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