1Cademy - Solving a Word Problem for the Past Cost of a Car

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Number Problems

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Solving a Word Problem for the Past Cost of a Car

Apply the seven-step problem-solving strategy to a real-world word problem that asks to find an unknown past value based on a known present value and a related comparative phrase.

Problem: According to the National Automobile Dealers Association, the average cost of a car in $2014$ was $\$28{,}400$ . This was $\$1{,}600$ less than six times the cost in $1975$ . What was the average cost of a car in $1975$ ?

Read the problem carefully. The goal is to find a past cost using a present cost.
Identify what to find: the average cost of a car in $1975$ .
Name the unknown: Let $c$ = the average cost of a car in $1975$ . The phrase " $\$1{,}600$ less than six times the cost in $1975$ " tells us that the cost in $2014$ can be expressed as $6c - 1{,}600$ .
Translate into an equation: The average cost in $2014$ was $\$28{,}400$ , so: $6c - 1{,}600 = 28{,}400$
Solve the equation. Combine the two operations by first adding $1{,}600$ to both sides: $6c = 30{,}000$ Divide both sides by $6$ : $c = 5{,}000$
Check: Is $\$1{,}600$ less than six times $\$5{,}000$ equal to $\$28{,}400$ ? $6(5{,}000) - 1{,}600 \stackrel{?}{=} 28{,}400$ $30{,}000 - 1{,}600 \stackrel{?}{=} 28{,}400$ $28{,}400 = 28{,}400 \checkmark$
Answer: The average cost of a car in $1975$ was $\$5{,}000$ .

This problem highlights translating comparative language into an equation. The expression $6c - 1{,}600$ accurately captures the relationship, which directly translates into a two-step equation.

Updated 2026-05-02

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OpenStax Intermediate Algebra

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