2009 AMC 12B Problems/Problem 10"

Revision as of 21:02, 24 June 2020

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 $12$ -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?

$\mathrm{(A)}\ \frac 12\qquad \mathrm{(B)}\ \frac 58\qquad \mathrm{(C)}\ \frac 34\qquad \mathrm{(D)}\ \frac 56\qquad \mathrm{(E)}\ \frac {9}{10}$

Solution

Solution 1

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 $\frac 23 \cdot \left(1 - \frac {15}{60}\right) = \frac 23 \cdot \frac 34 = \boxed{\frac 12}$ . The answer is $\mathrm{(A)}$ .

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 $x:yz$ , where $x$ is a number from 1 to 12 and $y$ and $z$ are digits, where $y<6$ . There are 8 values of $x$ that will display the correct time: 2, 3, 4, 5, 6, 7, 8, and 9. There are five values of $y$ that will display the correct time: 0, 2, 3, 4, and 5. There are nine values of $z$ that will display the correct time: 0, 2, 3, 4, 5, 6, 7, 8, and 9. Therefore there are $8\cdot 5\cdot 9=40\cdot 9=360$ correct times.

Therefore the required fraction is $\frac{360}{720}=\frac{1}{2}\Rightarrow \boxed{\mathrm{(A)}}$ .

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 1: / Line 1: @@
-#redirect [[2009 AMC 10B Problems/Problem 19]]
+{{duplicate|[[2009 AMC 10B Problems|2009 AMC 10B #19]] and [[2009 AMC 12B Problems|2009 AMC 12B #10]]}}
+== Problem ==
+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?
+<math>\mathrm{(A)}\ \frac 12\qquad
+\mathrm{(B)}\ \frac 58\qquad
+\mathrm{(C)}\ \frac 34\qquad
+\mathrm{(D)}\ \frac 56\qquad
+\mathrm{(E)}\ \frac {9}{10}</math>
+== Solution ==
+=== Solution 1 ===
+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>.
+=== 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 <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{\mathrm{(A)}}</math>.
+== See also ==
+{{AMC10 box|year=2009|ab=B|num-b=18|num-a=20}}
+{{AMC12 box|year=2009|ab=B|num-b=9|num-a=11}}
+{{MAA Notice}}

Page

Toolbox

Search