AMC 10 · 2025 · #1

Grade 8 rate-ratio

Archetype Meeting Time with Piecewise Rates └ Sub-archetype Staggered Start or Schedule Table

ratedimensional-analysislinear-equations-one-var convert-to-algebra ↑ Prerequisites: rate

📏 Medium solution 💡 1 insight

Problem

Andy rides north from Mathville at 8 mph starting at 1{:}30. One hour later, at 2{:}30, Betsy rides east from the same point at 12 mph. Find the clock time when the two riders are the same straight-line distance from Mathville.

Pick an answer.

(A)

$3{:}30$

(B)

$3{:}45$

(C)

$4{:}00$

(D)

$4{:}15$

(E)

$4{:}30$

How to solve

Strategy Introduce a Variable

Both riders travel at constant speed, and the question compares their distances at one shared instant, so name a single time variable (Tool #4) that ties the two motions together. Tool #8 (Analyze the Units) turns each rider's speed and elapsed time into a distance, and Tool #13 (Convert to Algebra) sets the two distance expressions equal and solves the resulting linear equation. The perpendicular directions matter only to justify that straight-line distance equals distance ridden; no Pythagorean work is needed.

1STEP 1

Distance equals rate times time

Each rider heads straight out, so distance from Mathville equals distance ridden — speed times time spent riding.

distance = rate × time

2STEP 2

Name the time and write both distances

Let t be hours since Andy's 1{:}30 start. Andy rides 8t miles; Betsy, gone since 2{:}30, rides 12(t-1) miles (valid for t at least 1).

Andy = 8t, Betsy = 12(t-1)

3STEP 3

Set the distances equal and solve

Set 8t = 12(t-1). Expand: 8t = 12t - 12, so 4t = 12 and t = 3 — 3 hours after 1{:}30, i.e. 4{:}30, choice (E).

8t = 12(t-1) → 8t = 12t - 12 → 4t = 12 → t = 3 → 1{:}30 + 3{:}00 = 4{:}30

Answer

4{:}30

Plug t = 3 back in: Andy has ridden 8 × 3 = 24 miles, and Betsy has ridden 12 × (3-1) = 12 × 2 = 24 miles. Both are exactly 24 miles from Mathville, so the distances truly match at 4{:}30. The answer also makes sense directionally: Betsy is faster but starts an hour behind, so she needs some time to catch Andy's lead, landing later in the afternoon rather than right after she departs.

💡Key takeaway

When two things move at steady speeds, write each distance as speed times time with one shared clock, then set the distances equal and solve for the time.

Distance equals rate times time
Name the time and write both distances
Set the distances equal and solve

Distance equals rate times time

Name the time and write both distances

Set the distances equal and solve

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