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 $N$

Answer: E

Concepts:arithmetic sequence distance rate and time

Difficulty rating: 1810

Small Hint:

After $t$ whole hours, sum the second man’s hourly distances as an arithmetic progression

Big Hint:

Set that sum plus $4t$ equal to $72$

Solution:

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.

1957 AMC 12 Problem 39

Answer: E

Difficulty rating: 1810

Small Hint:

Big Hint:

Solution:

Problem 39 in Other Years