One side of the rectangle has been reduced by 20%, and the other has been reduced by 40%.

One side of the rectangle has been reduced by 20%, and the other has been reduced by 40%. It turned out that the perimeter decreased by 25%. How many times is the rectangle longer than its width?

Let us denote the lengths of the sides of the rectangle through x and y.
The perimeter of this rectangle is 2 * (x + y).
After one of the sides of this rectangle was reduced by 20%, and the other was reduced by 40%, the perimeter of the resulting rectangle was:
2 * (x – (20/100) * x + y – (40/100) * y) = 2 * (x – (2/10) * x + y – (4/10) * y) = 2 * (x – 0.2 * x + y – 0.4 * y) = 2 * (0.8 * x + 0.6 * y).
According to the condition of the problem, the perimeter of the resulting rectangle decreased by 25%, that is, it was 100 – 25 = 75% of the perimeter of the original rectangle, therefore, we can draw up the following equation:
2 * (0.8 * x + 0.6 * y) = 2 * (x + y) * 0.75.
We solve the resulting equation:
0.8 * x + 0.6 * y = 2 * (x + y) * 0.75 / 2;
0.8 * x + 0.6 * y = (x + y) * 0.75;
0.8 * x + 0.6 * y = 0.75 * x + 0.75 * y;
0.8 * x – 0.75 * x = 0.75 * y – 0.6 * y;
0.05 * x = 0.15 * y;
x = (0.15 / 0.05) * y;
x = (15/5) * y;
x = 3 * y.
Therefore, one side of the original rectangle is 3 times the second side of this rectangle.

Answer: The length of the original rectangle is 3 times the width of this rectangle.