Difference between revisions of "2023 AMC 10A Problems/Problem 1"
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<cmath>\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27</cmath> | <cmath>\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27</cmath> | ||
| − | ==Solution== | + | ==Solution 1== |
This is a Distance=Time<math>\times</math>Speed so let <math>x</math> be the time it takes to meet. We can write the following equation: | This is a Distance=Time<math>\times</math>Speed so let <math>x</math> be the time it takes to meet. We can write the following equation: | ||
<cmath>12x+18x=45</cmath> | <cmath>12x+18x=45</cmath> | ||
Revision as of 14:41, 9 November 2023
Cities 
and 
are 
miles apart. Alicia lives in and Beth lives in . Alicia bikes towards at 18 miles per hour. Leaving at the same time, Beth bikes toward at 12 miles per hour. How many miles from City will they be when they meet?
![\[\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27\]](http://latex.artofproblemsolving.com/c/8/f/c8fdd15175bf5df101619ac5340f97dca9e75ac1.png)
Solution 1
This is a Distance=Time
Speed so let 
be the time it takes to meet. We can write the following equation:
![\[12x+18x=45\]](http://latex.artofproblemsolving.com/0/2/6/026f6ef1d38420553db4da955222cbf7eeb0952c.png)
Solving gives is 
. The 
is Alicia so 
~zhenghua
See Also
| 2022 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 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. 
