Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution)
m (→‎Solution)
 
(2 intermediate revisions by 2 users not shown)
Line 2: Line 2:
  
 
<math>\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</math>
 
<math>\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</math>
 +
 +
== Problem 24 ==
 +
 +
A man walked a certain distance at a constant rate. If he had gone <math>\textstyle\frac{1}{2}</math> mile per hour faster, he would have walked the distance in four-fifths of the time; if he had gone <math>\textstyle\frac{1}{2}</math> mile per hour slower, he would have been <math>2\textstyle\frac{1}{2}</math> hours longer on the road. The distance in miles he walked was
 +
 +
<math>\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</math> 
  
 
==Solution==
 
==Solution==
 
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.
  
−
<math>\frac{4t}{5}(x+\frac{1}{2})=xt=(t+\frac{1}{2})(x-\frac{1}{2})</math> where <math>x</math> is the man's rate and <math>t</math> is the time it takes him.   
+
<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.   
  
 
Looking at the first two parts of the equations,  
 
Looking at the first two parts of the equations,  
  
−
<math>\frac{4t}{5}(x+\frac{1}{2})=xt</math>
+
<cmath>\frac{4t}{5}(x+\frac{1}{2})=xt</cmath>
−
 
 
−
we note that we can solve for <math>x</math>.
 
  
−
Solving for <math>x</math>, we get <math>x=2</math>
+
we note that we can solve for <math>x</math>. Solving for <math>x</math>, we get <math>x=2.</math>
  
 
Now we look at the last two parts of the equation:
 
Now we look at the last two parts of the equation:
  
−
<math>xt=(t+\frac{1}{2})(x-\frac{1}{2})</math>
+
<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 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 want the find the distance so we just need  to find <math>xt</math> which is <math>\boxed{\mathrm{(B) \ } 15}</math>
+
-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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1972_AHSME_Problems/Problem_24&oldid=198600"