A cylindrical quarter has a $\frac{15}{16}$ inch diameter and a $\frac{1}{16}$ inch height. What would be the number of inches in the height of a coin whose volume is exactly four times that of the given quarter and whose diameter equals $1 \frac{1}{8}$ inches? Express your answer as a common fraction.

oliviapow06 Feb 25, 2021

#1**+1 **

The volume is proportional to $r^2$, so we multiply by $\left(\frac{\frac{9}{8}}{\frac{15}{16}}\right)^2=\left(\frac{6}{5}\right)^2=\frac{36}{25}$. It's proportional to the height itself, so we want to multiply by $\frac{4}{\frac{36}{25}}=\frac{25}{9}$. So $\frac{1}{16}*\frac{25}{9}=\boxed{\frac{25}{144}}$.

thedudemanguyperson Feb 25, 2021