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 ==
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==Solution 1==
===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.  
  
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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===
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==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 04:55, 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.

Problem

It takes Anna $30$ minutes to walk uphill $1$ km from her home to school, but it takes her only $10$ minutes to walk from school to her home along the same route. What is her average speed, in km/hr, for the round trip?

$\mathrm{(A) \ } 3\qquad \mathrm{(B) \ } 3.125\qquad \mathrm{(C) \ } 3.5\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } 4.5$

Solution 1

Since she walked $1$ km to school and $1$ km back home, her total distance is $1+1=2$ km.

Since she spent $30$ minutes walking to school and $10$ minutes walking back home, her total time is $30+10=40$ minutes = $\frac{40}{60}=\frac{2}{3}$ hours.

Therefore her average speed in km/hr is $\frac{2}{\frac{2}{3}}=\boxed{\mathrm{(A)}\ 3}$.

Solution 2

The average speed of two speeds that travel the same distance is the harmonic mean of the speeds, or $\dfrac{2}{\dfrac{1}{x}+\dfrac{1}{y}}=\dfrac{2xy}{x+y}$ (for speeds $x$ and $y$). Mary's speed going to school is $2\,\text{km/hr}$, and her speed coming back is $6\,\text{km/hr}$. Plugging the numbers in, we get that the average speed is $\dfrac{2\times 6\times 2}{6+2}=\dfrac{24}{8}=\boxed{\mathrm{(A)}\ 3}$.

See Also

2003 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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