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Difference between revisions of "1970 AHSME Problems/Problem 32"

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== Solution ==
 
== Solution ==
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<math>\fbox{C}</math>
 
  
 
Let <math>x</math> be half the circumference of the track. They first meet after <math>B</math> has run <math>100</math> yards, meaning that in the time <math>B</math> has run <math>100</math> yards, <math>A</math> has run <math>x-100</math> yards. The second time they meet is when <math>A</math> is 60 yards before he completes the lap. This means that in the time that <math>A</math> has run <math>2x-60</math> yards, <math>B</math> has run <math>x+60</math> yards.  
 
Let <math>x</math> be half the circumference of the track. They first meet after <math>B</math> has run <math>100</math> yards, meaning that in the time <math>B</math> has run <math>100</math> yards, <math>A</math> has run <math>x-100</math> yards. The second time they meet is when <math>A</math> is 60 yards before he completes the lap. This means that in the time that <math>A</math> has run <math>2x-60</math> yards, <math>B</math> has run <math>x+60</math> yards.  
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Because they r;;'''';';''';';';un at uniform speeds, we can write the equation
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Because they run at uniform speeds, we can write the equation
 
<cmath> \frac{100}{x-100}=\frac{x+60}{2x-60} .</cmath>
 
<cmath> \frac{100}{x-100}=\frac{x+60}{2x-60} .</cmath>
 
Cross multiplying,
 
Cross multiplying,

Latest revision as of 10:28, 30 October 2024

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

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

See also

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

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