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

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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>
 
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>
 
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>.
 
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>.
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==Alternate Solution==
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Note that <math>A</math> travels <math>60-12=48</math> miles in the time it takes <math>B</math> to travel <math>60+12=72</math> miles. Thus, <math>B</math> travels <math>72-48=24</math> more miles in the given time, meaning <math>\frac{24\text{miles}}{4\text{miles}/\text{hour}} = 6 \text{hours}</math> have passed, as <math>B</math> goes <math>4</math> miles per hour faster. Thus, <math>A</math> travels <math>48</math> miles per <math>6</math> hours, or <math>8</math> miles per hour.
  
 
==See Also==
 
==See Also==
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[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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[[Category:Rate Problems]]
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{{MAA Notice}}

Latest revision as of 18:03, 9 June 2017

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 $a=\boxed{\textbf{(B) }8 \text{ mph}}$.

Alternate Solution

Note that $A$ travels $60-12=48$ miles in the time it takes $B$ to travel $60+12=72$ miles. Thus, $B$ travels $72-48=24$ more miles in the given time, meaning $\frac{24\text{miles}}{4\text{miles}/\text{hour}} = 6 \text{hours}$ have passed, as $B$ goes $4$ miles per hour faster. Thus, $A$ travels $48$ miles per $6$ hours, or $8$ miles per hour.

See Also

1950 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 27 		Followed by
Problem 29
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All AHSME Problems and Solutions

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