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1972 AHSME Problems/Problem 24

Revision as of 20:06, 2 August 2015 by Npip99 (talk | contribs) (→‎Solution)

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.

$\frac{4t}{5}(x+\frac{1}{2})=xt=(t+\frac{1}{2})(x-\frac{1}{2})$ 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:

$xt=(t+\frac{1}{2})(x-\frac{1}{2})$

we note that we can solve for $t$ and we get $t=\frac{15}{2}$

We want the find the distance so we just need to find $xt$ which is $\boxed{\mathrm{(B) \ } 15}$

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