Difference between revisions of "2002 AMC 12A Problems/Problem 11"
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Substituting <math>t=\frac ds</math> and dividing both sides by <math>d</math>, we get <math>\frac 2s = \frac 1{40} + \frac 1{60}</math>, hence <math>s=\boxed{48}</math>. | Substituting <math>t=\frac ds</math> and dividing both sides by <math>d</math>, we get <math>\frac 2s = \frac 1{40} + \frac 1{60}</math>, hence <math>s=\boxed{48}</math>. | ||
− | (Note that this approach would work even if the time by which he is late was different from the time by which he is early in the other case - we would simply take a | + | (Note that this approach would work even if the time by which he is late was different from the time by which he is early in the other case - we would simply take a weighted sum in step two, and hence obtain a weighted harmonic mean in step three.) |
==See Also== | ==See Also== |
Revision as of 00:31, 20 January 2014
- The following problem is from both the 2002 AMC 12A #11 and 2002 AMC 10A #12, so both problems redirect to this page.
Problem
Mr. Earl E. Bird gets up every day at 8:00 AM to go to work. If he drives at an average speed of 40 miles per hour, he will be late by 3 minutes. If he drives at an average speed of 60 miles per hour, he will be early by 3 minutes. How many miles per hour does Mr. Bird need to drive to get to work exactly on time?
Solution
Solution 1
Let the time he needs to get there in be t and the distance he travels be d. From the given equations, we know that and . Setting the two equal, we have and we find of an hour. Substituting t back in, we find . From , we find that r, and our answer, is .
Solution 2
Since either time he arrives at is 3 minutes from the desired time, the answer is merely the harmonic mean of 40 and 60. The harmonic mean of a and b is . In this case, a and b are 40 and 60, so our answer is , so .
Solution 3
A more general form of the argument in Solution 2, with proof:
Let be the distance to work, and let be the correct average speed. Then the time needed to get to work is .
We know that and . Summing these two equations, we get: .
Substituting and dividing both sides by , we get , hence .
(Note that this approach would work even if the time by which he is late was different from the time by which he is early in the other case - we would simply take a weighted sum in step two, and hence obtain a weighted harmonic mean in step three.)
See Also
2002 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 10 |
Followed by Problem 12 |
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 | |
All AMC 12 Problems and Solutions |
2002 AMC 10A (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 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.