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

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==Solution==
 
==Solution==
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Let the speed of boy <math>A</math> be <math>a</math>, and the speed of boy <math>B</math> be <math>b</math>. Notice that <math>A</math> travels <math>4</math> miles per hour slower than boy <math>B</math>, so we can replace <math>b</math> with <math>a+4</math>.
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Now let us see the distances that the boys each travel. Boy <math>A</math> travels <math>60-12=48</math> miles, and boy <math>B</math> travels <math>60+12=72</math> miles. Now, we can use <math>d=rt</math> to make an equation, where we set the time to be equal: <cmath>\frac{48}{a}=\frac{72}{a+4}</cmath>
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Cross-multiplying gives <math>48a+192=72a</math>. Isolating the variable <math>a</math>, we get the equation <math>24a=192</math>, so <math>a=\boxed{\textbf{(B) }8 \text{ mph}}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 20:35, 9 April 2013

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

Let the speed of boy $A$ be $a$, and the speed of boy $B$ be $b$. Notice that $A$ travels $4$ miles per hour slower than boy $B$, so we can replace $b$ with $a+4$.

Now let us see the distances that the boys each travel. Boy $A$ travels $60-12=48$ miles, and boy $B$ travels $60+12=72$ miles. Now, we can use $d=rt$ to make an equation, where we set the time to be equal: \[\frac{48}{a}=\frac{72}{a+4}\] Cross-multiplying gives $48a+192=72a$. Isolating the variable $a$, we get the equation $24a=192$, so $a=\boxed{\textbf{(B) }8 \text{ mph}}$.

See Also

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