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: | ||
− | <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{ | + | <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>. |
== 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 23:47, 24 March 2015
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
, where
and
are relatively prime positive integers. Find
.
Solution
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
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.