Ratio problem , removing trees in a neighborhood

Question

Im working on some ratio problems in the algebra section of my book.

The problem is

$99 $% of the trees in our neighborhood are eucalyptus trees. The town planning commission wants to get rid of some of these trees because they spread too quickly. However, the people in my neighborhood like the trees. The commission argues that their new eucalyptus tree removal plan will cut down so few eucalyptus trees that $98$% of the trees in our neighborhood will be eucalyptus trees. If the plan only involves removing eucalyptus trees, what percent of the existing trees in my neighborhood would the plan remove?

Im not sure if Im misunderstanding the problem, but isn't the answer just 1%? We are just going from 99% eucalyptus trees to 98% eucalyptus trees. So we are removing $1$% of the trees in the neighborhood no? Any hints would be appreciated.

Answer 1

Let the total number of trees be $T$ and suppose that this is made up of eucalyptus trees $E$ and other kinds of trees, $O$. Then $E+O=T$, and since 99% of the trees are eucalyptus trees, $\frac{E}{E+O}=0.99$. Therefore $$0.99E+0.99O=E$$ $$0.01E=0.99O$$ $$E=99O.$$ Suppose we reduce the number of eucalyptus trees by some amount $x$ and that the remaining number of eucalyptus trees are now 98% of the total trees. Then we would have $\frac{E-x}{(E-x)+O}=0.98$. Therefore, $$0.02(E-x)=0.98O$$ $$0.02E-0.02x=0.98O$$ $$0.02x=0.02E-0.98O$$ $$x=E-50(0.98)O$$ $$x=E-49O$$ $$x=E-49(\frac{E}{99})$$ $$x=\frac{50}{99}E \approx 0.505050...\times E.$$ So we end up removing a little over half of the eucalyptus trees. EDIT: Forgot to mention - removing 50.505050...% of the eucalyptus trees means that we are removing $0.50505050....\times 0.99T$ which is $0.5T$ or half the trees in the neighborhood.