Difference between revisions of "2002 AMC 12A Problems/Problem 11"
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<math>\text{(A)}\ 45 \qquad \text{(B)}\ 48 \qquad \text{(C)}\ 50 \qquad \text{(D)}\ 55 \qquad \text{(E)} 58</math> | <math>\text{(A)}\ 45 \qquad \text{(B)}\ 48 \qquad \text{(C)}\ 50 \qquad \text{(D)}\ 55 \qquad \text{(E)} 58</math> | ||
− | + | ==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 <math>d=\left(t+\frac{1}{20}\right)40</math> and <math>d=\left(t-\frac{1}{20}\right)60</math>. Setting the two equal, we have <math>40t+2=60t-3</math> and we find <math>t=\frac{1}{4}</math> of an hour. Substituting t back in, we find <math>d=12</math>. From <math>d=rt</math>, we find that r, and our answer, is <math>\boxed{\text{(B)}\ 48 }</math>. | 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 <math>d=\left(t+\frac{1}{20}\right)40</math> and <math>d=\left(t-\frac{1}{20}\right)60</math>. Setting the two equal, we have <math>40t+2=60t-3</math> and we find <math>t=\frac{1}{4}</math> of an hour. Substituting t back in, we find <math>d=12</math>. From <math>d=rt</math>, we find that r, and our answer, is <math>\boxed{\text{(B)}\ 48 }</math>. | ||
− | + | ==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 <math>\frac{2}{\frac{1}{a}+\frac{1}{b}}=\frac{2ab}{a+b}</math>. In this case, a and b are 40 and 60, so our answer is <math>\frac{4800}{100}=48</math>, so <math>\boxed{\text{(B)}\ 48}</math>. | 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 <math>\frac{2}{\frac{1}{a}+\frac{1}{b}}=\frac{2ab}{a+b}</math>. In this case, a and b are 40 and 60, so our answer is <math>\frac{4800}{100}=48</math>, so <math>\boxed{\text{(B)}\ 48}</math>. | ||
− | + | ==Solution 3== | |
A more general form of the argument in Solution 2, with proof: | A more general form of the argument in Solution 2, with proof: |
Revision as of 21:28, 1 August 2011
- 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 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 $t+\frac 3{60} = \fracd{40}$ (Error compiling LaTeX. Unknown error_msg) 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 weighed sum in step two, and hence obtain a weighed 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 |