Difference between revisions of "2003 AMC 12A Problems/Problem 4"
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<math> \mathrm{(A) \ } 3\qquad \mathrm{(B) \ } 3.125\qquad \mathrm{(C) \ } 3.5\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } 4.5 </math> | <math> \mathrm{(A) \ } 3\qquad \mathrm{(B) \ } 3.125\qquad \mathrm{(C) \ } 3.5\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } 4.5 </math> | ||
| − | + | ==Solution 1== | |
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Since she walked <math>1</math> km to school and <math>1</math> km back home, her total distance is <math>1+1=2</math> km. | Since she walked <math>1</math> km to school and <math>1</math> km back home, her total distance is <math>1+1=2</math> km. | ||
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Therefore her average speed in km/hr is <math>\frac{2}{\frac{2}{3}}=\boxed{\mathrm{(A)}\ 3}</math>. | Therefore her average speed in km/hr is <math>\frac{2}{\frac{2}{3}}=\boxed{\mathrm{(A)}\ 3}</math>. | ||
| − | + | ==Solution 2== | |
The average speed of two speeds that travel the same distance is the [[harmonic mean]] of the speeds, or <math>\dfrac{2}{\dfrac{1}{x}+\dfrac{1}{y}}=\dfrac{2xy}{x+y}</math> (for speeds <math>x</math> and <math>y</math>). Mary's speed going to school is <math>2\,\text{km/hr}</math>, and her speed coming back is <math>6\,\text{km/hr}</math>. Plugging the numbers in, we get that the average speed is <math>\dfrac{2\times 6\times 2}{6+2}=\dfrac{24}{8}=\boxed{\mathrm{(A)}\ 3}</math>. | The average speed of two speeds that travel the same distance is the [[harmonic mean]] of the speeds, or <math>\dfrac{2}{\dfrac{1}{x}+\dfrac{1}{y}}=\dfrac{2xy}{x+y}</math> (for speeds <math>x</math> and <math>y</math>). Mary's speed going to school is <math>2\,\text{km/hr}</math>, and her speed coming back is <math>6\,\text{km/hr}</math>. Plugging the numbers in, we get that the average speed is <math>\dfrac{2\times 6\times 2}{6+2}=\dfrac{24}{8}=\boxed{\mathrm{(A)}\ 3}</math>. | ||
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== See Also == | == See Also == | ||
{{AMC10 box|year=2003|ab=A|num-b=3|num-a=5}} | {{AMC10 box|year=2003|ab=A|num-b=3|num-a=5}} | ||
Latest revision as of 03:55, 7 November 2020
- The following problem is from both the 2003 AMC 12A #4 and 2003 AMC 10A #4, so both problems redirect to this page.
Contents
Problem
It takes Anna 
minutes to walk uphill 
km from her home to school, but it takes her only 
minutes to walk from school to her home along the same route. What is her average speed, in km/hr, for the round trip?
Solution 1
Since she walked km to school and km back home, her total distance is 
km.
Since she spent minutes walking to school and minutes walking back home, her total time is 
minutes = 
hours.
Therefore her average speed in km/hr is 
.
Solution 2
The average speed of two speeds that travel the same distance is the harmonic mean of the speeds, or 
(for speeds 
and 
). Mary's speed going to school is 
, and her speed coming back is 
. Plugging the numbers in, we get that the average speed is 
.
See Also
| 2003 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 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 | ||
| 2003 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 3 |
Followed by Problem 5 |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
