2025 AMC 10A Problems/Problem 1: Difference between revisions

Revision as of 13:13, 6 November 2025

Problem

Andy and Betsy both live in Mathville. Andy leaves Mathville on his bicycle at $1:30$ , traveling due northat a steady $8$ mile per hour. Betsy leaves on her bicycle from the same point at $2:30$ , traveling due east at a steady $12$ miles per hour. At what time will they be exactly the same distance from their common starting point?

$\textbf{(A) } {3:30}\qquad\textbf{(B) } {3:45}\qquad\textbf{(C) } {4:00}\qquad\textbf{(D) } {4:15}\qquad\textbf{(E) } {4:30}$

Solution 1

We can see that Betsy travles 1 hour after Andy started. We have $lcm(8, 12)=24$ now we can find the time traveled $\frac{24}{8} = 3 \text{ hours}$

Now we have time $1{:}30 + 3{:}00 = \boxed{\textbf{(E) } 4{:}30}$

-Boywithnuke(Goal: 10 followers)

Solution 2

$h$ hours after Betsy left, Andy has traveled $8(h+1)$ miles, and Betsy has traveled $12h$ miles. We are told these are equal, so $8h+8=12h$ . Solving, we get $h=2$ , so Andy and Betsy will be exactly the same distance from their common starting point two hours after Betsy leaves, or $\text{(E) }4:30$ .

~mithu542

Solution 3

we can just use all the answer choices that we are given in order to see that $\text{(E) }4:30$ . is the correct answer.

~vgarg

@@ Line 17: / Line 17: @@
 ~mithu542
+==Solution 3==
+we can just use all the answer choices that we are given in order to see that <imath>\text{(E) }4:30</imath>. is the correct answer.
+~vgarg

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