AMC 10 · 2017 · #12

Grade 7 rate-ratio

Archetype Arithmetic Expression Decomposition └ Sub-archetype Successive Multiplicative Steps

percentageratio-proportionrate convert-to-algebradimensional-analysis ↑ Prerequisites: percentage

📏 Medium solution 💡 2 insights

Problem

Elmer's new car goes 50% farther on each liter of fuel than his old car. But the new car's diesel costs 20% more per liter than the old car's gasoline. For a long trip of fixed distance, by what percent is the new car's fuel cost lower than the old car's?

Pick an answer.

(A)

$20\%$

(B)

$26\tfrac23\%$

(C)

$27\tfrac79\%$

(D)

$33\tfrac13\%$

(E)

$66\tfrac23\%$

AMC 10 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Analyze the Units

Tool #8 (Analyze the Units): the quantities are a rate (km per liter) and a price (dollars per liter), and the goal is dollars per trip; tracking units shows cost = distance × price ÷ efficiency, which organizes the whole problem. Tool #4 (Introduce a Variable): no numbers are given, so I name the old efficiency, the old price, and the distance with letters and let them cancel. Tool #9 (Solve an Easier Related Problem): because the answer can't depend on the missing numbers, I may pick convenient values to double-check the algebra.

1STEP 1

Name the unknowns

No numbers given, so use letters: old car is E km/liter at price P; new car becomes 1.5E km/liter and 1.2P per liter; trip distance D.

old: E, P new: 1.5E, 1.2P distance D

2STEP 2

Cost of a trip from units

Track units: D km at E km/liter uses D÷E liters, and liters × price gives dollars, so any trip costs distance ÷ efficiency × price.

cost=D/efficiency×price=(D·price)/efficiency

3STEP 3

Write both costs

Old car costs $\frac{D·P}{E}$ ; new car costs $\frac{D·1.2P}{1.5E}$ . The shared $\frac{D·P}{E}$ cancels, so the cost ratio is just $\frac{1.2}{1.5}$ .

cost_new=1.2/1.5·(D · P)/E=1.2/1.5 cost_old

4STEP 4

Turn the ratio into a percent saved

The ratio is $\frac{1.2}{1.5}$ = $\frac{4}{5}$ = 0.8, so the new car costs 80% of the old; paying 80% means saving the leftover 20%, choice (A).

1.2/1.5=0.8 → save 100%-80%=20% → (A)

Answer

20%

Use concrete numbers as a check. Say the old car gets 30 km per liter and gasoline is $1 per liter, with a 300 km trip. Old car: 300 ÷ 30=10 liters at $1=$10. New car: 45 km per liter and diesel at $1.20, so 300 ÷ 45=

\frac{20}{3}

liters at $1.20=$8. Going from $10 to $8 is a $2 saving on $10, which is 20% — matching choice (A). It also makes sense that the answer is positive but modest: the 50% efficiency gain outweighs the 20% price bump, but not hugely.

💡Key takeaway

Trip cost is price divided by efficiency, so the new car costs $\frac{1.2}{1.5}$ =0.8 of the old — paying 80% means saving 20%, choice (A).

Name the unknowns
Cost of a trip from units
Write both costs
Turn the ratio into a percent saved

Name the unknowns

Cost of a trip from units

Write both costs

Turn the ratio into a percent saved

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