An exercise on the linear function

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A square $ABCD$ has a side that measures $5$. Denote by $y$ the area of the rectangle $A'B'C'D'$ that is obtained by decreasing the sides $AB$ and $CD$ by $x\%$ and leaving the measurements of the sides $BC$ and $AD$ unchanged. Express $y$ as a function of $x$, graph the function keeping in mind the limitations imposed by the problem.

My solution: The rectangle $A'B'C'D'$ has the sides $B'C'$ and $A'D'$ of length $5$ while the sides $A'B'$ and $C'D'$ have a length decreased by a percentage fraction $x$ i.e.: $$|A'B'|=|C'D'|=5-5x=5(1-x)$$ The area is: $$y=5\cdot 5(1-x)=25(1-x)=25-25x$$

After $0 \le x \le 1$. The solution of the book, instead, is $$y=25-\frac 14 x$$ Where is my mistake?

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$x$ is percentage, therefore:

$|A'B'|=|C'D'|=5(1-x/100)$

This will give you the correct answer.