Difference between revisions of "2002 AMC 12A Problems/Problem 11"
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===Solution 2=== | ===Solution 2=== | ||
| − | Since either time he arrives at is <math>3</math> minutes from the desired time, the answer is merely the [[harmonic mean]] of 40 and 60. | + | Since either time he arrives at is <math>3</math> minutes from the desired time, the answer is merely the [[harmonic mean]] of 40 and 60. |
| − | 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> | + | 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{\textbf{(B) }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 weighted sum in step two, and hence obtain a weighted harmonic mean in step three.) | (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.) | ||
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| + | ===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, <math>\frac x{40} -3 = y</math> and <math>\frac x{60} +3 = y</math> | ||
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| + | Solving, we get x = 720 and y = 15. | ||
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| + | 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>. | ||
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| + | ~MathKatana | ||
== Video Solution == | == Video Solution == | ||
https://youtu.be/ZpYOnfqm5Bc | https://youtu.be/ZpYOnfqm5Bc | ||
| + | |||
| + | ==Video Solution by Daily Dose of Math== | ||
| + | |||
| + | https://youtu.be/S17zidbQYWo | ||
| + | |||
| + | ~Thesmartgreekmathdude | ||
==See Also== | ==See Also== | ||
Latest revision as of 11:34, 10 August 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
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
Video Solution by Daily Dose of Math
~Thesmartgreekmathdude
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. 
