The volume of a cylinder is given by cross-sectional area $\times$ length, so $\pi r^2\times \text{length}$.

Working in mm, the original cylinder of sugar has volume $\pi \left(\dfrac{30}{2}\right)^2\times 40mm^3$

**Finding the volume of sugar as a number**

$\pi \left(\dfrac{30}{2}\right)^2\times 40=\pi\dfrac{30^2}{4}\times40=\pi\times900\times10=9000\pi$.

The cylinder of candy floss has volume $\pi \left(\dfrac{1}{2}\right)^2 x$, so $$\begin{align}\pi \left(\frac{1}{2}\right)^2 x=&9000\pi\\

\Rightarrow\pi\times\frac{1}{4}x=&9000\pi\\

\Rightarrow x=&9000\times4\\

\Rightarrow x=&36000\end{align}$$So $x=36000$mm, or $36$m.

**Using algebra to find the length directly**

The cylinder of candy floss has volume $\pi \left(\dfrac{1}{2}\right)^2 x$, so $$\begin{align}\pi \left(\frac{1}{2}\right)^2 x&=\pi \left(\frac{30}{2}\right)^2\times 40\\

\Rightarrow 1^2x&=30^2\times40\\

\Rightarrow x&=36000\end{align}$$

So $x=36000$mm, or $36$m.

**Using scale factors of enlargement**

The ratio of the diameters of the cylinders is 1:30,

so the ratio of the cross-sectional areas is 1:900,

so the height of the thin cylinder will need to 900 times longer than the height of the fat cylinder.

Therefore the height of the thin cylinder is 36m long.

Working in mm, the original cylinder of sugar has volume $\pi \left(\dfrac{30}{2}\right)^2\times 40mm^3$

$\pi \left(\dfrac{30}{2}\right)^2\times 40=\pi\dfrac{30^2}{4}\times40=\pi\times900\times10=9000\pi$.

The cylinder of candy floss has volume $\pi \left(\dfrac{1}{2}\right)^2 x$, so $$\begin{align}\pi \left(\frac{1}{2}\right)^2 x=&9000\pi\\

\Rightarrow\pi\times\frac{1}{4}x=&9000\pi\\

\Rightarrow x=&9000\times4\\

\Rightarrow x=&36000\end{align}$$So $x=36000$mm, or $36$m.

The cylinder of candy floss has volume $\pi \left(\dfrac{1}{2}\right)^2 x$, so $$\begin{align}\pi \left(\frac{1}{2}\right)^2 x&=\pi \left(\frac{30}{2}\right)^2\times 40\\

\Rightarrow 1^2x&=30^2\times40\\

\Rightarrow x&=36000\end{align}$$

So $x=36000$mm, or $36$m.

The ratio of the diameters of the cylinders is 1:30,

so the ratio of the cross-sectional areas is 1:900,

so the height of the thin cylinder will need to 900 times longer than the height of the fat cylinder.

Therefore the height of the thin cylinder is 36m long.

You can find more short problems, arranged by curriculum topic, in our short problems collection.