Difference between revisions of "2002 AMC 10A Problems/Problem 12"
(New page: == Problem == Mr. 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...) |
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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== | + | |
| + | ==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 <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-2</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-2</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== | + | ===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>. | ||
Revision as of 21:55, 26 December 2008
Problem
Mr. 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 
.
See Also
| 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 | ||
