2025 AMC 10B Problem 7

1. Draw a number line along the fence. Put the locked gate at $0$ and let east be positive.
- The box is at $x$.
- The east gate is $9$ meters past the box, at $x + 9$.
- The west gate is at $-8$.
- Frances is $15$ meters south of $0$.

2. The west route. Frances walks slanted to the west gate: $8$ across and $15$ up, so by the Pythagorean theorem the distance is $\sqrt{8^2 + 15^2} = \sqrt{289} = 17$. Then she walks along the fence from $-8$ to $x$, which is $x + 8$ meters.
$$\text{West total} = 17 + x + 8 = x + 25.$$

3. The east route. She walks slanted to the east gate: $x + 9$ across and $15$ up, a distance of $\sqrt{15^2 + (x+9)^2}$. Then she walks back $9$ meters to the box.
$$\text{East total} = \sqrt{225 + (x+9)^2} + 9.$$

4. Set the two totals equal. Take $9$ from both sides to get the square root alone:
$$\sqrt{225 + (x+9)^2} = x + 16.$$

5. Square both sides. The left becomes $225 + x^2 + 18x + 81$. The right becomes $x^2 + 32x + 256$. The $x^2$ terms cancel:
$$306 + 18x = 256 + 32x \quad\Longrightarrow\quad 50 = 14x \quad\Longrightarrow\quad x = \frac{50}{14} = \frac{25}{7} = 3\tfrac47.$$

The answer is $\boxed{\textbf{(C)}\ 3\frac47}$.

Picking one landmark (the locked gate) and measuring every position from it turns a wordy story into two distance formulas. The Pythagorean theorem gives the slanted walks, and spotting $8$-$15$-$17$ keeps one side free of square roots. The $x^2$ terms cancel after squaring, so what looks like a quadratic is really a simple linear equation.