Difference between revisions of "1972 AHSME Problems/Problem 24"
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<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> | ||
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| + | == 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. | ||
| − | < | + | <cmath>\frac{4t}{5}(x+\frac{1}{2})=xt=(t+\frac{1}{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, | ||
| − | < | + | <cmath>\frac{4t}{5}(x+\frac{1}{2})=xt</cmath> |
| − | |||
| − | |||
| − | 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: | ||
| − | < | + | <cmath>xt=(t+\frac{1}{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> |
| − | + | -edited for readability | |
Revision as of 09:42, 29 January 2021
A man walked a certain distance at a constant rate. If he had gone 
mile per hour faster, he would have walked the distance in four-fifths of the time; if he had gone mile per hour slower, he would have been 
hours longer on the road. The distance in miles he walked was
Problem 24
A man walked a certain distance at a constant rate. If he had gone mile per hour faster, he would have walked the distance in four-fifths of the time; if he had gone mile per hour slower, he would have been hours longer on the road. The distance in miles he walked was
Solution
We can make three equations out of the information, and since the distances are the same, we can equate these equations.
![\[\frac{4t}{5}(x+\frac{1}{2})=xt=(t+\frac{1}{2})(x-\frac{1}{2})\]](http://latex.artofproblemsolving.com/3/5/7/3573c16a0322aa82385bf2f68cefe7ac797a8a40.png)
where 
is the man's rate and 
is the time it takes him.
Looking at the first two parts of the equations,
![\[\frac{4t}{5}(x+\frac{1}{2})=xt\]](http://latex.artofproblemsolving.com/d/3/8/d386dfc3a1c75abf6edd86ea2f2d76d1269b3543.png)
we note that we can solve for . Solving for , we get 
Now we look at the last two parts of the equation:
![\[xt=(t+\frac{1}{2})(x-\frac{1}{2})\]](http://latex.artofproblemsolving.com/d/c/2/dc2b9474d9526f9a1d9c93b021cd91c904ec168b.png)
we note that we can solve for and we get 
We want the find the distance, which is 
-edited for readability
