Difference between revisions of "2021 Fall AMC 10A Problems/Problem 11"
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Since Emily takes 210 steps to walk along with the boat and 42 steps to walk opposite the boat, that means it takes her 5x longer to walk the length of a stationary boat at <math>v_e-v_s</math> compared to <math>v_e+v_s</math>. | Since Emily takes 210 steps to walk along with the boat and 42 steps to walk opposite the boat, that means it takes her 5x longer to walk the length of a stationary boat at <math>v_e-v_s</math> compared to <math>v_e+v_s</math>. | ||
This means that <math>5(v_e-v_s)=v_e+v_s \rightarrow v_s = \frac{2v_e}{3}</math> | This means that <math>5(v_e-v_s)=v_e+v_s \rightarrow v_s = \frac{2v_e}{3}</math> | ||
− | As Emily takes 210 steps to walk the length of the boat at a speed of <math>v_e- \frac{2v_e}{3}=\frac{v_e}{3}</math>, she must take 1/3rd of the time to walk the length of the boat at a speed of <math>v_e</math>, so our answer is 210/3 | + | As Emily takes 210 steps to walk the length of the boat at a speed of <math>v_e- \frac{2v_e}{3}=\frac{v_e}{3}</math>, she must take 1/3rd of the time to walk the length of the boat at a speed of <math>v_e</math>, so our answer is <math>210/3 \rightarrow \boxed{A) 70}</math> |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2021 Fall|ab=A|num-b=10|num-a=12}} | {{AMC10 box|year=2021 Fall|ab=A|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 23:47, 27 November 2021
Contents
Problem
Emily sees a ship traveling at a constant speed along a straight section of a river. She walks parallel to the riverbank at a uniform rate faster than the ship. She counts equal steps walking from the back of the ship to the front. Walking in the opposite direction, she counts
steps of the same size from the front of the ship to the back. In terms of Emily's equal steps, what is the length of the ship?
Solution 1 (One Variable)
Let be the length of the ship.
Then, in the time that Emily walks
steps, the ship moves
steps.
Also, in the time that Emily walks
steps, the ship moves
steps.
Since the ship and Emily both travel at some constant rate,
. Dividing both sides by
and cross multiplying, we get
, so
, and
.
~ihatemath123
Solution 2 (Two Variables)
Let the speed at which Emily walks be steps per hour. Let the speed at which the ship is moving be
. Walking in the direction of the ship, it takes her
steps, or
hours, to travel. We can create an equation:
where
is the length of the ship. Walking in the opposite direction of the ship, it takes her
steps, or
hour. We can create a similar equation:
Now we have two variables and two equations. We can equate the expressions for
and solve for
:
Therefore, we have
.
~LucaszDuzMatz (Solution)
~Arcticturn (Minor Edits)
Solution 3 (Three Variables)
Suppose that Emily and the ship take steps simultaneously such that Emily's steps cover a greater length than the ship's steps.
Let be the length of the ship,
be Emily's step length, and
be the ship's step length. We wish to find
When Emily walks from the back of the ship to the front, she walks a distance of and the front of the ship moves a distance of
We have
for this scenario, which rearranges to
When Emily walks in the opposite direction, she walks a distance of
and the back of the ship moves a distance of
We have
for this scenario, which rearranges to
We multiply
by
and then add
to the result:
from which
~Steven Chen (www.professorchenedu.com)
~MRENTHUSIASM
Solution 4 (Relative Speeds)
Call the speed of the boat and the speed of Emily
.
Consider the scenario when Emily is walking along with the boat. Relative to an observer on the boat, her speed is
.
Consider the scenario when Emily is walking in the opposite direction. Relative to an observer on the boat, her speed is
Since Emily takes 210 steps to walk along with the boat and 42 steps to walk opposite the boat, that means it takes her 5x longer to walk the length of a stationary boat at
compared to
.
This means that
As Emily takes 210 steps to walk the length of the boat at a speed of
, she must take 1/3rd of the time to walk the length of the boat at a speed of
, so our answer is
See Also
2021 Fall AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.