1970 AHSME Problems/Problem 32

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 $2 \cdot 240 = 480, \text{or } \fbox{C}$ .

1970 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 31	Followed by Problem 33
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1970 AHSME Problems/Problem 32

Problem

Solution

See also