Difference between revisions of "2023 AMC 12A Problems/Problem 1"
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This is a Distance=Time<math>\times</math>Speed problem, 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 problem, 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> | ||
− | Solving gives us <math>x=1.5</math>. The <math>18x</math> is Alicia so <math>18\times1.5=\boxed{\ | + | Solving gives us <math>x=1.5</math>. The <math>18x</math> is Alicia so <math>18\times1.5=\boxed{\textbf{(E) 27}}</math> |
~zhenghua | ~zhenghua |
Revision as of 20:08, 9 November 2023
Contents
Problem
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?
Solution 1
This is a Distance=TimeSpeed problem, so let be the time it takes to meet. We can write the following equation: Solving gives us . The is Alicia so
~zhenghua
Solution 2
We can solve this by first adding 18 and 12 mph, which gets us 30mph. Their collective speed is 30mph, which means they will mean when they travel 45 miles at 30mph which would take 90 minutes. Then 18 miles per hour for 90 minutes is 27 miles. (E)
~walmartbrian
See also
2023 AMC 12A (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 12 Problems and Solutions |
2023 AMC 10A (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 |
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