Difference between revisions of "2003 AMC 12A Problems/Problem 4"
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+ | {{duplicate|[[2003 AMC 12A Problems|2003 AMC 12A #4]] and [[2003 AMC 10A Problems|2003 AMC 10A #4]]}} | ||
== Problem == | == Problem == | ||
It takes Mary <math>30</math> minutes to walk uphill <math>1</math> km from her home to school, but it takes her only <math>10</math> minutes to walk from school to her home along the same route. What is her average speed, in km/hr, for the round trip? | It takes Mary <math>30</math> minutes to walk uphill <math>1</math> km from her home to school, but it takes her only <math>10</math> minutes to walk from school to her home along the same route. What is her average speed, in km/hr, for the round trip? | ||
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== Solution == | == Solution == | ||
+ | ===Solution 1=== | ||
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. | ||
Since she spent <math>30</math> minutes walking to school and <math>10</math> minutes walking back home, her total time is <math>30+10=40</math> minutes = <math>\frac{40}{60}=\frac{2}{3}</math> hours. | Since she spent <math>30</math> minutes walking to school and <math>10</math> minutes walking back home, her total time is <math>30+10=40</math> minutes = <math>\frac{40}{60}=\frac{2}{3}</math> hours. | ||
− | Therefore her average speed in km/hr is <math>\frac{2}{\frac{2}{3}}= | + | 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>. | ||
== See Also == | == See Also == | ||
− | + | {{AMC10 box|year=2003|ab=A|num-b=3|num-a=5}} | |
{{AMC12 box|year=2003|ab=A|num-b=3|num-a=5}} | {{AMC12 box|year=2003|ab=A|num-b=3|num-a=5}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] |
Revision as of 16:23, 29 July 2011
- The following problem is from both the 2003 AMC 12A #4 and 2003 AMC 10A #4, so both problems redirect to this page.
Problem
It takes Mary 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
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 |