AMC 10 · 2022 · #4
Grade 6 arithmeticProblem
In some countries, automobile fuel efficiency is measured in liters per 100 kilometers while other countries use miles per gallon. Suppose that 1 kilometer equals m miles, and 1 gallon equals l liters. Which of the following gives the fuel efficiency in liters per 100 kilometers for a car that gets x miles per gallon?
Pick an answer.
AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: Fuel efficiency: $x$ miles per gallon (mi/gal); $1$ km $= m$ miles, so $1$ mile $= \dfrac{1}{m}$ km; $1$ gallon $= l$ liters, so $1$ liter $= \dfrac{1}{l}$ gallon; Answer choices: (A) $\dfrac{x}{100lm}$, (B) $\dfrac{xlm}{100}$, (C) $\dfrac{lm}{100x}$, (D) $\dfrac{100}{xlm}$, (E) $\dfrac{100lm}{x}$
Unknowns: The fuel efficiency expressed in liters per $100$ kilometers (L / $100$ km)
Understand
Restated: 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.
Givens: Fuel efficiency: $x$ miles per gallon (mi/gal); $1$ km $= m$ miles, so $1$ mile $= \dfrac{1}{m}$ km; $1$ gallon $= l$ liters, so $1$ liter $= \dfrac{1}{l}$ gallon; Answer choices: (A) $\dfrac{x}{100lm}$, (B) $\dfrac{xlm}{100}$, (C) $\dfrac{lm}{100x}$, (D) $\dfrac{100}{xlm}$, (E) $\dfrac{100lm}{x}$
Plan
Primary tool: #9 Solve an Easier Related Problem
Secondary: #8 Analyze the Units, #3 Eliminate Possibilities
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.
Execute — Answer: E
6.RP.A.3 Step 1 - Pick easy numbers.
- Set $x = 30$ mpg (a normal car), $m = 0.5$ (pretend $1$ km $= 0.5$ mi for round arithmetic), and $l = 4$ (pretend $1$ gal $= 4$ L).
- Use these to compute the true L per $100$ km, then see which formula reproduces that number.
💡 Concrete numbers turn an abstract variable hunt into a normal Grade 6 ratio question — "how many liters to go $100$ km?".
6.RP.A.3 Step 2 - Trace units to compute the actual L / $100$ km with these test values.
- The car gets $30$ mi/gal.
- $100$ km $= 100 \times 0.5 = 50$ miles.
- To go $50$ miles, the car uses $50/30 = 5/3$ gallons, which is $5/3 \times 4 = 20/3$ liters.
💡 Following the chain miles $\to$ gallons $\to$ liters is exactly the Grade 6 rate-chain skill — same idea as a recipe scaling problem.
6.EE.A.2 Step 3 - Now plug $x = 30$, $l = 4$, $m = 0.5$ into each choice.
- (E) $\dfrac{100lm}{x} = \dfrac{100 \cdot 4 \cdot 0.5}{30} = \dfrac{200}{30} = \dfrac{20}{3}$.
- Match!
- Check a few others for safety: (B) $\dfrac{xlm}{100} = \dfrac{30 \cdot 4 \cdot 0.5}{100} = \dfrac{60}{100} = 0.6$ — wrong.
- (D) $\dfrac{100}{xlm} = \dfrac{100}{60} \approx 1.67$ — wrong.
- Only (E) lands on $20/3$.
💡 Treating each formula as a numerical expression and evaluating is Grade 6 "write and evaluate expressions with letters" — concrete numbers expose the right formula instantly.
6.RP.A.3 Step 4 - Confirm symbolically.
- Starting from $x$ mi/gal and tracking units: in $1$ gallon ($= l$ liters) the car covers $x$ miles ($= x/m$ km).
- So per liter, the car covers $\dfrac{x/m}{l} = \dfrac{x}{lm}$ km.
- Flip to get liters per km, multiply by $100$ for liters per $100$ km.
💡 Units cancel like factors — once each letter's unit role is named, the algebra writes itself and lands on the same answer the numbers gave.
6.RP.A.3 Pick easy numbers. Set $x = 30$ mpg (a normal car), $m = 0.5$ (pretend $1$ km $= 6.RP.A.3 Trace units to compute the actual L / $100$ km with these test values. The car g 6.EE.A.2 Now plug $x = 30$, $l = 4$, $m = 0.5$ into each choice. (E) $\dfrac{100lm}{x} = 6.RP.A.3 Confirm symbolically. Starting from $x$ mi/gal and tracking units: in $1$ gallon Review
Reasonableness: 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) $\dfrac{100lm}{x}$ matches every direction check.
Alternative: Tool #3 (Eliminate Possibilities) alone. Use direction-checks: $x$ must be downstairs ($\to$ eliminate (B)); $100$ must be upstairs since we want per-$100$-km, not per-$1$-km ($\to$ eliminate (A), (C), (D)). Only (E) has $100$ on top and $x$ on the bottom.
CCSS standards used (min grade 6)
6.RP.A.3Use ratio and rate reasoning to solve real-world problems (Chaining the unit conversions miles $\to$ km and gallons $\to$ liters to convert mi/gal into L per $100$ km.)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Substituting concrete test values $x = 30$, $l = 4$, $m = 0.5$ into each candidate formula and comparing to the true value $\dfrac{20}{3}$.)
⭐ This AMC 10 problem only needs Grade 6 unit-rate reasoning — try friendly numbers and the formula that lands on $\dfrac{20}{3}$ wins, which is $\dfrac{100lm}{x}$.
⭐ This AMC 10 problem only needs Grade 6 unit-rate reasoning — try friendly numbers and the formula that lands on $\dfrac{20}{3}$ wins, which is $\dfrac{100lm}{x}$.
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