AMC 10 · 2023 · #1

Grade 6 rate-ratio
ratelinear-equations-one-varunit-conversion dimensional-analysisidentify-subproblems ↑ Prerequisites: ratemulti-digit-arithmetic
📏 Short solution 💡 2 insights
Problem
Two cyclists start at the same time from two cities 45 miles apart and pedal toward each other. One goes 18 miles per hour, the other 12. Find how far from the first city they meet.

Pick an answer.

(A)
20
(B)
24
(C)
25
(D)
26
(E)
27
How to solve
Strategy Draw a Diagram

The problem describes positions on a road and motion along that road — Tool #1 (Draw a Diagram) is the natural lead. A simple labeled segment from A to B with arrows showing the two bikers makes the structure obvious: the gap of 45 miles closes at the combined speed 18 + 12 = 30 mph. Tool #8 (Analyze the Units) is the verification companion — tracking miles, mph, and hours through every multiplication makes sure the final number is in miles, which is what the problem asks for. Algebra (Tool #13) would also work but the diagram-plus-units path is faster and more visual for a Grade 6 rate concept.

1STEP 1

Draw the picture

They ride toward each other.

A → … 12 mph B, |AB| = 45
2STEP 2

How fast the gap closes

Facing each other, the speeds add.

18 + 12 = 30 mph
3STEP 3

Find the meeting time

Divide the distance by that speed.

t = (45 mi)/(30 mi/hr) = 1.5 hr
4STEP 4

Find one rider's distance

Multiplying gives 27 miles.

d_A = 18 mi/hr × 1.5 hr = 27 mi → (E)
Answer
27
Cross-check from Beth's side. In 1.5 hours Beth covers 12 × 1.5 = 18 miles from B toward A. Alicia's 27 miles plus Beth's 18 miles is 27 + 18 = 45 miles — the full distance, so they really do meet at that spot. Also, since Alicia is the faster rider, the meeting point should sit past the midpoint (22.5 mi) on the B side; 27 > 22.5 confirms this.
💡Key takeaway

This AMC 12 problem only needs Grade 6 unit rates you already know — speeds add when two riders close in, then distance equals speed times time!