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Difference between revisions of "1972 AHSME Problems/Problem 24"

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m (→‎Solution)
 
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We can make three equations out of the information, and since the distances are the same, we can equate these equations.
 
We can make three equations out of the information, and since the distances are the same, we can equate these equations.
  
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<cmath>\frac{4t}{5}(x+\frac{1}{2})=xt=(t+\frac{1}{2})(x-\frac{1}{2})</cmath>
+
<cmath>\frac{4t}{5}(x+\frac{1}{2})=xt=(t+\frac{5}{2})(x-\frac{1}{2})</cmath>
 
where <math>x</math> is the man's rate and <math>t</math> is the time it takes him.   
 
where <math>x</math> is the man's rate and <math>t</math> is the time it takes him.   
  
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Now we look at the last two parts of the equation:
 
Now we look at the last two parts of the equation:
  
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<cmath>xt=(t+\frac{1}{2})(x-\frac{1}{2})</cmath>
+
<cmath>xt=(t+\frac{5}{2})(x-\frac{1}{2})</cmath>
  
 
we note that we can solve for <math>t</math> and we get <math>t=\frac{15}{2}.</math>  We want the find the distance, which is <math>xt= \boxed{15}.</math>
 
we note that we can solve for <math>t</math> and we get <math>t=\frac{15}{2}.</math>  We want the find the distance, which is <math>xt= \boxed{15}.</math>
  
 
-edited for readability
 
-edited for readability

Latest revision as of 21:27, 25 September 2023

A man walked a certain distance at a constant rate. If he had gone $\textstyle\frac{1}{2}$ mile per hour faster, he would have walked the distance in four-fifths of the time; if he had gone $\textstyle\frac{1}{2}$ mile per hour slower, he would have been $2\textstyle\frac{1}{2}$ hours longer on the road. The distance in miles he walked was

$\textbf{(A) }13\textstyle\frac{1}{2}\qquad \textbf{(B) }15\qquad \textbf{(C) }17\frac{1}{2}\qquad \textbf{(D) }20\qquad \textbf{(E) }25$

Problem 24

A man walked a certain distance at a constant rate. If he had gone $\textstyle\frac{1}{2}$ mile per hour faster, he would have walked the distance in four-fifths of the time; if he had gone $\textstyle\frac{1}{2}$ mile per hour slower, he would have been $2\textstyle\frac{1}{2}$ hours longer on the road. The distance in miles he walked was

$\textbf{(A) }13\textstyle\frac{1}{2}\qquad \textbf{(B) }15\qquad \textbf{(C) }17\frac{1}{2}\qquad \textbf{(D) }20\qquad \textbf{(E) }25$

Solution

We can make three equations out of the information, and since the distances are the same, we can equate these equations.

\[\frac{4t}{5}(x+\frac{1}{2})=xt=(t+\frac{5}{2})(x-\frac{1}{2})\] where $x$ is the man's rate and $t$ is the time it takes him.

Looking at the first two parts of the equations,

\[\frac{4t}{5}(x+\frac{1}{2})=xt\]

we note that we can solve for $x$. Solving for $x$, we get $x=2.$

Now we look at the last two parts of the equation:

\[xt=(t+\frac{5}{2})(x-\frac{1}{2})\]

we note that we can solve for $t$ and we get $t=\frac{15}{2}.$ We want the find the distance, which is $xt= \boxed{15}.$

-edited for readability

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