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

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==Problem==
 
==Problem==
  
A watch loses <math>2\frac{1}{2}</math> minutes per day. It is set right at <math>1</math> P.M. on March 15. Let <math>n</math> be the positive correction, in minutes, to be added to te time shown by the watch at a given time. When the watch shows <math>9</math> A.M. on March 21, <math>n</math> equals:
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A watch loses <math>2\frac{1}{2}</math> minutes per day. It is set right at <math>1</math> P.M. on March 15. Let <math>n</math> be the positive correction, in minutes, to be added to the time shown by the watch at a given time. When the watch shows <math>9</math> A.M. on March 21, <math>n</math> equals:
  
 
<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>
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==See Also==
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{{AHSME box|year=1964|num-b=39|after=Last Problem}}
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{{MAA Notice}}
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[[Category:Introductory Algebra Problems]]

Latest revision as of 22:22, 24 July 2019

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:

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


See Also

1964 AHSME (ProblemsAnswer KeyResources)
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


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