1972 AMC 12 Problem 24

Question

A man walked a certain distance at a constant rate. If he had gone

\frac12

mile per hour faster, he would have walked the distance in four-fifths of the time; if he had gone

\frac12

mile per hour slower, he would have been

2\frac12

hours longer on the road. The distance in miles he walked was:

13\frac12

15

17\frac12

20

25

Let the actual speed and time be

v

and

t

First use

vt=(v+\frac{1}{2})(\frac{4t}{5})

to determine

v

Let the actual speed and time be

v

and

t.

The faster case gives

vt=\left(v+\frac12\right)\frac{4t}{5},

so

v=2.

The slower case then gives

2t=\frac32\left(t+\frac52\right),

whence

t=\frac{15}{2}.

The distance was

vt=2\cdot\frac{15}{2}=15.

Therefore, the correct answer is B.

Problem 24 in other years