Results 1 to 5 of 5

Math Help - Percent problem

  1. #1
    Member
    Joined
    Jul 2006
    Posts
    90

    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%??
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member eumyang's Avatar
    Joined
    Jan 2010
    Posts
    278
    Thanks
    1
    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}<br />
A &= l \times w \\<br />
x^2 &= (1.25x)(x)(1 - n) \\<br />
x^2 &= 1.25x^2(1 - n) \\<br />
1 &= 1.25(1 - n) \\<br />
1 &= 1.25 - 1.25n \\<br />
-0.25 &= -1.25n \\<br />
0.2 &= n<br />
\end{aligned}<br />
    0.2 is 20%, so there needs to be a 20% decrease in the width.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jul 2006
    Posts
    90
    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??
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member eumyang's Avatar
    Joined
    Jan 2010
    Posts
    278
    Thanks
    1
    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?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jul 2006
    Posts
    90
    That makes sense! Thanks!!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. percent problem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: December 3rd 2009, 01:16 PM
  2. percent word problem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: October 28th 2009, 03:10 PM
  3. Percent word problem help
    Posted in the Algebra Forum
    Replies: 4
    Last Post: October 1st 2009, 04:00 PM
  4. Interst Problem (Percent)
    Posted in the Math Topics Forum
    Replies: 4
    Last Post: August 23rd 2008, 02:12 PM
  5. percent problem
    Posted in the Algebra Forum
    Replies: 2
    Last Post: April 29th 2008, 02:48 PM

Search Tags


/mathhelpforum @mathhelpforum