Difference between revisions of "2023 AMC 10A Problems/Problem 1"

Revision as of 15:39, 9 November 2023

Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B$ . Alicia bikes towards $B$ at 18 miles per hour. Leaving at the same time, Beth bikes toward $A$ at 12 miles per hour. How many miles from City $A$ will they be when they meet?

$\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27$

Solution

This is a Distance=Time $\times$ Speed so let $x$ be the time it takes to meet. We can write the following equation: $12x+18x=45$ Solving gives is $x=1.5$ . The $18x$ is Alicia so $18\times1.5=\boxed{\text{(E) 27}}$

~zhenghua

@@ Line 1: / Line 1: @@
-I’ll admit that I don’t have any jokes on calculus... but at least I know my limits.
+Cities <math>A</math> and <math>B</math> are <math>45</math> miles apart. Alicia lives in <math>A</math> and Beth lives in <math>B</math>. Alicia bikes towards <math>B</math> at 18 miles per hour. Leaving at the same time, Beth bikes toward <math>A</math> at 12 miles per hour. How many miles from City <math>A</math> will they be when they meet?
+<cmath>\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27</cmath>
+==Solution==
+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>
+Solving gives is <math>x=1.5</math>. The <math>18x</math> is Alicia so <math>18\times1.5=\boxed{\text{(E) 27}}</math>
+~zhenghua

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Difference between revisions of "2023 AMC 10A Problems/Problem 1"

Revision as of 15:39, 9 November 2023

Solution