AMC 8 · 2015 · #17

Grade 8 rate-ratioalgebra

Archetype Arithmetic Expression Decomposition └ Sub-archetype Linear Rate Accumulation

ratesystems-of-equationsunit-conversion convert-to-algebraidentify-subproblems ↑ Prerequisites: fraction-arithmeticlinear-equations-two-var

📏 Medium solution 💡 3 insights

Problem

Jeremy's dad drives him to school in 20 minutes during rush hour traffic. On a clear day his dad drives 18 mph faster and the trip takes only 12 minutes. The route is the same both days. How long is the route in miles?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Find a Variable

Two unknowns are floating around — the distance d and the rush-hour speed v — but the problem only asks for d. Tool #5 (Find a Variable) says: name both, write what each scenario tells you (d = v · t₁ and d = (v+18) · t₂), then use the shared distance to eliminate v and solve for d. Tool #8 (Analyze the Units) handles the trap that times are in minutes while speeds are in mph: convert 20 min and 12 min to hours first so every r · t product comes out in miles.

1STEP 1

Convert both times to hours to match the mph units: 20 min = $\frac{1}{3}$ hr and 12 min = $\frac{1}{5}$ hr.

t₁ =

\frac{20}{60}

\frac{1}{3}

hr, t₂ =

\frac{12}{60}

\frac{1}{5}

2STEP 2

Name the unknowns: let d be the distance and v the rush-hour speed, so the clear-day speed is v + 18.

d = miles to school, v = rush-hour speed (mph), v+18 = clear-day speed

3STEP 3

Use distance = speed × time for each day: rush hour gives d = $\frac{v}{3}$ , clear day gives d = $\frac{v+18}{5}$ , same d both times.

d =

\frac{v}{3}

, d =

\frac{v+18}{5}

4STEP 4

Set the two equal: from v = 3d and 5d = v + 18, substituting gives 5d = 3d + 18, so 2d = 18.

\frac{v}{3}

\frac{v+18}{5}

→ 5v = 3v + 54 → v = 27, d =

\frac{27}{3}

= 9

5STEP 5

Read off the answer: the distance to school is d = 9 miles, which is choice (D).

d = 9 mi → (D)

Answer

Check both scenarios with d = 9 miles. Rush hour: v = 27 mph for

\frac{1}{3}

hr gives 27 ×

\frac{1}{3}

= 9 miles. Clear day: v+18 = 45 mph for

\frac{1}{5}

hr gives 45 ×

\frac{1}{5}

= 9 miles. Both match, and 27 mph in traffic plus 45 mph on a clear road are realistic driving speeds — not a sanity-check red flag.

💡Key takeaway

Name what you don't know with a letter, write one equation per scenario, and let the matching distance do the work — that is the Grade 8 simultaneous-equations move.

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