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Difference between revisions of "2014 AIME I Problems/Problem 4"
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== Problem 4 ==
 
== Problem 4 ==
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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 <math>20</math> miles per hour, and Steve rides west at <math>20</math> 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 <math>1</math> minute to go past Jon. The westbound train takes <math>10</math> times as long as the eastbound train to go past Steve. The length of each train is <math>\tfrac{m}{n}</math> miles, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
  
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== Solution ==
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== Solution 1 ==
 +
For the purposes of this problem, we will use miles and minutes as our units; thus, the bikers travel at speeds of <math>\dfrac{1}{3}</math> mi/min.
  
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Let <math>d</math> be the length of the trains, <math>r_1</math> be the speed of train 1 (the faster train), and <math>r_2</math> 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 <math>r_1 - \dfrac{1}{3}</math>. Similarly, the second train has to cover a distance equal to its own length, at a rate of <math>r_2 + \dfrac{1}{3}</math>. Since the times are equal and <math>d = rt</math>, we have that <math>\dfrac{d}{r_1 - \dfrac{1}{3}} = \dfrac{d}{r_2 + \dfrac{1}{3}}</math>. Solving for <math>r_1</math> in terms of <math>r_2</math>, we get that <math>r_1 = r_2 + \dfrac{2}{3}</math>.
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 +
Now, let's examine the times it takes the trains to pass Steve. This time, we augment train 1's speed by <math>\dfrac{1}{3}</math>, and decrease train 2's speed by <math>\dfrac{1}{3}</math>. Thus, we have that <math>\dfrac{d}{r_2 - \dfrac{1}{3}} = 10\dfrac{d}{r_1 + \dfrac{1}{3}}</math>.
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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:
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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{13}{27}</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>.
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==Solution 2==
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Using a similar approach to Solution 1, let the speed of the east bound train be <math>a</math> and the speed of the west bound train be <math>b</math>.
 +
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So <math>a-20=b+20</math> and <math>a+20=10(b-20)</math>.
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From the first equation, <math>a=b+40</math>. Substituting into the second equation,
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<cmath>b+60=10b-200</cmath>
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<cmath>260=9b</cmath>
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<cmath>b=\frac{260}{9}\text{ mph}</cmath>
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This means that
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<cmath>a=\frac{260}{9}+40=\frac{620}{9}\text{ mph}</cmath>
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Checking, we get that the common difference in Jon's speed and trains' speeds is <math>\frac{440}{9}</math> and the differences for Steve is <math>\frac{800}{9}</math> and <math>\frac{80}{9}</math>.
 +
 +
This question assumes the trains' lengths in MILES:
 +
<cmath>\frac{440}{9}\cdot \frac{1}{60}=\frac{440}{540}=\frac{22}{27}\text{ miles}</cmath>
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Adding up, we get <math>22+27=\boxed{049}</math>.
 +
 +
==Solution 3==
 +
Let the length of the trains be <math>L</math>, let the rate of the westward train be <math>W_{R}</math>, ;et the rate of the eastward train be <math>E_{R}</math>, and let the time it takes for the eastward train to pass Steve be <math>E_{T}</math>.
 +
 +
We have that
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<math>L=(\frac{1}{60})(W_{R}+20)</math>
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<math>L=(\frac{1}{60})(E_{R}-20)</math>.
 +
 +
Adding both of the equations together, we get that
 +
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<math>2L=\frac{W_{R}}{60}+\frac{E_{R}}{60}\implies 120L=W_{R}+E_{R}</math>.
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Now, from the second part of the problem, we acquire that
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<math>L=(E_{T})(E_{R}+20)</math>
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<math>L=(10E_{T})(W_{R}-20)</math>
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Dividing the second equation by the first, we get that...
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<math>1=\frac{10(W_{R}-20)}{E_{R}+20}\implies E_{R}+20=10W_{R}-200\implies E_{R}+220=10W_{R}\implies E_{R}=10W_{R}-220</math>.
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Now, substituting into the <math>120L=W_{R}+E_{R}</math>.
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<math>120L=W_{R}+(10W_{R}-220)\implies 120L= 11W_{R}-220\implies W_{R}=\frac{120L+220}{11}</math>.
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Finally, plugging this back into our very first equation..
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<math>L=(\frac{1}{60})((\frac{120L+220}{11})+20)\implies 660L=120L+440\implies 540L=440\implies L=\frac{22}{27}</math>.
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Hence, the answer is <math>22+27=\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}}

Latest revision as of 12:59, 30 November 2021

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 $\tfrac{m}{n}$ miles, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution 1

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 $\dfrac{13}{27}$ for $r_2$ and solving for $d$, we get that $d = \dfrac{22}{27}$, and our answer is $27 + 22 = \boxed{049}$.

Solution 2

Using a similar approach to Solution 1, let the speed of the east bound train be $a$ and the speed of the west bound train be $b$.

So $a-20=b+20$ and $a+20=10(b-20)$.

From the first equation, $a=b+40$. Substituting into the second equation, \[b+60=10b-200\] \[260=9b\] \[b=\frac{260}{9}\text{ mph}\] This means that \[a=\frac{260}{9}+40=\frac{620}{9}\text{ mph}\] Checking, we get that the common difference in Jon's speed and trains' speeds is $\frac{440}{9}$ and the differences for Steve is $\frac{800}{9}$ and $\frac{80}{9}$.

This question assumes the trains' lengths in MILES: \[\frac{440}{9}\cdot \frac{1}{60}=\frac{440}{540}=\frac{22}{27}\text{ miles}\] Adding up, we get $22+27=\boxed{049}$.

Solution 3

Let the length of the trains be $L$, let the rate of the westward train be $W_{R}$, ;et the rate of the eastward train be $E_{R}$, and let the time it takes for the eastward train to pass Steve be $E_{T}$.

We have that

$L=(\frac{1}{60})(W_{R}+20)$

$L=(\frac{1}{60})(E_{R}-20)$.

Adding both of the equations together, we get that

$2L=\frac{W_{R}}{60}+\frac{E_{R}}{60}\implies 120L=W_{R}+E_{R}$.

Now, from the second part of the problem, we acquire that

$L=(E_{T})(E_{R}+20)$

$L=(10E_{T})(W_{R}-20)$

Dividing the second equation by the first, we get that... $1=\frac{10(W_{R}-20)}{E_{R}+20}\implies E_{R}+20=10W_{R}-200\implies E_{R}+220=10W_{R}\implies E_{R}=10W_{R}-220$.

Now, substituting into the $120L=W_{R}+E_{R}$.

$120L=W_{R}+(10W_{R}-220)\implies 120L= 11W_{R}-220\implies W_{R}=\frac{120L+220}{11}$.

Finally, plugging this back into our very first equation..

$L=(\frac{1}{60})((\frac{120L+220}{11})+20)\implies 660L=120L+440\implies 540L=440\implies L=\frac{22}{27}$.

Hence, the answer is $22+27=\boxed{049}$.

See also

2014 AIME I (Problems • Answer Key • Resources)
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