Difference between revisions of "2002 AMC 12A Problems/Problem 11"
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Solving, we get x = 720 and y = 15. | Solving, we get x = 720 and y = 15. | ||
− | We divide x by y to get the average speed, <math>\frac 720{15} = 48</math>. Therefore, the answer is <math> | + | We divide x by y to get the average speed, <math>\frac {720}{15} = 48</math>. Therefore, the answer is <math>\boxed{\textbf{(B) }48}</math>. |
~MathKatana | ~MathKatana |
Revision as of 23:19, 31 January 2024
- The following problem is from both the 2002 AMC 12A #11 and 2002 AMC 10A #12, so both problems redirect to this page.
Contents
[hide]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 and the distance he travels be . 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 , our answer, is .
Solution 2
Since either time he arrives at is minutes from the desired time, the answer is merely the harmonic mean of 40 and 60. 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.)
Solution 3
Let x be equal to the total amount of distance he needs to cover. Let y be equal to the amount of time he would travel correctly. Setting up a system of equations, and
Solving, we get x = 720 and y = 15.
We divide x by y to get the average speed, . Therefore, the answer is .
~MathKatana
Video Solution
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.