A logistics company offers two freight pricing models for oversized cargo. Model X charges… — GMAT

A logistics company offers two freight pricing models for oversized cargo. Model X charges a fixed monthly account fee of $\$8$ plus $\$0.60$ per kilometer traveled. Model Y charges a fixed monthly account fee of $\$54$ plus $\$0.20$ per kilometer traveled. For how many kilometers in a month will the total monthly charge under Model X be exactly $20\%$ less than the total monthly charge under Model Y?

Answer: 80 kilometers

Answer

80 kilometers

The total monthly cost under Model X is

C_X = 8 + 0.60k

and under Model Y is

C_Y = 54 + 0.20k

. The condition that Model X is 20% less than Model Y means

C_X = 0.80 C_Y

. Substituting the expressions gives

8 + 0.60k = 0.80(54 + 0.20k) = 43.2 + 0.16k

. Subtracting

0.16k

and

8

from both sides yields

0.44k = 35.2

, which simplifies to

k = 80

Step-by-Step Solution

Define variables and establish linear cost equations for both models.

Model X cost:

C_X = 8 + 0.60k

; Model Y cost:

C_Y = 54 + 0.20k

, where

k

is kilometers traveled.

Linear modeling translates flat fees and variable rates into algebraic expressions.

Formulate the linear equation based on the condition that Model X is 20% less than Model Y.

C_X = 0.80 C_Y \implies 8 + 0.60k = 0.80(54 + 0.20k)

Being 20% less than a base value means taking 80% (or 0.80) of that value.

Expand and simplify the algebraic equation.

8 + 0.60k = 43.2 + 0.16k \implies 0.44k = 35.2

Distributing 0.80 across

(54 + 0.20k)

yields

43.2 + 0.16k

, and subtracting

0.16k

and

8

isolates

k

on one side.

Calculate the value of

k

k = \frac{35.2}{0.44} = 80

Dividing

35.2

0.44

gives the exact number of kilometers required.

Key Concept

Linear Equations in One and Two Variables

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