AMC 10 · 2022 · #4

Grade 6 arithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Unit Conversion Chain

unit-conversionratefraction-arithmetic dimensional-analysiseasier-related-problem ↑ Prerequisites: rateunit-conversion

📏 Short solution 💡 2 insights

Problem

Convert a car's fuel efficiency from x miles per gallon into liters per 100 kilometers, given 1 kilometer = m miles and 1 gallon = l liters. Pick the matching formula.

Pick an answer.

(A)

$\frac{x}{100lm}$

(B)

$\frac{xlm}{100}$

(C)

$\frac{lm}{100x}$

(D)

$\frac{100}{xlm}$

(E)

$\frac{100lm}{x}$

AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Solve an Easier Related Problem

All three letters are abstract conversion factors — easy to confuse. Tool #9 (Solve an Easier Related Problem) replaces them with friendly concrete numbers (x=30 mpg, l=4, m=0.6), gets a numerical answer in L per 100 km, then plugs the same numbers into each choice to see which matches. Tool #8 (Analyze the Units) then confirms the same answer symbolically — miles cancel with miles, gallons cancel with gallons — and tool #3 (Eliminate Possibilities) finishes by ruling out the other four choices by their numerical mismatch.

1STEP 1

Pick friendly numbers — x = 30 mpg, m = 0.5, l = 4 — then compute the true L per 100 km and see which formula reproduces it.

x = 30, m = 0.5, l = 4

2STEP 2

Trace units: 100 km = 50 mi; at 30 mi/gal that is $\frac{50}{30}$ = $\frac{5}{3}$ gal, and $\frac{5}{3}$ × 4 = $\frac{20}{3}$ L per 100 km.

100 km = 50 mi; (50 mi)/(30 mi/gal) =

\frac{5}{3}

gal =

\frac{20}{3}

3STEP 3

Plug the numbers into each choice: only (E) gives $\frac{200}{30}$ = $\frac{20}{3}$ ; (B) gives 0.6 and (D) gives 1.67, both wrong.

(E):

\frac{100 · 4 · 0.5}{30}

\frac{20}{3}

✓

4STEP 4

Confirm symbolically: 1 gallon (l L) covers x mi = $\frac{x}{m}$ km, so per liter $\frac{x}{lm}$ km; flip and ×100 to land on (E).

(x mi)/(1 gal) = (

\frac{x}{m}

km)/(l L) =

\frac{x}{lm}

km/L →

\frac{lm}{x}

L/km →

\frac{100lm}{x}

L/100 km → (E)

Answer

\frac{100lm}{x}

Direction check. A higher x (more efficient car) should mean fewer liters per 100 km, so x belongs in the denominator. Higher l (bigger gallons = more liters per gallon) should make L/100 km bigger, so l belongs in the numerator. Higher m (one km contains more miles, so distances in km feel "longer") also pushes L/100 km up — m in the numerator. The factor of 100 scales from per-km to per-100-km. Choice (E)

\frac{100lm}{x}

matches every direction check.

💡Key takeaway

This AMC 10 problem only needs Grade 6 unit-rate reasoning — try friendly numbers and the formula that lands on $\frac{20}{3}$ wins, which is $\frac{100lm}{x}$ .

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