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Difference between revisions of "1989 AHSME Problems/Problem 12"

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== Solution 2 ==  
 
== Solution 2 ==  
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Since the westbound vehicles are equally spaced and move at a constant speed, the eastbound driver passes one westbound vehicle every <math>\frac{5}{20} = \frac{1}{4}</math> minute, or every <math>\frac{1}{240}</math> hour.  Let <math>t = 0</math> hours be the time eastbound driver passes its first westbound vehicle.  Then the eastbound driver will pass its second westbound vehicle at <math>t = \frac{1}{240}</math> hour.  In this time, both the eastbound and second westbound vehicles traveled <math>60 \cdot \frac{1}{240} = \frac{1}{4}</math> miles, so they were <math>\frac{1}{2}</math> mile apart at <math>t = 0</math>.  Since the first westbound and eastbound vehicles were at the same position at <math>t = 0</math>, the first and second westbound vehicles were <math>\frac {1}{2}</math> mile apart.  Therefore, all westbound vehicles are <math>\frac{1}{2}</math> mile apart, and a <math>100</math>-mile section must contain <math>\frac{100}{\frac{1}{2}} = 200</math> westbound vehicles.   
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Since the westbound vehicles are equally spaced and move at a constant speed, the eastbound driver passes one westbound vehicle every <math>\frac{5}{20} = \frac{1}{4}</math> minute, or every <math>\frac{1}{240}</math> hour.  Let <math>t = 0</math> hours be the time eastbound driver passes its first westbound vehicle.  Then the eastbound driver will pass its second westbound vehicle at <math>t = \frac{1}{240}</math> hour.  In this time, the eastbound and second westbound vehicles each traveled <math>60 \cdot \frac{1}{240} = \frac{1}{4}</math> miles, so they were <math>\frac{1}{2}</math> mile apart at <math>t = 0</math>.  Since the first westbound and eastbound vehicles were at the same position at <math>t = 0</math>, the first and second westbound vehicles were <math>\frac {1}{2}</math> mile apart.  Therefore, all pairs of consecutive westbound vehicles are <math>\frac{1}{2}</math> mile apart, and a <math>100</math>-mile section must contain <math>\frac{100}{\frac{1}{2}} = \boxed{(\textbf{C})\ 200}</math> westbound vehicles.   
  
 
-j314andrews
 
-j314andrews

Latest revision as of 01:27, 11 July 2025

Problem

The traffic on a certain east-west highway moves at a constant speed of 60 miles per hour in both directions. An eastbound driver passes 20 west-bound vehicles in a five-minute interval. Assume vehicles in the westbound lane are equally spaced. Which of the following is closest to the number of westbound vehicles present in a 100-mile section of highway?

$\textrm{(A)}\ 100\qquad\textrm{(B)}\ 120\qquad\textrm{(C)}\ 200\qquad\textrm{(D)}\ 240\qquad\textrm{(E)}\ 400$

Solution 1

At the beginning of the five-minute interval, say the eastbound driver is at the point $x=0$, and at the end of the interval at $x=5$, having travelled five miles. Because both lanes are travelling at the same speed, the last westbound car to be passed by the eastbound driver was just west of the position $x=10$ at the start of the five minutes. The first westbound car to be passed was just east of $x=0$ at that time. Therefore, the eastbound driver passed all of the cars initially in the stretch of road between $x=0$ and $x=10$. That makes $20$ cars in ten miles, so we estimate $200$ cars in a hundred miles.

[asy] dot((0,0));dot((25,0));dot((50,0)); draw((15,5)--(5,5),EndArrow); draw((45,5)--(35,5),EndArrow); draw((0,0)--(50,0),dashed); draw((0,-5)--(10,-5),EndArrow); label("$0$",(0,0),W); label("$10$",(50,0),E); label("$5$",(25,0),S); [/asy]

Solution 2

Since the westbound vehicles are equally spaced and move at a constant speed, the eastbound driver passes one westbound vehicle every $\frac{5}{20} = \frac{1}{4}$ minute, or every $\frac{1}{240}$ hour. Let $t = 0$ hours be the time eastbound driver passes its first westbound vehicle. Then the eastbound driver will pass its second westbound vehicle at $t = \frac{1}{240}$ hour. In this time, the eastbound and second westbound vehicles each traveled $60 \cdot \frac{1}{240} = \frac{1}{4}$ miles, so they were $\frac{1}{2}$ mile apart at $t = 0$. Since the first westbound and eastbound vehicles were at the same position at $t = 0$, the first and second westbound vehicles were $\frac {1}{2}$ mile apart. Therefore, all pairs of consecutive westbound vehicles are $\frac{1}{2}$ mile apart, and a $100$-mile section must contain $\frac{100}{\frac{1}{2}} = \boxed{(\textbf{C})\ 200}$ westbound vehicles.

-j314andrews

See also

1989 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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
All AHSME Problems and Solutions

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