# percentage problem from an elementary algebra book

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• April 9th 2010, 08:35 PM
dreamcatcher
percentage problem from an elementary algebra book
The new price of a car is 25% higher than the old price. The old price is what percent lower than the new price. what formulas would I use, and how would I write out the equation showing the method of solving this problem. The back of the book says that the answer is 20%. please help.
• April 9th 2010, 08:44 PM
maddas
If A is x% more than B, this means that A = (1+x/100)B. In the same way, if A is x% less than B, this means that A = (1- x/100)B.

So if the price of the new car is N, and the price of the old car is O, then N = (1+25/100)O. You are asked to find x such that O is x% less than N, or O = (1- x/100)N. Substitute for N to get O = (1-x/100)(1+25/100)O or 1 = (1-x/100)(1+25/100). Now solve for x.
• April 9th 2010, 09:03 PM
dreamcatcher
ok, so I substituted 125 for N and got 100 for O which works, but I don't know how to solve for x in this step: 1=(1-x/100)(1+25/100)
• April 9th 2010, 09:19 PM
maddas
$1=(1-\frac x{100})(1+\frac{25}{100})$

$1=(1-\frac x{100})(1+\frac14)$

$1=(1-\frac x{100})\frac54$

$\frac45=(1-\frac x{100})\frac54\cdot\frac45$

$\frac45=1-\frac x{100}$

$\frac45-1=-\frac x{100}$

$1-\frac45 = \frac x{100}$

$100 - \frac{400}5 = x$

$100 - \frac{80\cdot 5}5 = x$

$100 - 80 = x$

$x =20$
• April 9th 2010, 09:27 PM
dreamcatcher
Wow, Thanks!