Percentage of rectangle's sides and perimeter Imagine a rectangle, then shorten its sides. Shorten it in length by 10%, and 20% in width. After that, the rectangle's perimeter was lowered by 12%. By how many percent would the original rectangle's perimeter be smaller if we were to lower its length by 20%, and width by 10%?
 A: Hint:
$$0.88(2w+2l) = 2(0.8w+0.9l)$$
$$\Leftrightarrow 1.76w+1.76l = 1.6w+1.8l$$
$$\Leftrightarrow 0.16w = 0.04l$$
$$\Leftrightarrow w = 4l$$
A: Let the original rectangle be $x \times y$ such that its perimeter is $p_0 = 2x + 2y$.

After the 1st change, the new rectangle is $0.9x \times 0.8y$ and its perimeter = $p_1 = 2(0.9x + 0.8 y) = 1.8x + 1.6y$
The percentage change in perimeter $= - 12\%$ 
$= \dfrac {p_1 – p_0}{p_0} \times 100\%$ 
$= \dfrac {1.8x + 1.6y – (2x + 2y)}{2x + 2y} \times 100\%$
$ = \dfrac {– 0.2x – 0.4y}{2x + 2y} \times 100% = - 12\%$
∴ $0.1x + 0.2y = 0.12x + 0.12y$
After simplification, we get $x = 4y$. This means $p_0 = 2((4y) + y) = 10y$ 
In the 2nd change, the perimeter $= p_2 = 2(0.8x + 0.9y) = 2(0.8(4y) + 0.9y) = 8.2y$
The new percentage change in perimeter = $\dfrac {p_2 - p_0}{p_0} \times 100 \% = \dfrac {8.2y – 10 y}{10y} \times 100\% 
= - 18\% $ (minus means lowered)
A: Denote length,width and perimeter by $ (x,y,P) $ respectively. 
Solving the two simultaneous equations
$$ x+y=P, \quad  0.9 x + 0.8 y  = 0.88 P,   $$
you get $$x= 0.8 P, \quad y= 0.2 P $$
Plugging into the new perimeter
$$ 0.8 \,* \, 0.8 P + 0.9 \,*\, 0.2 P =0.82  P. \, $$
That makes $ 100-82= 18$ % reduction in perimeter.
