1989 AHSME Problems/Problem 12

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

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, both the eastbound and second westbound vehicles 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 westbound vehicles are $\frac{1}{2}$ mile apart, and a $100$ -mile section must contain $\frac{100}{\frac{1}{2}} = 200$ westbound vehicles.

-j314andrews

1989 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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1989 AHSME Problems/Problem 12

Contents

Problem

Solution 1

Solution

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