Percent problem

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July 6th 2010, 11:14 AM
sarahh

Percent problem

The length of one side of a square is increased by 25%. By what percent would the length of the adjacent side have to be decreased so that the area of the new figure (which will be rectangular) is the same as the area of the original square?

I'm not sure how they got the answer as 20%??
July 6th 2010, 12:00 PM
eumyang

length of side of square = x
percent decrease = n

area of rectangle = length * width
length of rectangle = x + .25x = 1.25x ("25% increase")
width of rectangle = x - nx = x(1 - n) ("some percent decrease")
area of rectangle = area of square = x^2

So:
$\begin{aligned} A &= l \times w \\ x^2 &= (1.25x)(x)(1 - n) \\ x^2 &= 1.25x^2(1 - n) \\ 1 &= 1.25(1 - n) \\ 1 &= 1.25 - 1.25n \\ -0.25 &= -1.25n \\ 0.2 &= n \end{aligned} $
0.2 is 20%, so there needs to be a 20% decrease in the width.
July 6th 2010, 12:15 PM
sarahh

Hi eumyang, your worked out solution makes sense but I don't understand why we set the initial length and width of the rectangle equal to the "values" you proposed??
July 6th 2010, 12:46 PM
eumyang

If something is increased by 25%, you're taking a number and adding 25% of it. Say the number is 40. If you're increasing 40 by 25%, you take 25% of 40, which is 10 (0.25 * 40 = 10), and add it to 40 to get 50. In other words, 40 + 10 = 50.

Now, if you take a number x and increase it by 25%, do the same thing. 25% of x = 0.25x, and add it to the number: x + 0.25x, or 1.25x. Does that help?
July 6th 2010, 01:23 PM
sarahh

That makes sense! Thanks!!