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2017 AMC 10B Problem 12

P11-15~3 minPrint
Ratios, Percents & AveragesTest small/specific values or special cases to find or verify the answer

Elmer's new car gives 50%50\% better fuel efficiency. However, the new car uses diesel fuel, which is 20%20\% more expensive per liter than the gasoline the old car used. By what percent will Elmer save money if he uses his new car instead of his old car for a long trip?

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Key insight

Try real numbers: a trip needing 3 liters in the old car needs 2 in the new one. At 1.20 dollars instead of 1, cost drops from 3.00 to 2.40.

Solution

Percent problems with no actual numbers become easy once you pick convenient numbers of your own.

  1. Say the old car goes 1010 km on one liter, gasoline costs $1\$1 per liter, and the trip is 3030 km.
  2. Old car. It uses 30÷10=330 \div 10 = 3 liters. Cost: 3⋅$1=$3.003 \cdot \$1 = \$3.00.
  3. New car. 50%50\% better efficiency means it goes 50%50\% farther per liter: 1515 km per liter. It uses 30÷15=230 \div 15 = 2 liters.
  4. Diesel is 20%20\% more expensive: $1.20\$1.20 per liter. Cost: 2⋅$1.20=$2.402 \cdot \$1.20 = \$2.40.
  5. Savings: $3.00−$2.40=$0.60\$3.00 - \$2.40 = \$0.60.
  6. As a percent of the old cost: 0.603.00=15=20%\frac{0.60}{3.00} = \frac{1}{5} = 20\%.

The answer is (A) 20%\boxed{\textbf{(A)}\ 20\%}.

Why this works

Percent changes combine by multiplying, not by adding or subtracting. Going 1.51.5 times as far per liter means using 23\frac{2}{3} as much fuel, and paying 1.21.2 times as much per liter; together the cost is 23⋅1.2=0.8\frac23 \cdot 1.2 = 0.8 times the old cost, a 20%20\% saving. Picking real numbers first protects you from turning a factor upside down.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Pick convenient numbers. Let the old car travel 1010 km per liter and let gasoline cost $1\$1 per liter, and take a 3030 km trip.

Old car: uses 30/10=330/10 = 3 liters at $1\$1 each, so the trip costs $3\$3.

New car: 50%50\% better efficiency means 1515 km per liter, so it uses 30/15=230/15 = 2 liters. Diesel costs 20%20\% more, $1.20\$1.20 per liter, so the trip costs 2⋅1.20=$2.402 \cdot 1.20 = \$2.40.

Savings: $3−$2.40=$0.60\$3 - \$2.40 = \$0.60, which is 0.603=20%\frac{0.60}{3} = 20\% of the old cost.

The answer is (A) 20%\boxed{\textbf{(A)}\ 20\%}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.