Question

test prep a truck rental company rents a truck for a one - time fee of $25 plus $1.50 per mile traveled. kelly has $80 she can spend on the rental truck. written as a fraction, what is the greatest number of miles that she can travel? enter the correct answer in the box.

Explanation:

Step1: Define variables and cost equation

Let \( m \) be the number of miles. The total cost \( C \) is the one - time fee plus the cost per mile, so \( C = 25+1.50m \). Kelly has \( \$80 \) to spend, so we set up the inequality \( 25 + 1.50m\leqslant80 \).

Step2: Solve the inequality for \( m \)

Subtract 25 from both sides: \( 1.50m\leqslant80 - 25 \).
\( 80-25 = 55 \), so \( 1.50m\leqslant55 \).
To solve for \( m \), divide both sides by \( 1.50 \). \( m\leqslant\frac{55}{1.50} \).
Simplify \( \frac{55}{1.50}=\frac{55}{\frac{3}{2}}=55\times\frac{2}{3}=\frac{110}{3}\approx36.67 \). Since the number of miles must be a whole number (and we want the greatest number of miles she can travel), we take the floor of \( \frac{110}{3} \), but the question asks for the answer as a fraction, so \( m=\frac{110}{3} \) (or we can also note that if we solve \( 25 + 1.5m=80 \) for \( m \) (since we want the maximum number of miles, we can set the cost equal to the money she has), \( 1.5m=80 - 25=55 \), \( m=\frac{55}{1.5}=\frac{55\times2}{3}=\frac{110}{3} \)).

Answer:

\(\frac{110}{3}\) (or approximately \( 36\frac{2}{3} \))