SOLUTION: The width of a rectangle is decreased by 20%. By what percent would the length have to be increased for the area to remain the same?
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Question 70119: The width of a rectangle is decreased by 20%. By what percent would the length have to be increased for the area to remain the same? Answer by Edwin McCravy(20055) (Show Source):
The width of a rectangle is decreased by 20%.
By what percent would the length have to be increased.
for the area to remain the same?
Let old width = W
Let old length = L
Old area = LW
Let p = the decimal representation of the desired percent
increase in L.
New width = W - .20W = 1.00W - .20W = .80W = .8W
New length = L + pL
New area = (L + pL)(.8W) = .8W(L + pL) = .8LW + .8pLW
New Area = Old Area
.8LW + .8pLW = LW
Divide every term by LW
.8 + .8p = 1
.8p = 1 - .8
.8p = .2
p = .2/.8
p = .25 = 25%
Edwin