AMC 10 · 2024 · #2

Grade 8 algebrarate-ratio

Archetype Arithmetic Expression Decomposition └ Sub-archetype Linear Rate Accumulation

linear-equations-two-varsystems-of-equationsrateformula-substitution convert-to-algebraidentify-subproblems ↑ Prerequisites: linear-equations-one-varmulti-digit-arithmetic

📏 Short solution 💡 2 insights

Problem

A hiking-time model is T = aL + bG, where L is trail length in miles, G is altitude gain in feet, and T is time in minutes. Two different trails both take 69 minutes: one is 1.5 mi with 800 ft gain, the other is 1.2 mi with 1100 ft gain. Using the same model, find T for a trail that is 4.2 mi long with 4000 ft gain.

Pick an answer.

(A)

240

(B)

246

(C)

252

(D)

258

(E)

264

AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Convert to Algebra

The problem hands us a formula with two unknown constants and two data points, which is the textbook setup for Tool #13 (Convert to Algebra) — translate each trail into one linear equation, then solve the 2× 2 system. Tool #7 (Identify Subproblems) splits the work into two clean parts: (i) recover the constants a and b from Trails 1 and 2, then (ii) plug them into T = aL + bG for the target trail. The two scenarios share the same T = 69, so subtracting (or equating) the equations cancels T immediately and gives a one-line relation between a and b.

1STEP 1

Subproblem 1: plug each trail's (L, G, T) into T = aL + bG to get one linear equation per trail.

1.5a + 800b = 69 & (Trail 1) ; 1.2a + 1100b = 69 & (Trail 2)

2STEP 2

Both equations equal the same time, so subtract them; the constant cancels, leaving a = 1000b.

1.5a + 800b = 1.2a + 1100b → 0.3a = 300b → a = 1000b

3STEP 3

Substitute a = 1000b into Trail 1: 2300b = 69, giving b = $\frac{3}{100}$ and a = 30.

1.5(1000b) + 800b = 69 → 2300b = 69 → b =

\frac{3}{100}

, a = 1000b = 30

4STEP 4

Subproblem 2: plug the constants and the new trail's (4.2, 4000) into the model to get T = 246.

T = 30(4.2) +

\frac{3}{100}

(4000) = 126 + 120 = 246 → (B)

Answer

246

Sanity-check the constants on Trail 2: 1.2(30) + 1100(0.03) = 36 + 33 = 69. Matches. The numbers also pass a units check: a = 30 min/mile and b = 0.03 min/ft, so a longer, steeper trail naturally takes more time. For the target, 4.2 mi is about 3× longer than Trail 1 and 4000 ft is 5× steeper, so a time of 246 ≈ 3.5 × 69 is in the right ballpark.

💡Key takeaway

Two trails with the same time give two equations in the same two unknowns — solve the little system once, and the model is yours to use on any new trail.

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