Question

Difficulty: MediumSystems of Linear Equations

A logistics company operates two types of delivery trucks: Model A and Model B. Model A consumes 0.250.25 liters of fuel per mile and incurs $1.50\$1.50 in maintenance costs per mile. Model B consumes 0.400.40 liters of fuel per mile and incurs $1.00\$1.00 in maintenance costs per mile. On a given day, the fleet consumed a total of 300300 liters of fuel and incurred a total of $1,100\$1,100 in maintenance costs. What was the total number of miles traveled by the Model A trucks on that day?

Answer: 400 miles

Answer

The Model A trucks traveled a total of 400 miles on that day.
Let xx be the number of miles driven by Model A trucks and yy be the number of miles driven by Model B trucks. Set up two equations: 0.25x+0.40y=3000.25x + 0.40y = 300 (fuel) and 1.50x+1.00y=11001.50x + 1.00y = 1100 (maintenance). Substituting y=11001.5xy = 1100 - 1.5x into the fuel equation yields 0.25x+0.40(11001.5x)=3000.25x + 0.40(1100 - 1.5x) = 300, which simplifies to 0.35x+440=300-0.35x + 440 = 300, leading directly to x=400x = 400.

Step-by-Step Solution

1
Formulate the linear system
0.25x+0.40y=3000.25x + 0.40y = 300 and 1.50x+1.00y=11001.50x + 1.00y = 1100, where xx is miles driven by Model A and yy is miles driven by Model B.
Translates the rates of fuel consumption and maintenance per mile into total daily figures.
2
Express yy in terms of xx
y=11001.5xy = 1100 - 1.5x
Isolating yy makes substitution into the fuel equation straightforward.
3
Substitute yy into the first equation and solve for xx
0.25x+0.40(11001.5x)=300    0.35x+440=300    0.35x=140    x=4000.25x + 0.40(1100 - 1.5x) = 300 \implies -0.35x + 440 = 300 \implies -0.35x = -140 \implies x = 400
Reduces the system to a single linear equation in xx to find the distance for Model A.

Key Concept

Setting up and solving a two-variable system of linear equations via substitution or elimination.
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