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Difficulty: MediumLinear Inequalities in One Variable

A truck rental company charges a daily fee of 45.00plus45.00 plus 0.75 per mile driven. A driver rents a truck for one day, has a budget of at most $150.00, and is required to drive at least 60 miles for a delivery. What is the maximum number of additional miles the driver can drive beyond the required 60 miles without exceeding the budget?

Answer: 80 miles

Answer

80
To find the maximum number of additional miles, we set up the inequality 45+0.75(60+a)≤15045 + 0.75(60 + a) \leq 150, where aa represents the number of additional miles driven. Distributing 0.750.75 yields 45+45+0.75a≤15045 + 45 + 0.75a \leq 150, which simplifies to 90+0.75a≤15090 + 0.75a \leq 150. Subtracting 9090 from both sides gives 0.75a≤600.75a \leq 60. Finally, dividing by 0.750.75 gives a≤80a \leq 80. The maximum value of aa is therefore 80.

Step-by-Step Solution

1
Set up the inequality representing the total budget constraint.
45+0.75(60+a)≤15045 + 0.75(60 + a) \leq 150, where aa is the number of additional miles.
The daily fee is 45,theper−milerateis45, the per-mile rate is 0.75, the driver must drive at least 60 miles plus aa additional miles, and the total cost cannot exceed the $150 budget.
2
Simplify the expression by distributing 0.750.75 and combining constant terms.
90+0.75a≤15090 + 0.75a \leq 150
0.75×60=450.75 \times 60 = 45, and adding the daily fee of 4545 gives 9090.
3
Isolate the variable term by subtracting 9090 from both sides.
0.75a≤600.75a \leq 60
This determines the remaining budget available for the additional miles.
4
Solve for aa by dividing both sides by 0.750.75.
a≤80a \leq 80
Dividing 6060 by 0.750.75 gives the maximum number of additional miles.

Key Concept

Solving multi-step linear inequalities in context
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Topics

Linear Inequalities in One Variable

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