1963 AMC 12 Problem 20

Question

Two men at points

R

and

S,

76

miles apart, set out at the same time to walk towards each other. The man at

R

walks uniformly at

4\dfrac12

miles per hour; the man at

S

walks at

3\dfrac14

miles per hour for the first hour, at

3\dfrac34

miles per hour for the second hour, and so on, in arithmetic progression. If the men meet

x

miles nearer

R

than

S

in an integral number of hours, then

x

is:

10

8

6

4

2

Let the integral meeting time be

h

hours and sum the second man’s hourly distances

Their combined distance equation simplifies to

h^2+30h-304=0

In

h

hours the first man walks

\frac{9h}{2}.

The second walks an arithmetic-series total

\begin{aligned} &\frac h2\left(2\cdot\frac{13}{4} +(h-1)\frac12\right)\\ &\qquad=3h+\frac{h^2}{4}. \end{aligned}

Their distances sum to

76,

giving

h^2+30h-304=0,

whose positive root is

h=8.

They walk

36

and

40

miles, so the meeting point is

4

miles nearer

R

than

S.

Thus, the correct answer is D.

Problem 20 in other years