Difference between revisions of "1950 AHSME Problems/Problem 28"

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==Problem==
 
==Problem==
 
 
Two boys <math>A</math> and <math>B</math> start at the same time to ride from Port Jervis to Poughkeepsie, <math>60</math> miles away. <math>A</math> travels <math>4</math> miles an hour slower than <math>B</math>. <math>B</math> reaches Poughkeepsie and at once turns back meeting <math>A</math> <math>12</math> miles from Poughkeepsie. The rate of <math>A</math> was:
 
Two boys <math>A</math> and <math>B</math> start at the same time to ride from Port Jervis to Poughkeepsie, <math>60</math> miles away. <math>A</math> travels <math>4</math> miles an hour slower than <math>B</math>. <math>B</math> reaches Poughkeepsie and at once turns back meeting <math>A</math> <math>12</math> miles from Poughkeepsie. The rate of <math>A</math> was:
  
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\textbf{(D)}\ 16\text{ mph} \qquad
 
\textbf{(D)}\ 16\text{ mph} \qquad
 
\textbf{(E)}\ 20\text{ mph}</math>
 
\textbf{(E)}\ 20\text{ mph}</math>
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==Solution==
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{{solution}}
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==See Also==
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{{AHSME box|year=1950|num-b=27|num-a=29}}
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[[Category:Introductory Algebra Problems]]

Revision as of 14:20, 17 April 2012

Problem

Two boys $A$ and $B$ start at the same time to ride from Port Jervis to Poughkeepsie, $60$ miles away. $A$ travels $4$ miles an hour slower than $B$. $B$ reaches Poughkeepsie and at once turns back meeting $A$ $12$ miles from Poughkeepsie. The rate of $A$ was:

$\textbf{(A)}\ 4\text{ mph}\qquad \textbf{(B)}\ 8\text{ mph} \qquad \textbf{(C)}\ 12\text{ mph} \qquad \textbf{(D)}\ 16\text{ mph} \qquad \textbf{(E)}\ 20\text{ mph}$

Solution

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See Also

1950 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 27
Followed by
Problem 29
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