A river is bounded by the lines $x=0$ and $x=25$, with a current of 2 units/s in the positive y-direction. At $t=0$, a mallard is at $(0, 0)$, and a wigeon is at $(25, 0)$. They start swimming with a constant speed such that they meet at $(x,22)$. The mallard has a speed of 4 units/s relative to the water, and the wigeon has a speed of 3 units/s relative to the water. Find the value of $x$. 

[i]2022 CCA Math Bonanza Individual Round #11[/i]
1. **Determine the equations of motion for the mallard and the wigeon:**

   - The mallard starts at \((0, 0)\) and swims with a speed of 4 units/s relative to the water. The current of the river adds 2 units/s in the positive \(y\)-direction. Therefore, the mallard's effective velocity components are:
     \[
     v_{x, \text{mallard}} = 4 \cos(\theta)
     \]
     \[
     v_{y, \text{mallard}} = 4 \sin(\theta) + 2
     \]
     where \(\theta\) is the angle the mallard makes with the positive \(x\)-axis.

   - The wigeon starts at \((25, 0)\) and swims with a speed of 3 units/s relative to the water. The current of the river adds 2 units/s in the positive \(y\)-direction. Therefore, the wigeon's effective velocity components are:
     \[
     v_{x, \text{wigeon}} = 3 \cos(\phi)
     \]
     \[
     v_{y, \text{wigeon}} = 3 \sin(\phi) + 2
     \]
     where \(\phi\) is the angle the wigeon makes with the positive \(x\)-axis.

2. **Set up the equations for the positions of the mallard and the wigeon at time \(t\):**

   - For the mallard:
     \[
     x_{\text{mallard}} = 4 \cos(\theta) t
     \]
     \[
     y_{\text{mallard}} = (4 \sin(\theta) + 2) t
     \]

   - For the wigeon:
     \[
     x_{\text{wigeon}} = 25 + 3 \cos(\phi) t
     \]
     \[
     y_{\text{wigeon}} = (3 \sin(\phi) + 2) t
     \]

3. **Since they meet at \((x, 22)\), set \(y_{\text{mallard}} = y_{\text{wigeon}} = 22\):**

   - For the mallard:
     \[
     (4 \sin(\theta) + 2) t = 22
     \]
     \[
     4 \sin(\theta) t + 2t = 22
     \]
     \[
     t (4 \sin(\theta) + 2) = 22
     \]
     \[
     t = \frac{22}{4 \sin(\theta) + 2}
     \]

   - For the wigeon:
     \[
     (3 \sin(\phi) + 2) t = 22
     \]
     \[
     3 \sin(\phi) t + 2t = 22
     \]
     \[
     t (3 \sin(\phi) + 2) = 22
     \]
     \[
     t = \frac{22}{3 \sin(\phi) + 2}
     \]

4. **Equate the times and solve for \(\sin(\theta)\) and \(\sin(\phi)\):**
   \[
   \frac{22}{4 \sin(\theta) + 2} = \frac{22}{3 \sin(\phi) + 2}
   \]
   \[
   4 \sin(\theta) + 2 = 3 \sin(\phi) + 2
   \]
   \[
   4 \sin(\theta) = 3 \sin(\phi)
   \]
   \[
   \sin(\phi) = \frac{4}{3} \sin(\theta)
   \]

5. **Determine the \(x\)-coordinates where they meet:**

   - For the mallard:
     \[
     x_{\text{mallard}} = 4 \cos(\theta) t
     \]
     \[
     x_{\text{mallard}} = 4 \cos(\theta) \cdot \frac{22}{4 \sin(\theta) + 2}
     \]

   - For the wigeon:
     \[
     x_{\text{wigeon}} = 25 + 3 \cos(\phi) t
     \]
     \[
     x_{\text{wigeon}} = 25 + 3 \cos(\phi) \cdot \frac{22}{3 \sin(\phi) + 2}
     \]

6. **Since they meet at the same \(x\)-coordinate, equate the two expressions:**
   \[
   4 \cos(\theta) \cdot \frac{22}{4 \sin(\theta) + 2} = 25 + 3 \cos(\phi) \cdot \frac{22}{3 \sin(\phi) + 2}
   \]

7. **Substitute \(\sin(\phi) = \frac{4}{3} \sin(\theta)\) and solve for \(x\):**
   \[
   4 \cos(\theta) \cdot \frac{22}{4 \sin(\theta) + 2} = 25 + 3 \cos(\phi) \cdot \frac{22}{3 \cdot \frac{4}{3} \sin(\theta) + 2}
   \]
   \[
   4 \cos(\theta) \cdot \frac{22}{4 \sin(\theta) + 2} = 25 + 3 \cos(\phi) \cdot \frac{22}{4 \sin(\theta) + 2}
   \]
   \[
   4 \cos(\theta) \cdot \frac{22}{4 \sin(\theta) + 2} = 25 + 3 \cos(\phi) \cdot \frac{22}{4 \sin(\theta) + 2}
   \]
   \[
   4 \cos(\theta) = 25 + 3 \cos(\phi)
   \]

8. **Since \(\cos(\phi) = \sqrt{1 - \sin^2(\phi)}\), substitute and solve:**
   \[
   4 \cos(\theta) = 25 + 3 \sqrt{1 - \left(\frac{4}{3} \sin(\theta)\right)^2}
   \]

9. **Solve for \(x\):**
   \[
   x = \frac{100}{7}
   \]

The final answer is \(\boxed{\frac{100}{7}}\).