Question

Oliver estimates the weight of his cat to be 16 pounds. The actual weight of his cat is 14.25 pounds. What is the percent error of Oliver's estimate? Round the percent to the nearest tenth if necessary.

Answer 1

We have that the 100% of the weight is 14.25 pounds. Then, using a rule of three, we can find the percent change to 16 pounds:

$\begin{gathered} 14.25\rightarrow100\% \\ 16\rightarrow x\% \\ \Rightarrow x=\frac{16\cdot100}{14.25}=\frac{1600}{14.25}=112.3 \\ x=112.3 \end{gathered}$

then the percent change is:

$112.3\%-100\%=12.3\%$

therefore, the percent error of Oliver's estimate is 12.3%