Question

kathleen and arnob both run from the park entrance along a loop. kathleen starts walking from the park entrance and gets a 5-mile head start on arnob. the graph shows how far they have both traveled.

how many minutes does it take arnob to catch up to kathleen?

• 9.8
• 10
• 73.5
• 75

Explanation:

Set up the system of linear equations

The graph shows distance \(y\) (in miles) versus time \(x\) (in minutes).
For Kathleen (blue line starting at \(y = 5\)):

$$y = \frac{1}{15}x + 5$$

For Arnob (red line starting at \(y = 0\)):

$$y = \frac{2}{15}x$$

Solve for the intersection point

To find when Arnob catches up to Kathleen, set their distances equal:

$$\frac{2}{15}x = \frac{1}{15}x + 5$$

Subtract \(\frac{1}{15}x\) from both sides:

$$\frac{1}{15}x = 5$$

Multiply by 15:

$$x = 75$$

Verify the solution

At \(x = 75\) minutes:
Kathleen's distance: \(y = \frac{1}{15}(75) + 5 = 5 + 5 = 10\) miles.
Arnob's distance: \(y = \frac{2}{15}(75) = 10\) miles.
The distances are equal, confirming it takes 75 minutes.

Answer:

• 9.8
• 10
• 73.5
• 75 (Correct answer)