Difference between revisions of "2002 AMC 12A Problems/Problem 11"
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(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.) | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/ZpYOnfqm5Bc | ||
==See Also== | ==See Also== |
Revision as of 20:44, 25 August 2022
- 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. 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.)
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.