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Difference between revisions of "2014 AIME I Problems/Problem 4"

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Multiplying this out and simplifying, we get that <math>r_1 = 10r_2 - \dfrac{11}{3}</math>. Since we now have 2 expressions for <math>r_1</math> in terms of <math>r_2</math>, we can set them equal to each other:
 
Multiplying this out and simplifying, we get that <math>r_1 = 10r_2 - \dfrac{11}{3}</math>. Since we now have 2 expressions for <math>r_1</math> in terms of <math>r_2</math>, we can set them equal to each other:
  
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<math>r_2 + \dfrac{2}{3} = 10r_2 - \dfrac{11}{3}</math>. Solving for <math>r_2</math>, we get that <math>r_2 = \dfrac{13}{27}</math>. Since we know that it took train 2 1 minute to pass Jon, we know that <math>1 = \dfrac{d}{r_2 + \dfrac{1}{3}}</math>. Plugging in <math>\dfrac{11}{3}</math> for <math>r_2</math> and solving for <math>d</math>, we get that <math>d = \dfrac{22}{27}</math>, and our answer is <math>27 + 22 = 049</math>.
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<math>r_2 + \dfrac{2}{3} = 10r_2 - \dfrac{11}{3}</math>. Solving for <math>r_2</math>, we get that <math>r_2 = \dfrac{13}{27}</math>. Since we know that it took train 2 1 minute to pass Jon, we know that <math>1 = \dfrac{d}{r_2 + \dfrac{1}{3}}</math>. Plugging in <math>\dfrac{11}{3}</math> for <math>r_2</math> and solving for <math>d</math>, we get that <math>d = \dfrac{22}{27}</math>, and our answer is <math>27 + 22 = \boxed{049}</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2014|n=I|num-b=3|num-a=5}}
 
{{AIME box|year=2014|n=I|num-b=3|num-a=5}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 11:55, 15 March 2014

Problem 4

Jon and Steve ride their bicycles along a path that parallels two side-by-side train tracks running the east/west direction. Jon rides east at $20$ miles per hour, and Steve rides west at $20$ miles per hour. Two trains of equal length, traveling in opposite directions at constant but different speeds each pass the two riders. Each train takes exactly $1$ minute to go past Jon. The westbound train takes $10$ times as long as the eastbound train to go past Steve. The length of each train is $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

For the purposes of this problem, we will use miles and minutes as our units; thus, the bikers travel at speeds of $\dfrac{1}{3}$ mi/min.

Let $d$ be the length of the trains, $r_1$ be the speed of train 1 (the faster train), and $r_2$ be the speed of train 2.

Consider the problem from the bikers' moving frame of reference. In order to pass Jon, the first train has to cover a distance equal to its own length, at a rate of $r_1 - \dfrac{1}{3}$. Similarly, the second train has to cover a distance equal to its own length, at a rate of $r_2 + \dfrac{1}{3}$. Since the times are equal and $d = rt$, we have that $\dfrac{d}{r_1 - \dfrac{1}{3}} = \dfrac{d}{r_2 + \dfrac{1}{3}}$. Solving for $r_1$ in terms of $r_2$, we get that $r_1 = r_2 + \dfrac{2}{3}$.

Now, let's examine the times it takes the trains to pass Steve. This time, we augment train 1's speed by $\dfrac{1}{3}$, and decrease train 2's speed by $\dfrac{1}{3}$. Thus, we have that $\dfrac{d}{r_2 - \dfrac{1}{3}} = 10\dfrac{d}{r_1 + \dfrac{1}{3}}$.

Multiplying this out and simplifying, we get that $r_1 = 10r_2 - \dfrac{11}{3}$. Since we now have 2 expressions for $r_1$ in terms of $r_2$, we can set them equal to each other:

$r_2 + \dfrac{2}{3} = 10r_2 - \dfrac{11}{3}$. Solving for $r_2$, we get that $r_2 = \dfrac{13}{27}$. Since we know that it took train 2 1 minute to pass Jon, we know that $1 = \dfrac{d}{r_2 + \dfrac{1}{3}}$. Plugging in $\dfrac{11}{3}$ for $r_2$ and solving for $d$, we get that $d = \dfrac{22}{27}$, and our answer is $27 + 22 = \boxed{049}$.

See also

2014 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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All AIME Problems and Solutions

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