Difference between revisions of "2008 AIME I Problems/Problem 12"

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== Solution ==
 
== Solution ==
Let <math>n</math> be the number of car lengths that separate the cars.  Then their speed is <math>< 15n</math>, and the distance between the cars (front to front) is <math>4(n + 1)</math>.  Hence, the number of cars that pass in an hour is <math>\frac {15,000n\frac {meters}{hour}}{4(n + 1)\frac {meters}{car}} = \frac {15,000}{4(n + 1)}\frac {cars}{hour}</math>.  We wish to maximize this.  Observe that as <math>n</math> gets larger, <math>n + 1</math> gets less and less significant, so we take the limit as <math>n</math> approaches infinity
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Let <math>n</math> be the number of car lengths that separate the cars.  Then their speed is <math>< 15n</math>, and the distance between the cars (front to front) is <math>4(n + 1)</math>.  Hence, the number of cars that pass in an hour is <math>\frac {15,000n\frac {meters}{hour}}{4(n + 1)\frac {meters}{car}} = \frac {15,000n}{4(n + 1)}\frac {cars}{hour}</math>.  We wish to maximize this.  Observe that as <math>n</math> gets larger, <math>n + 1</math> gets less and less significant, so we take the limit as <math>n</math> approaches infinity
 
<cmath>
 
<cmath>
 
\lim_{n\rightarrow \infty}\frac {15,000n}{4(n + 1)} = \lim_{n\rightarrow \infty}\frac {15,000n}{4n} = \frac {15,000}{4} = 3750
 
\lim_{n\rightarrow \infty}\frac {15,000n}{4(n + 1)} = \lim_{n\rightarrow \infty}\frac {15,000n}{4n} = \frac {15,000}{4} = 3750

Revision as of 18:12, 23 March 2008

Problem

On a long straight stretch of one-way single-lane highway, cars all travel at the same speed and all obey the safety rule: the distance form the back of the car ahead to the front of the car behind is exactly one car length for each 15 kilometers per hour of speed or fraction thereof (Thus the front of a car traveling 52 kilometers per hour will be four car lengths behind the back of the car in front of it.) A photoelectric eye by the side of the road counts the number of cars that pass in one hour. Assuming that each car is 4 meters long and that the cars can travel at any speed, let $M$ be the maximum whole number of cars that can pass the photoelectric eye in one hour. Find the quotient when $M$ is divided by $10$.

Solution

Let $n$ be the number of car lengths that separate the cars. Then their speed is $< 15n$, and the distance between the cars (front to front) is $4(n + 1)$. Hence, the number of cars that pass in an hour is $\frac {15,000n\frac {meters}{hour}}{4(n + 1)\frac {meters}{car}} = \frac {15,000n}{4(n + 1)}\frac {cars}{hour}$. We wish to maximize this. Observe that as $n$ gets larger, $n + 1$ gets less and less significant, so we take the limit as $n$ approaches infinity \[\lim_{n\rightarrow \infty}\frac {15,000n}{4(n + 1)} = \lim_{n\rightarrow \infty}\frac {15,000n}{4n} = \frac {15,000}{4} = 3750\] Now, as the speeds are clearly finite, we can never actually reach $3750$. However, we can stop the camera after the $3750$th car has passed, but not the space behind it, so $3750$ cars is possible. Hence, our answer is $\boxed {375}$.

See also

2008 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions