National 5 Maths
Percentages

In this 3-mark question, we were told that \(80\%\) of the total equals \(480\,000\small.\)

Probably the simplest way to work out what \(100\%\) equals is to lay our working out as follows.

As this is a Paper 1 (non-calculator) question, we need to think of an "easy" number that is a factor of both \(80\) and \(100\small.\) There are two good options that keep the arithmetic simple: \(10\) or \(20\small.\)

We will use \(20\%\) for the middle step. As an exercise, you might like to try the question again for yourself, but using \(10\%\) instead.

$$ \begin{eqnarray} 80\%\ &=& 480\,000 \\[6pt] \small\div\textsf{4} &\ & \small \div \textsf{4} \normalsize\\[6pt] 20\%\ &=& 120\,000 \\[6pt] \small \times \textsf{5} &\ & \small \times \textsf{5} \normalsize\\[6pt] 100\%\ &=& 600\,000 \\[6pt] \end{eqnarray} $$

So the total number of available tickets was \(600\,000\small.\)

This 3-mark non-calculator question is very similar to example 1 above.

The best common factor of \(75\) and \(100\) to use for the middle step is \(25\small.\) To keep the arithmetic as simple as possible, we should usually try to think of the highest common factor.

$$ \begin{eqnarray} 75\%\ &=& 2400 \\[6pt] \small\div\textsf{3} &\ & \small \div \textsf{3} \normalsize\\[6pt] 25\%\ &=& 800 \\[6pt] \small \times \textsf{4} &\ & \small \times \textsf{4} \normalsize\\[6pt] 100\%\ &=& 3200 \\[6pt] \end{eqnarray} $$

So the band sold \(3200\) tickets for their Glasgow gig.

This is a reverse percentage question.

We think of the original price as \(100\%\).

So the sale price is \(100\%-20\%=80\%\) of the original price.

Method 1:

Create an equation. Let's use \(x\) for the original price.

$$ \begin{eqnarray} \small\textsf{sale price}\normalsize &=& 80\% \small\textsf{ of original price}\normalsize \\[6pt] 48 &=& 0.8x \\[6pt] x &=& \frac{48}{0.8} \\[6pt] x &=& \frac{480}{8}\:\small\textsf{ (x10 top and bottom)}\normalsize \\[6pt] x &=& 60 \end{eqnarray} $$

So the original price was £60.

Method 2:

This method doesn't use algebra. Some people find it easier.

Because this is a non-calculator question, we don't use 1% for the middle step. Instead, we choose a nice, round number that is a factor of both 80 and 100. We used 10%, but 20% would also work well.

$$ \begin{eqnarray} 80\%\ \small\textsf{of the original price}\normalsize &=& £48 \\[6pt] \small\div\textsf{8} &\ & \small \div \textsf{8} \normalsize\\[6pt] 10\%\ \small\textsf{of the original price}\normalsize &=& £6 \\[6pt] \small \times \textsf{10} &\ & \small \times \textsf{10} \normalsize\\[6pt] 100\%\ \small\textsf{of the original price}\normalsize &=& £60 \\[6pt] \end{eqnarray} $$

Again, we think of the original number as \(100\%\).

This has increased by \(15\%\), so it's now \(100\%+15\%=115\%\) of the original number.

Method 1:

Create an equation. Let's use \(x\) for the original number.

$$ \begin{eqnarray} \small\textsf{new number}\normalsize &=& 115\% \small\textsf{ of original number}\normalsize \\[6pt] 161 &=& 1.15x \\[6pt] x &=& \frac{161}{1.15} \:\small\textsf{ (not suitable for Paper 1)}\normalsize \\[6pt] x &=& 140 \end{eqnarray} $$

So the firm had 140 clients before their marketing campaign.

Method 2:

$$ \begin{eqnarray} 115\%\ \small\textsf{of the original number}\normalsize &=& 161 \\[6pt] \small\div\textsf{115} &\ & \small \div \textsf{115} \normalsize\\[6pt] 1\%\ \small\textsf{of the original number}\normalsize &=& 1.4 \\[6pt] \small \times \textsf{100} &\ & \small \times \textsf{100} \normalsize\\[6pt] 100\%\ \small\textsf{of the original number}\normalsize &=& 140 \\[6pt] \end{eqnarray} $$

As always, we think of the original value as \(100\%\).

This has increased by \(8\%\), so it's now \(100\%+8\%=108\%\) of the original value.

Method 1:

Create an equation. Let's use \(x\) for the original number.

$$ \begin{eqnarray} \small\textsf{new value}\normalsize &=& 108\% \small\textsf{ of original value}\normalsize \\[6pt] 94\,500 &=& 1.08x \\[6pt] x &=& \frac{94\,500}{1.08} \:\small\textsf{ (not suitable for Paper 1)}\normalsize \\[6pt] x &=& 87\,500 \end{eqnarray} $$

So Nadim paid £87 500 for his flat.

Method 2:

$$ \begin{eqnarray} 108\%\ \small\textsf{of the original value}\normalsize &=& £94\,500 \\[6pt] \small\div\textsf{108} &\ & \small \div \textsf{108} \normalsize\\[6pt] 1\%\ \small\textsf{of the original value}\normalsize &=& £875 \\[6pt] \small \times \textsf{100} &\ & \small \times \textsf{100} \normalsize\\[6pt] 100\%\ \small\textsf{of the original value}\normalsize &=& £87\,500\\[6pt] \end{eqnarray} $$

For a \(2.5\%\) annual increase, we multiply by \(100\%+2.5\%=102.5\%\)

As a decimal, \(102.5\%=102.5 \div 100=1.025\)

So the final balance will be \( 20\small\ \normalsize 000\times 1.025^3 = £21\small\ \normalsize 537.81\)

The question asks for the interest, so this is \(£21\small\ \normalsize 537.81-£20\small\ \normalsize 000=£1537.81\)

For a \(15\%\) annual decrease, we multiply by \(100\%-15\%=85\%=0.85\)

So the value after four years will be \( 25\small\ \normalsize 000\times 0.85^4 = £13\small\ \normalsize 050.15\)

Rounded to 3 significant figures, this is \(£13\small\ \normalsize 100\).

\(15\%\) decrease: \(100\%-15\%=85\%=0.85\)

So the number after three years will be \( 964\times 0.85^3 = 592.0165\)

Rounded to the nearest \(10\small,\) this is \(590\) pupils.

\(15\%\) increase: \(100\%+15\%=115\%=1.15\)

So the number after three years will be \( 80\small\ \normalsize 000\times 1.15^3 = 121\small\ \normalsize 670\)

\(4\%\) increase: \(100\%+4\%=104\%=1.04\)

So the value after two years will be \( 250\small\ \normalsize 000\times 1.04^2 = £270\small\ \normalsize 400\)

We think of the original price as \(100\%\).

So the discounted price is \(100\%-30\%=70\%\) of the price before discount.

Method 1:

Create an equation. Let's use \(x\) for the original price.

$$ \begin{eqnarray} \small\textsf{discounted price}\normalsize &=& 70\% \small\textsf{ of original price}\normalsize \\[6pt] 16.10 &=& 0.7x \\[6pt] x &=& \frac{16.1}{0.7} \\[6pt] x &=& \frac{161}{7}\:\small\textsf{ (x10 top and bottom)}\normalsize \\[6pt] x &=& 23 \end{eqnarray} $$

So the original price was £23.

Method 2:

This method is more commonly used than the first.

10 is a nice, round number that is a factor of both 70 and 100, so let's use 10% for the middle step.

$$ \begin{eqnarray} 70\%\ \small\textsf{of the original price}\normalsize &=& £16.10 \\[6pt] \small\div\textsf{7} &\ & \small \div \textsf{7} \normalsize\\[6pt] 10\%\ \small\textsf{of the original price}\normalsize &=& £2.30 \\[6pt] \small \times \textsf{10} &\ & \small \times \textsf{10} \normalsize\\[6pt] 100\%\ \small\textsf{of the original price}\normalsize &=& £23 \\[6pt] \end{eqnarray} $$

\(11\%\) depreciation: \(100\%-11\%=89\%=0.89\)

\(6\%\) depreciation: \(100\%-6\%=94\%=0.94\)

So the value after three years will be \( 20\small\ \normalsize 000\times 0.89\times 0.94^2 = £15\small\ \normalsize 728.08\)

Note: This is a silly level of accuracy, but the question didn't tell us to round it, so we should leave it as it is.

\(26\%\) decrease: \(100\%-26\%=74\%=0.74\)

So the value after three years will be \( 460\times 0.74^3 = £186.40\small,\) rounded to the nearest penny.

We think of the cost last year as \(100\%\).

This has increased by \(16\%\), so it's now \(100\%+16\%=116\%\) of last year's cost.

Method 1:

Create an equation. Let's use \(x\) for the original cost.

$$ \begin{eqnarray} \small\textsf{new cost}\normalsize &=& 116\% \small\textsf{ of original cost}\normalsize \\[6pt] 278.40 &=& 1.16x \\[6pt] x &=& \frac{278.40}{1.16}\normalsize \\[6pt] x &=& 240 \end{eqnarray} $$

So Charley's car insurance cost £240.00 last year.

Method 2:

$$ \begin{eqnarray} 116\%\ \small\textsf{of last year's cost}\normalsize &=& £278.40 \\[6pt] \small\div\textsf{116} &\ & \small \div \textsf{116} \normalsize\\[6pt] 1\%\ \small\textsf{of last year's cost}\normalsize &=& £2.40 \\[6pt] \small \times \textsf{100} &\ & \small \times \textsf{100} \normalsize\\[6pt] 100\%\ \small\textsf{of last year's cost}\normalsize &=& £240 \\[6pt] \end{eqnarray} $$