R9 - Percentage changes and financial mathematics

Every percentage compares with a base quantity. You will calculate percentage changes, reverse them to recover an original value, use simple interest and compare the complete costs of financial offers.

Comparing quantities per hundred

If a price rises by 20% and then falls by 20%, is it back where it started? Try £100: it rises to £120, but 20% of £120 is £24. The final price is £96. Understanding what counts as 100% explains the surprise.

You already know

A percentage is a fraction with denominator 100. For example, $35\%=35/100=0.35$ , so 14 out of 40 is $14/40=0.35=35\%$ .

The base is the quantity you choose to represent the whole hundred parts.

Percentage

A percentage is a number of parts per hundred. When expressing one quantity as a percentage of another, the second quantity is the base representing 100%. A comparison can exceed 100% if the first quantity is larger.

Write down the base before dividing. Comparable measurements must use common units: $300\,\mathrm g$ as a percentage of $1.2\,\mathrm{kg}$ is $300/1200\times100\%=25\%$ .

Percentage comparison and change

\begin{aligned} \text{percentage of base}&=\frac{\text{compared quantity}}{\text{base quantity}}\times100\%\\ \text{percentage increase}&=\frac{\text{new}-\text{original}}{\text{original}}\times100\%\\ \text{percentage decrease}&=\frac{\text{original}-\text{new}}{\text{original}}\times100\% \end{aligned}

Use a non-zero base. For these ordinary change calculations, original quantities are positive; choose increase or decrease to match the direction.

Not on the formula sheet: learn it.

“Percentage of” uses the whole new amount; “percentage increase” uses only the amount added.

Worked example 1: a comparison above 100%

Question. A ribbon length grows from $12\,\mathrm{cm}$ to $18\,\mathrm{cm}$ . Express the new length as a percentage of the original and calculate the percentage increase.

Plan. Use the original length as base, with a different numerator for each request.

\begin{aligned} \text{new as percentage of original}&=\frac{18}{12}\times100\%=150\% && \text{compare whole lengths}\\ \text{increase}&=18-12=6\,\mathrm{cm} && \text{find the amount added}\\ \text{percentage increase}&=\frac6{12}\times100\%=50\% && \text{compare the change} \end{aligned}

Why this step? The original is 100%, so the additional 50% gives a new amount of 150%.

Answer. 150% of the original; a 50% increase.

Check. Half of $12\,\mathrm{cm}$ is $6\,\mathrm{cm}$ , and $12+6=18$ .

Change only the direction: go from the same longer ribbon back to the shorter one. The base changes.

Worked example 2: reversing the comparison

Question. A ribbon shrinks from $18\,\mathrm{cm}$ to $12\,\mathrm{cm}$ . Express the new length as a percentage of the original and calculate the percentage decrease.

Plan. Use 18 as the original base for both calculations.

\begin{aligned} \text{new as percentage of original}&=\frac{12}{18}\times100\%=66\frac23\% && \text{compare lengths}\\ \text{decrease}&=18-12=6\,\mathrm{cm} && \text{find the amount lost}\\ \text{percentage decrease}&=\frac6{18}\times100\%=33\frac13\% && \text{use the new base} \end{aligned}

Why this step? The same $6\,\mathrm{cm}$ change is a smaller fraction of 18 than of 12.

Answer. $66\frac23\%$ of the original; a $33\frac13\%$ decrease.

Check. $66\frac23+33\frac13=100$ .

Rates already written as percentages can have a percentage-point change. Moving from 40% to 50% adds 10 percentage points, but its relative increase is $(50-40)/40\times100\%=25\%$ . Also, £75 is 125% of £60, while £60 is 80% of £75: order matters.

Watch out

Wrong: an increase from 12 to 18 is 150%. That is the new amount as a percentage of the original. Correct: the increase is $(18-12)/12\times100\%=50\%$ .

Use the same two requests for another decrease.

Multipliers for increases and decreases

A multiplier combines the original whole with the fraction added or removed, so one multiplication gives the new value.

For a 15% increase, keep 100% and add 15%: $115\%=1.15$ . For a 15% decrease, keep $85\%=0.85$ . In fraction form, a 20% decrease retains $4/5$ , so $75\times4/5=60$ .

Percentage change multipliers

\begin{aligned} \text{new value}&=\text{original}\times\left(1+\frac r{100}\right) && r\%\text{ increase}\\ \text{new value}&=\text{original}\times\left(1-\frac r{100}\right) && r\%\text{ decrease} \end{aligned}

For ordinary positive quantities, use $r\ge0$ for an increase and $0\le r\le100$ for a decrease that leaves a non-negative amount.

Not on the formula sheet: learn it.

An increase may exceed 100%: a 125% increase has multiplier $2.25$ , taking 40 to 90. Separate the retained whole from the added fraction.

Worked example 3: an increase

Question. Increase a price of £80 by 15%.

Plan. Convert the retained 115% to a decimal multiplier.

\begin{aligned} \text{multiplier}&=1+\frac{15}{100}=1.15 && \text{keep the original and add}\\ \text{new price}&=80\times1.15=92 && \text{apply to the original} \end{aligned}

Why this step? Multiplying by $0.15$ would find only the increase, rather than the final price.

Answer. £92.

Check. $0.15\times80=12$ , and $80+12=92$ .

Change only the direction of change: the same price and percentage now decrease.

Worked example 4: a decrease

Question. Decrease a price of £80 by 15%.

Plan. Find the retained 85%, then multiply the original price.

\begin{aligned} \text{multiplier}&=1-\frac{15}{100}=0.85 && \text{remove the discount fraction}\\ \text{new price}&=80\times0.85=68 && \text{find the remaining amount} \end{aligned}

Why this step? A positive price after a discount must be less than the original.

Answer. £68.

Check. $80-12=68$ , using the same £12 change.

Read multipliers backwards too: $1.08$ means an 8% increase; $0.92$ means an 8% decrease; $1.8$ means an 80% increase. A decrease beyond 100% would make a positive original negative, so it needs a suitable context rather than an ordinary sale-price interpretation.

Watch out

Wrong: a 15% discount on £80 gives $80\times0.15=£12$ as the sale price. £12 is the discount. Correct: subtract it from £80, or multiply by $0.85$ , giving £68.

Change only the discount rate.

Recovering original values and identifying the base

To recover an original value, undo the multiplication that produced the changed value.

The original is the 100% base. A changed price might be 80% or 115% of that base; being the only given price does not make it 100%.

Worked example 5: reversing a discount

Question. A price after a 20% discount is £64. Find the original price.

Plan. Treat £64 as 80% of the original and divide by $0.8$ .

\begin{aligned} 0.8P&=64 && \text{write the forward change}\\ P&=64\div0.8=80 && \text{undo multiplication} \end{aligned}

Why this step? Adding 20% of £64 would use the discounted price as the base, which is the wrong whole.

Answer. £80.

Check. $80\times0.8=64$ ; the discount is £16.

Change only the direction of the original percentage change. The final price stays £64.

Worked example 6: reversing an increase

Question. A price after a 20% increase is £64. Find the original price, to the nearest penny.

Plan. Treat £64 as 120% of the original and divide by $1.2$ .

\begin{aligned} 1.2P&=64 && \text{write the forward increase}\\ P&=64\div1.2=\frac{160}{3} && \text{recover the base}\\ P&\approx53.33 && \text{round only at the end} \end{aligned}

Why this step? The original must be smaller than the increased value; division by $1.2$ gives that direction.

Answer. £53.33 to the nearest penny.

Check. Using the unrounded value, $(64/1.2)\times1.2=64$ .

The same method reverses a 15% increase to £230: $230/1.15=£200$ . You can also recover 1% by dividing by 115, then multiply by 100. These reverse-percentage methods are not on the formula sheet: derive them by undoing the taught percentage multiplier.

Equal rates in opposite directions do not cancel because their bases differ. From £120 back to £100 requires a decrease of $20/120\times100\%=16\frac23\%$ . Likewise, 90 to 72 is a 20% decrease, but 72 to 90 is a 25% increase.

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R9 - Percentage changes and financial mathematics

Comparing quantities per hundred

Multipliers for increases and decreases

Recovering original values and identifying the base

Continue this lesson.

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