Difference between revisions of "2021 Fall AMC 10A Problems/Problem 11"
(Created solution for 2021 AMC 10A Fall #11.) |
m (Added See Also) |
||
| Line 5: | Line 5: | ||
<cmath>d = 42 + s = \boxed{\textbf{(A) } 70} </cmath> | <cmath>d = 42 + s = \boxed{\textbf{(A) } 70} </cmath> | ||
~LucaszDuzMatz | ~LucaszDuzMatz | ||
| + | ==See Also== | ||
| + | {{AMC10 box|year=2021 Fall|ab=A|num-b=10|num-a=12}} | ||
| + | {{MAA Notice}} | ||
Revision as of 20:41, 22 November 2021
Solution
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
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. 
