1957 AMC 12 Problem 39

Question

Two men set out at the same time to walk towards each other from

M

and

N,

72

miles apart. The first man walks at the rate of

4

mph. The second man walks

2

miles the first hour,

2\frac12

miles the second hour,

3

miles the third hour, and so on in arithmetic progression. Then the men will meet:

in

7

hours

in

8\frac14

hours

nearer

M

than

N

nearer

N

than

M

midway between

M

and

N

After

t

whole hours, sum the second man’s hourly distances as an arithmetic progression

Set that sum plus

4t

equal to

72

In

t

hours, the second man’s hourly distances form an arithmetic progression with first term

2

and last term

2+\frac{t-1}{2}.

His distance is

\frac t2\left(4+\frac{t-1}{2}\right) =\frac{t(t+7)}4.

Together the men cover

4t+\frac{t(t+7)}4=72,

or

t^2+23t-288=0.

The positive root is

t=9.

The first man then walks

4(9)=36

miles, exactly half the original distance.

Therefore, the correct answer is E.

Problem 39 in other years