SAT Mathematics Practice Test 3

Let $\mathit{l}$ and $\mathit{w}$ be the length and width, respectively, of the original rectangle.
The area of the original rectangle is $\mathit{A}=\mathit{l}\mathit{w}$.
The rectangle is altered by increasing its length by 10 percent and decreasing its width by $\mathit{p}$ percent
thus, the length of the altered rectangle is $1.1\mathit{l}$,
and the width of the altered rectangle is $\left(1-\frac{\mathit{p}}{100}\right)\mathit{w}$
The alterations decrease the area by12 percent, so the area of the altered rectangle is $\left(1-0.12\right)\mathit{A}=0.88\mathit{A}.$
The  altered rectangle is the product of its length and width, so $0.88\mathit{A}=\left(1.1\mathit{l}\right)\left(1-\frac{\mathit{p}}{100}\right)\mathit{w}$
Since $\mathit{A}=\mathit{l}\mathit{w}$ his last equation can be rewritten as from which it follows that $0.88\mathit{A}\left(1.1\right)\left(1-\frac{\mathit{P}}{100}\right),$ o $0.8=\left(1-\frac{\mathit{P}}{100}\right),$
Therefore, $\frac{\mathit{P}}{100}=0.2$ and so the value of p is 20.
Choice A is incorrect and may be the result of confusing the 12 percent decrease in area with the percent decrease in width.
Choice B is incorrect because decreasing the width by 15 percent results in a 6.5 percent decrease in area, not a 12 percent decrease.
Choice D is incorrect and may be the result of adding the percent's given in the question $\left(10+12\right)$.