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2020 AMC 10A Problem 4

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2020 AMC 10A Problem 4AMC 10 Step by Step
Name:                             Date: September 13, 2026
Linear Equations & Word Problems · difficulty 1 of 5 · about 2 min · position P1-10

A driver travels for 22 hours at 6060 miles per hour, during which her car gets 3030 miles per gallon of gasoline. She is paid $0.50\$0.50 per mile, and her only expense is gasoline at $2.00\$2.00 per gallon. What is her net rate of pay, in dollars per hour, after this expense?

  1. A)

    2020

  2. B)

    2222

  3. C)

    2424

  4. D)

    2525

  5. E)

    2626

Answer

E

Key insight

Work per hour: 60 miles earns 30 dollars and burns 2 gallons costing 4 dollars, leaving 26 dollars.

Solution

Since the answer is a rate per hour, look at a single hour of driving.

In one hour she covers 6060 miles and earns 60⋅0.50=3060 \cdot 0.50 = 30 dollars.

Those 6060 miles use 6030=2\frac{60}{30} = 2 gallons of gasoline, costing 2⋅2.00=42 \cdot 2.00 = 4 dollars.

Net pay per hour: 30−4=2630 - 4 = 26 dollars.

The answer is (E) 26\boxed{\textbf{(E)}\ 26}.

Why this works

Everything in the problem scales linearly with time, so the two-hour duration is irrelevant to a per-hour rate. Choosing the unit of time matching the question avoids an extra division at the end. Keep income and expense in the same units (dollars per hour) before subtracting.

Alternative approach

Over the full trip: 120120 miles earn $60\$60; 120/30=4120/30 = 4 gallons cost $8\$8; net $52\$52 over 22 hours is $26\$26 per hour.

The trap

Forgetting to divide the two-hour net (52 dollars) by 2, or computing gas cost per mile incorrectly.

Common mistakes

  • Forgetting to divide the two-hour net (52 dollars) by 2, or computing gas cost per mile incorrectly.
  • Mixing up miles per gallon with gallons per mile and charging $60\$60 for gas.

Techniques

Set up the equation/formula and compute; no special trick needed