2008 AIME I Problems/Problem 12

Revision as of 13:32, 23 March 2008 by Azjps (talk | contribs) (solution by SmitSchu)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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 $\frac {15,000n\frac {meters}{hour}}{4(n + 1)\frac {meters}{car}} = \frac {15,000}{4(n + 1)}\frac {cars}{hour}$. 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 3750th 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