A circle and a square overlap such that the overlapping area is $50\%$ of the area of the circle, and is $25\%$ of the area of the square, as shown in the figure. Find the ratio of the area of the square outside the circle to the area of the whole figure.
[img]https://cdn.artofproblemsolving.com/attachments/e/2/c209a95f457dbf3c46f66f82c0a45cc4b5c1c8.png[/img]
1. Let the area of the circle be \( A_{\text{circle}} \). According to the problem, the overlapping area is \( 50\% \) of the area of the circle. Therefore, the overlapping area is:
   \[
   \text{Overlapping Area} = 0.5 \times A_{\text{circle}}
   \]

2. Let the area of the square be \( A_{\text{square}} \). The problem states that the overlapping area is \( 25\% \) of the area of the square. Therefore, the overlapping area is:
   \[
   \text{Overlapping Area} = 0.25 \times A_{\text{square}}
   \]

3. Equate the two expressions for the overlapping area:
   \[
   0.5 \times A_{\text{circle}} = 0.25 \times A_{\text{square}}
   \]

4. Solve for \( A_{\text{square}} \) in terms of \( A_{\text{circle}} \):
   \[
   A_{\text{square}} = 2 \times A_{\text{circle}}
   \]

5. Let \( A_{\text{circle}} = 2x \). Then, the area of the square is:
   \[
   A_{\text{square}} = 2 \times 2x = 4x
   \]

6. The area of the square outside the circle is the area of the square minus the overlapping area:
   \[
   \text{Area of square outside the circle} = A_{\text{square}} - \text{Overlapping Area} = 4x - x = 3x
   \]

7. The total area of the whole figure (the union of the circle and the square) is the sum of the area of the circle and the area of the square minus the overlapping area:
   \[
   \text{Total Area} = A_{\text{circle}} + A_{\text{square}} - \text{Overlapping Area} = 2x + 4x - x = 5x
   \]

8. The ratio of the area of the square outside the circle to the area of the whole figure is:
   \[
   \text{Ratio} = \frac{\text{Area of square outside the circle}}{\text{Total Area}} = \frac{3x}{5x} = \frac{3}{5}
   \]

The final answer is \(\boxed{\frac{3}{5}}\).