Let
l and
w be the length and width, respectively, of the original rectangle.
The area of the original rectangle is
A=lw.
The rectangle is altered by increasing its length by 10 percent and decreasing its width by
p percent
thus, the length of the altered rectangle is
1.1l,
and the width of the altered rectangle is
(1−)w The alterations decrease the area by12 percent, so the area of the altered rectangle is
(1−0.12)A=0.88A.The altered rectangle is the product of its length and width, so
0.88A=(1.1l)(1−)w Since
A=lw his last equation can be rewritten as
from which it follows that
0.88A(1.1)(1−), o
0.8=(1−), Therefore,
=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
(10+12).