# 1970 AHSME Problems/Problem 32

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## Problem $A$ and $B$ travel around a circular track at uniform speeds in opposite directions, starting from diametrically opposite points. If they start at the same time, meet first after $B$ has travelled $100$ yards, and meet a second time $60$ yards before $A$ completes one lap, then the circumference of the track in yards is $\text{(A) } 400\quad \text{(B) } 440\quad \text{(C) } 480\quad \text{(D) } 560\quad \text{(E) } 880$

## Solution $\fbox{C}$

Let $x$ be half the circumference of the track. They first meet after $B$ has run $100$ yards, meaning that in the time $B$ has run $100$ yards, $A$ has run $x-100$ yards. The second time they meet is when $A$ is 60 yards before he completes the lap. This means that in the time that $A$ has run $2x-60$ yards, $B$ has run $x+60$ yards. Because they run at uniform speeds, we can write the equation $$\frac{100}{x-100}=\frac{x+60}{2x-60} .$$ Cross multiplying, $$200x-6000=x^2-40x-6000$$ Adding $6000$ to both sides and simplifying, we have $$200x=x^2-40x$$ $$240x=x^2$$ $$x=240.$$ Because $x$ is only half of the circumference of the track, the answer we are looking for is $2 \cdot 240 = 480, \text{or } \fbox{C}$.

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