Percentage Unchanged
If the base of a rectangle is increased by 10% and the area is unchanged, by what percentage (exactly) is the width decreased by ?
![Percentage Unchanged Percentage Unchanged](/sites/default/files/styles/large/public/thumbnails/content-97-06-six6-rect.gif?itok=KQ6txxiP)
Answer: 9.09090...%
Using scale factors
$$\begin{align} \tfrac{?}{100}\times110 &= 100\\
?\times110&=10000\\
?\times11&=1000\\
? &= \tfrac{1000}{11} \text{ or } 90.90909...\end{align}$$ So the width is decreased by $100-90.90909...=9.090909...\text{ or }\frac{100}{11}=9\frac1{11}$
Using algebra
The area of the original rectangle was $wb$
![Percentage Unchanged Percentage Unchanged](/sites/default/files/styles/large/public/thumbnails/content-97-06-six6-percentage%252520unchanged%252520method%2525202.png?itok=4fyifkkY)
The area of the altered rectangle is $pw \times 1.1b$
![Percentage Unchanged Percentage Unchanged](/sites/default/files/styles/large/public/thumbnails/content-97-06-six6-percentage%252520unchanged%252520method%2525202%252520image%2525202.png?itok=FzlUXdnO)
Since the two areas are equal,
$1.1 pwb = wb$
so
$1.1 p = 1$
so $p = \frac{1}{1.1} = \frac{10}{11}$
So the width is $\frac{10}{11}$ of its original value, so it has been decreased by $\frac1{11}$, which as a percentage is $9$ and $\frac1{11}\%$, or $9.09\%$ to 2.d.p.