Difference between revisions of "2021 Fall AMC 10A Problems/Problem 11"
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<math>\textbf{(A) }70\qquad\textbf{(B) }84\qquad\textbf{(C) }98\qquad\textbf{(D) }105\qquad\textbf{(E) }126</math> | <math>\textbf{(A) }70\qquad\textbf{(B) }84\qquad\textbf{(C) }98\qquad\textbf{(D) }105\qquad\textbf{(E) }126</math> | ||
| − | ==Solution== | + | ==Solution 1== |
Let the speed at which Emily walks be 42 steps per hour. Let the speed at which the ship is moving be <math>s</math>. Walking in the direction of the ship, it takes her 210 steps, or 210/42 = 5 hours, to travel. We can create the equation: <cmath>d = (42-s)5</cmath> Where d is the length of the ship. Walking in the opposite direction of the ship, it takes her 42 steps, or 42/42 = 1 hour. We can create a similar equation: <cmath>d = (42+s)1</cmath> Now we have 2 variables and 2 equations, and we can solve for d. | Let the speed at which Emily walks be 42 steps per hour. Let the speed at which the ship is moving be <math>s</math>. Walking in the direction of the ship, it takes her 210 steps, or 210/42 = 5 hours, to travel. We can create the equation: <cmath>d = (42-s)5</cmath> Where d is the length of the ship. Walking in the opposite direction of the ship, it takes her 42 steps, or 42/42 = 1 hour. We can create a similar equation: <cmath>d = (42+s)1</cmath> Now we have 2 variables and 2 equations, and we can solve for d. | ||
<cmath>210-5s = 42 + s</cmath> | <cmath>210-5s = 42 + s</cmath> | ||
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~LucaszDuzMatz | ~LucaszDuzMatz | ||
| + | ==Solution 2== | ||
| + | Let <math>x</math> be the length of the ship. | ||
| + | Then, in the time that Emily walks <math>210</math> steps, the ship moves <math>210-x</math> steps. | ||
| + | Also, in the time that Emily walks <math>42</math> steps, the ship moves <math>x-42</math> steps. | ||
| + | Since the ship and Emily both travel at some constant rate, <math>\frac{210}{210-x} = \frac{42}{x-42}</math>. Dividing both sides by <math>42</math> and cross multiplying, we get <math>5(x-42) = 210-x</math>, so <math>6x = 420</math>, and <math>x = \boxed{70}</math>. | ||
| + | ~ihatemath123 | ||
==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 00:56, 23 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 tha 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
Let the speed at which Emily walks be 42 steps per hour. Let the speed at which the ship is moving be 
. Walking in the direction of the ship, it takes her 210 steps, or 210/42 = 5 hours, to travel. We can create the equation: ![\[d = (42-s)5\]](http://latex.artofproblemsolving.com/9/1/f/91f7d7a113ae1d49423fdf4dc29d99c6ee4c7cd5.png)
Where d is the length of the ship. Walking in the opposite direction of the ship, it takes her 42 steps, or 42/42 = 1 hour. We can create a similar equation: ![\[d = (42+s)1\]](http://latex.artofproblemsolving.com/e/e/a/eea581b75a83976ed940d5e83ad0784edbae0fd0.png)
Now we have 2 variables and 2 equations, and we can solve for d.
![\[210-5s = 42 + s\]](http://latex.artofproblemsolving.com/c/f/0/cf0778c9cdf9c0cac492a7fbd4bdce382a845f81.png)
![\[s = 28\]](http://latex.artofproblemsolving.com/7/2/0/720c9cb8e094b577c9ccdcab44264da4f8189442.png)
![\[d = 42 + s = \boxed{\textbf{(A) } 70}\]](http://latex.artofproblemsolving.com/e/d/3/ed348da71c524477b4005ac8a642f2e4d47aee75.png)
~LucaszDuzMatz
Solution 2
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
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. 
