National 5 Maths
Volume

Note that a cylinder behaves as if it is a prism, because it has a uniform circular cross-section.

Beware! The units are mismatched here. We will use 10 mm instead of 1 cm.

$$ \begin{eqnarray} V&=& Ah\\[6pt] &=& \pi r^{2}h\\[6pt] &=& \pi \times 6^2 \times 10\\[6pt] &=& 360 \pi\\[6pt] &\approx& 1131\ \small\textsf{mm³} \end{eqnarray} $$

This simple question just needs us to select the correct volume formula, substitute the values and calculate the answer.

$$ \begin{eqnarray} V&=& \small\frac{1}{3}\normalsize \pi r^{2}h\\[8pt] &=& \small\frac{1}{3}\normalsize \pi \times 3^{2} \times 5\\[8pt] &=& \small\frac{1}{3}\normalsize \pi \times 45\\[8pt] &=& 15 \pi\\[8pt] &\approx& 47.1\ \small\textsf{cm³} \end{eqnarray} $$

The diameter \(d=8\) so the radius \(r=4\small.\)

Half of \(\large\frac{4}{3}\) is \(\large\frac{2}{3}\) so we can use this formula for the hemisphere:

$$ \begin{eqnarray} V&=& \small\frac{2}{3}\normalsize\pi r^3\\[8pt] &=& \small\frac{2}{3}\normalsize\pi\times 4^3\\[8pt] &=& \small\frac{2}{3}\normalsize\pi\times 64\\[8pt] &=& \small\frac{128}{3}\normalsize\pi\ \small\textsf{cm³}\\[8pt] \end{eqnarray} $$

Note that \(\large\frac{128}{3}\normalsize\pi\) and \(\large\frac{128\pi}{3}\) are the same, so the answer may be written in either form.

Here, we are given the volume and we need to work backwards to find the radius. This involves multiplying both sides of the equation by \(3\) to get rid of the fraction, dividing both sides by \(4\pi\) and finally cube rooting.

$$ \begin{gather} V = \small\frac{4}{3}\normalsize\pi r^3\\[8pt] 9 = \small\frac{4}{3}\normalsize\pi r^3\\[8pt] 27 = 4\pi r^3\\[8pt] \small\frac{27}{4\pi}\normalsize = r^3\\[8pt] r = \sqrt[\leftroot{-1}\uproot{6}\scriptstyle 3]{\small\frac{27}{4\pi}\normalsize}\\[8pt] r\approx 1.290\\[12pt] \end{gather} $$

So the diameter is \(2\times 1.290 \approx 2.58\) metres.

Note that we are not told the shape of the base, but that doesn't matter. We know its area.

$$ \begin{eqnarray} V&=& \small\frac{1}{3}\normalsize Ah\\[8pt] 36&=& \small\frac{1}{3}\normalsize\times 12 h\\[8pt] 36&=& 4h\\[8pt] h&=& 9 \end{eqnarray} $$

So the height of the pyramid is 9 cm.

Fun fact: The shape of the carton is called a conical frustrum. You don't need to know that!

We need to subtract the volume of the small cone from the volume of the large cone.

Note that the diameters are given in the question, so we halve these to get the radii.

$$ \begin{eqnarray} \small\textsf{Volume of large cone}\normalsize &=& \small\frac{1}{3}\normalsize \pi r^{2}h\\[8pt] &=& \small\frac{1}{3}\normalsize \pi \times 16^{2} \times 24\\[8pt] &\approx& 6434.0\ \small\textsf{cm³}\\[12pt] \small\textsf{Volume of small cone}\normalsize &=& \small\frac{1}{3}\normalsize \pi r^{2}h\\[8pt] &=& \small\frac{1}{3}\normalsize \pi \times 9^{2} \times 13.5\\[8pt] &\approx& 1145.1\ \small\textsf{cm³}\\[8pt] \end{eqnarray} $$

The volume of the carton is therefore \(6434.0-1145.1=5288.9\) cm³.

Rounded to 2 significant figures, this is \(5300\) cm³.

The volume of the chocolate coating is the difference between the volumes of the big sphere (the whole sweet) and the small sphere (just the caramel centre).

Note that the radius of the whole sweet is \(24\div 2=12\) mm and the radius of the caramel filling is \(3\) mm less than this, so it is \(9\) mm.

$$ \begin{eqnarray} \small\textsf{Volume of whole sweet}\normalsize &=& \small\frac{4}{3}\normalsize \pi r^3\\[8pt] &=& \small\frac{4}{3}\normalsize \pi \times 12^3\\[8pt] &\approx& 7238.23\ \small\textsf{mm³}\\[12pt] \small\textsf{Volume of caramel centre}\normalsize &=& \small\frac{4}{3}\normalsize \pi r^3\\[8pt] &=& \small\frac{4}{3}\normalsize \pi \times 9^3\\[8pt] &\approx& 3053.63\ \small\textsf{mm³}\\[8pt] \end{eqnarray} $$

So the volume of the chocolate coating is \(7238.23-3053.63\approx 4184.6\) mm³.

Rounded to 3 significant figures, this is \(4180\) mm³.

This relatively simple 3-mark non-calculate question requires us to work backwards and then evaluate \(138 \div 12\small.\)

Note that if you are comfortable with writing \(138\) as \(138.0\) and using short division to divide by \(12\) then you could skip the step where we simplified the fraction.

$$ \begin{eqnarray} V&=& \small\frac{1}{3}\normalsize Ah\\[8pt] 138&=& \small\frac{1}{3}\normalsize\times 6 \times 6 \times h\\[8pt] 138&=& \small\frac{1}{3}\normalsize\times 36h\\[8pt] 138&=& 12h\\[8pt] h&=& \small\frac{138}{12}\normalsize\\[8pt] h&=& \small\frac{23}{2}\normalsize\\[8pt] \end{eqnarray} $$

So the height of the pyramid is \(23\div 2=11.5\) cm.

This non-calculator question combined the 3D coordinates topic and the volume topic.

Note that the radius of the hemisphere is 3 units.

Half of \(\large\frac{4}{3}\) is \(\large\frac{2}{3}\) so we can use this formula:

$$ \begin{eqnarray} V&=& \small\frac{2}{3}\normalsize\pi r^3\\[8pt] &=& \small\frac{2}{3}\normalsize\pi (3^3)\\[8pt] &=& \small\frac{2}{3}\normalsize\pi (27)\\[8pt] &=& 18\pi\ \small\textsf{cm³}\\[8pt] \end{eqnarray} $$

This question is similar to 2016 P2 Q7 above, except that the base in this question is a square.

We need to subtract the volume of the small pyramid from the volume of the large pyramid.

Note that the total height is \(60+48=108\) cm.

$$ \begin{eqnarray} \small\textsf{Volume of large pyramid}\normalsize &=& \small\frac{1}{3}\normalsize Ah\\[8pt] &=& \small\frac{1}{3}\normalsize \times 90^2 \times 108\\[8pt] &=& 291\small\,\normalsize 600\ \small\textsf{cm³}\\[12pt] \small\textsf{Volume of small pyramid}\normalsize &=& \small\frac{1}{3}\normalsize Ah\\[8pt] &=& \small\frac{1}{3}\normalsize \times 40^2 \times 48\\[8pt] &=& 25\small\,\normalsize 600\ \small\textsf{cm³}\\[8pt] \end{eqnarray} $$

So the volume of the concrete block is \(291\small\,\normalsize 600-25\small\,\normalsize 600=266\small\,\normalsize 000\) cm³.