A triangle has side lengths of $x,75,100$ where $x<75$ and altitudes of lengths $y,28,60$ where $y<28$. What is the value of $x+y$?

[i]2019 CCA Math Bonanza Team Round #2[/i]
1. **Understanding the problem**: We are given a triangle with side lengths \(x, 75, 100\) and altitudes \(y, 28, 60\) corresponding to these sides. We need to find the value of \(x + y\) given the constraints \(x < 75\) and \(y < 28\).

2. **Using the area relationship**: The area of a triangle can be expressed using any side and its corresponding altitude. Therefore, the product of a side and its corresponding altitude is constant for all sides of the triangle. This gives us the following relationships:
   \[
   \frac{1}{2} \times x \times 60 = \frac{1}{2} \times 75 \times 28 = \frac{1}{2} \times 100 \times y
   \]

3. **Simplifying the equations**: We can remove the \(\frac{1}{2}\) from all terms since it is common:
   \[
   x \times 60 = 75 \times 28 = 100 \times y
   \]

4. **Calculating the constant product**: First, calculate \(75 \times 28\):
   \[
   75 \times 28 = 75 \times (30 - 2) = 75 \times 30 - 75 \times 2 = 2250 - 150 = 2100
   \]
   Therefore, the constant product is 2100.

5. **Solving for \(x\)**: Using the equation \(x \times 60 = 2100\):
   \[
   x = \frac{2100}{60} = 35
   \]

6. **Solving for \(y\)**: Using the equation \(100 \times y = 2100\):
   \[
   y = \frac{2100}{100} = 21
   \]

7. **Summing \(x\) and \(y\)**: Now, we find \(x + y\):
   \[
   x + y = 35 + 21 = 56
   \]

The final answer is \(\boxed{56}\).