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

Difference between revisions of "1964 AHSME Problems/Problem 40"

(→‎See Also)
 
(2 intermediate revisions by one other user not shown)
Line 5: Line 5:
 
<math>\textbf{(A) }14\frac{14}{23}\qquad\textbf{(B) }14\frac{1}{14}\qquad\textbf{(C) }13\frac{101}{115}\qquad\textbf{(D) }13\frac{83}{115}\qquad \textbf{(E) }13\frac{13}{23}</math>
 
<math>\textbf{(A) }14\frac{14}{23}\qquad\textbf{(B) }14\frac{1}{14}\qquad\textbf{(C) }13\frac{101}{115}\qquad\textbf{(D) }13\frac{83}{115}\qquad \textbf{(E) }13\frac{13}{23}</math>
  
 +
==Solution==
 +
 +
From March 15 <math>1</math> P.M. on the watch to March 21 <math>9</math> A.M. on the watch, the watch passed <math>20 + 5 \times 24 = 140</math> hours.
 +
 +
Since <math>1</math> watch hour equals <math>\frac{24}{23 + \frac{57.5}{60}} = \frac{576}{575}</math> real hour, the difference between the watch time and the actual time passed is <math>140 \times \left( \frac{576}{575} - 1 \right) = \frac{28}{115}</math> hour <math>=14\frac{14}{23}</math> minutes.
  
 
==See Also==
 
==See Also==
−
{{AHSME box|year=1983|num-b=29|after=Last Problem}}
+
{{AHSME box|year=1964|num-b=39|after=Last Problem}}
  
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
 +
[[Category:Introductory Algebra Problems]]

Latest revision as of 19:25, 21 October 2021

Problem

A watch loses $2\frac{1}{2}$ minutes per day. It is set right at $1$ P.M. on March 15. Let $n$ be the positive correction, in minutes, to be added to the time shown by the watch at a given time. When the watch shows $9$ A.M. on March 21, $n$ equals:

$\textbf{(A) }14\frac{14}{23}\qquad\textbf{(B) }14\frac{1}{14}\qquad\textbf{(C) }13\frac{101}{115}\qquad\textbf{(D) }13\frac{83}{115}\qquad \textbf{(E) }13\frac{13}{23}$

Solution

From March 15 $1$ P.M. on the watch to March 21 $9$ A.M. on the watch, the watch passed $20 + 5 \times 24 = 140$ hours.

Since $1$ watch hour equals $\frac{24}{23 + \frac{57.5}{60}} = \frac{576}{575}$ real hour, the difference between the watch time and the actual time passed is $140 \times \left( \frac{576}{575} - 1 \right) = \frac{28}{115}$ hour $=14\frac{14}{23}$ minutes.

See Also

1964 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 39 		Followed by
Last Problem
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions


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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1964_AHSME_Problems/Problem_40&oldid=163888"