Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus
During AMC 10A/12A testing, the AoPS Wiki is in read-only mode and no edits can be made.

2003 AMC 12B Problems/Problem 11: Difference between revisions

← Older edit
Nathan wailes (talk | contribs)
2,960 edits
No edit summary
Gina (talk | contribs)
783 edits
No edit summary
 
Line 1: Line 1:
==Problem==
Cassandra sets her watch to the correct time at noon. At the actual time of 1:00 PM, she notices that her watch reads 12:57 and 36 seconds. Assuming that her watch loses time at a constant rate, what will be the actual time when her
Cassandra sets her watch to the correct time at noon. At the actual time of 1:00 PM, she notices that her watch reads 12:57 and 36 seconds. Assuming that her watch loses time at a constant rate, what will be the actual time when her
watch first reads 10:00 PM?
watch first reads 10:00 PM?
Line 5: Line 6:
\text {(A) 10:22 PM and 24 seconds}  \qquad \text {(B) 10:24 PM} \qquad \text {(C) 10:25 PM}  \qquad \text {(D) 10:27 PM}  \qquad \text {(E) 10:30 PM}  
\text {(A) 10:22 PM and 24 seconds}  \qquad \text {(B) 10:24 PM} \qquad \text {(C) 10:25 PM}  \qquad \text {(D) 10:27 PM}  \qquad \text {(E) 10:30 PM}  
</math>
</math>
==Solution==
For every <math>60</math> minutes that pass by in actual time, <math>57+\frac{36}{60}=57.6</math> minutes pass by on Cassandra's watch. When her watch first reads, 10:00 pm, <math>10(60)=600</math> minutes have passed by on her watch. Setting up a proportion,
<cmath>\frac{57.6}{60}=\frac{600}{x}</cmath>
where <math>x</math> is the number of minutes that have passed by in actual time. Solve for <math>x</math> to get <math>625</math> minutes, or <math>10</math> hours and <math>25</math> minutes <math>\Rightarrow \boxed{\textbf{(C)}\ \text{10:25 PM}}</math>.
==See Also==
{{AMC12 box|year=2003|ab=B|num-b=10|num-a=12}}
{{MAA Notice}}
{{MAA Notice}}

Latest revision as of 23:54, 4 January 2014

Problem

Cassandra sets her watch to the correct time at noon. At the actual time of 1:00 PM, she notices that her watch reads 12:57 and 36 seconds. Assuming that her watch loses time at a constant rate, what will be the actual time when her watch first reads 10:00 PM?

$\text {(A) 10:22 PM and 24 seconds} \qquad \text {(B) 10:24 PM} \qquad \text {(C) 10:25 PM} \qquad \text {(D) 10:27 PM} \qquad \text {(E) 10:30 PM}$

Solution

For every $60$ minutes that pass by in actual time, $57+\frac{36}{60}=57.6$ minutes pass by on Cassandra's watch. When her watch first reads, 10:00 pm, $10(60)=600$ minutes have passed by on her watch. Setting up a proportion,

\[\frac{57.6}{60}=\frac{600}{x}\]

where $x$ is the number of minutes that have passed by in actual time. Solve for $x$ to get $625$ minutes, or $10$ hours and $25$ minutes $\Rightarrow \boxed{\textbf{(C)}\ \text{10:25 PM}}$.

See Also

2003 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12B_Problems/Problem_11&oldid=58640"