1.6 - Reverse percentages

A sale label tells you today's price. Can you recover yesterday's? You will undo percentage increases and decreases, choose the correct base, and check that your answer recreates the given amount.

Recovering an amount before an increase

A ticket now costs $\pounds67.20$ after a $12\%$ increase. Was its old price $\pounds60$ , or should you subtract $12\%$ of the new price? Keep that question in mind.

You already know

An increase of $12\%$ uses multiplier $1.12$ . Multiplication and division undo each other: $6\times4=24$ means $24\div4=6$ .

The old price represents $100\%$ . The new price represents $112\%$ of that same old price. We know the result of multiplying; we need to work backwards.

Reverse percentage

A reverse percentage calculation finds an original amount from an amount after a stated percentage change. Finding a discount from an original price is a forward calculation instead.

You can divide by the multiplier in one step. Alternatively, divide by $112$ to find $1\%$ , then multiply by $100$ . Both routes recover the same whole.

Reversing an increase

For an increase of $p\%$ ,

\text{original}=\frac{\text{new amount}}{1+p/100}

The new amount must result from this percentage increase alone. Foundation and Higher: Not in the formulae sheet: learn it.

The arrows show why the reverse operation is division.

[DIAGRAM: asset_slug: reverse_percentage_arrows; caption: The forward multiplier and its inverse connect the same two prices; description: An original price of 60 pounds becomes 67.20 pounds by multiplication by 1.12. The reverse arrow divides 67.20 by 1.12 to recover 60. The original is labelled 100 percent and the new price 112 percent; file: reverse_percentage_arrows.png; recommended_method: matplotlib]
The forward multiplier and its inverse connect the same two prices

Worked example 1: recover the ticket price

Question. A ticket costs $\pounds67.20$ after a $12\%$ increase. Find its original price.

Plan. Identify the multiplier from the original price, then divide by it.

\begin{aligned} 100\%+12\%&=112\% &&\text{new percentage}\\ 1.12x&=67.20 &&\text{original price }x\\ x&=67.20\div1.12 &&\text{undo multiplication}\\ x&=60 &&\text{calculate} \end{aligned}

Why divide by $1.12$ ? The given price is $112\%$ , rather than $100\%$ , of the original.

Answer. The original price was $\pounds60$ .

Check. $0.12\times60=7.20$ , and $60+7.20=67.20$ .

Now change only the rate. A larger increase requires a different divisor, even though the new price is unchanged.

Worked example 2: change the increase rate

Question. A ticket costs $\pounds67.20$ after a $20\%$ increase. Find its original price.

Plan. Divide the new price by its new multiplier.

\begin{aligned} 100\%+20\%&=120\% &&\text{identify the whole}\\ x&=67.20\div1.20 &&\text{reverse the increase}\\ x&=56 &&\text{calculate} \end{aligned}

Why is the answer smaller this time? More growth reached the same final price, so the starting price was lower.

Answer. The original price was $\pounds56$ .

Check. $56\times1.20=67.20$ .

The word “original” is a reminder to label the amount that represents $100\%$ .

Watch out

The line $67.20\times0.88=59.136$ does not undo the $12\%$ rise. It removes $12\%$ of the new price. Use $67.20\div1.12=60$ .

Try the same rate with a different final price.

Recovering an amount before a decrease

A discount leaves a fraction of the old price. After $30\%$ off, the sale price is $70\%$ of the original. The percentage remaining is the useful one.

Reversing a decrease

For a decrease of $p\%$ , where $0\leq p<100$ ,

\text{original}=\frac{\text{new amount}}{1-p/100}

Foundation and Higher: Not in the formulae sheet: learn it.

Dividing by a positive number below $1$ makes the result larger. That fits a sale: the old price should exceed the discounted price. A $100\%$ reduction leaves zero whatever the old price was, so it cannot reveal a unique original.

Worked example 3: recover a sale price

Question. A coat costs $\pounds56$ after a $30\%$ discount. Find its original price.

Plan. Find the percentage remaining, then recover $100\%$ .

\begin{aligned} 100\%-30\%&=70\% &&\text{percentage remaining}\\ 0.70x&=56 &&\text{sale price relationship}\\ x&=56\div0.70=80 &&\text{recover the original} \end{aligned}

Why divide by $0.70$ ? The $\pounds56$ is the amount kept after the discount, not the amount removed.

Answer. The coat originally cost $\pounds80$ .

Check. $30\%$ of $\pounds80$ is $\pounds24$ , leaving $\pounds56$ .

Keep the sale price, but reduce the discount rate. The original price should now be closer to the sale price.

Worked example 4: a smaller discount

Question. A coat costs $\pounds56$ after a $20\%$ discount. Find its original price.

Plan. Divide by the fraction remaining after this discount.

\begin{aligned} 100\%-20\%&=80\% &&\text{amount kept}\\ x&=56\div0.80 &&\text{undo the multiplier}\\ x&=70 &&\text{calculate} \end{aligned}

Why is $0.20$ the wrong divisor? It describes the discount, while $56$ describes the price after discount.

Answer. The original price was $\pounds70$ .

Check. $70\times0.80=56$ .

Before using your calculator, write which percentage the given amount represents. That one line makes your method clear.

Watch out

The line $56\times1.20=67.20$ adds $20\%$ of the reduced price. It does not reverse the discount. The correct line is $56\div0.80=70$ .

Use the same remaining multiplier with a new sale price.

Checking the base and the inverse

A percentage always has a base: the amount it is a percentage of. Equal percentage rates can describe unequal cash amounts when their bases differ.

Here is a useful surprise. Doubling a price is a $100\%$ increase, but halving the new price is only a $50\%$ decrease. For example, $40\times2=80$ and $80\times0.5=40$ . Opposite rates need not undo each other; inverse multipliers do.

Inverse multiplier

If $\text{new}=m\times\text{original}$ and $m\ne0$ , then

\text{original}=\frac{\text{new}}m

The reverse multiplier is $1/m$ . Foundation and Higher: Not in the formulae sheet: learn it.

To test a proposed original, send it forwards through the stated change. A plausible size is helpful, but recreating the exact new amount is stronger evidence.

Worked example 5: test a proposed original

Question. An amount becomes $\pounds144$ after a $20\%$ increase. Sam says the original was $\pounds115.20$ . Is Sam correct? Find the original.

Plan. Apply the forward multiplier to Sam's answer, then use division to recover the correct value.

\begin{aligned} 115.20\times1.20&=138.24 &&\text{test Sam's amount}\\ 138.24&\ne144 &&\text{it fails the check}\\ x&=144\div1.20=120 &&\text{recover the original} \end{aligned}

Why use a forward check? Any true original must produce the given final amount under the stated change.

Answer. Sam is incorrect. The original was $\pounds120$ .

Check. $120\times1.20=144$ .

Keep the $20\%$ increase and the checking method. Change only the proposed original, so you can see what a successful check looks like.

Worked example 6: recognise a successful check

Question. An amount becomes $\pounds144$ after a $20\%$ increase. Jo says the original was $\pounds120$ . Is Jo correct?

Plan. Apply the stated increase to the proposed amount.

\begin{aligned} 120\times1.20&=144 &&\text{apply the increase}\\ 144\div1.20&=120 &&\text{confirm the inverse} \end{aligned}

Why does this settle it? The nonzero multiplier has just one inverse, and Jo's amount recreates the given price.

Answer. Jo is correct: the original was $\pounds120$ .

Check. The added amount is $\pounds24$ , which is $20\%$ of $\pounds120$ .

We can now settle the opening ticket puzzle. $\pounds60$ grows to $\pounds67.20$ ; subtracting $12\%$ of $\pounds67.20$ uses the wrong base.

Watch out

The proposed inverse $1.20^{-1}=0.80$ is false. Check the product: $1.20\times0.80=0.96$ , not $1$ . The inverse is $1\div1.20$ .

Make the final decision by checking the given candidate.

Quick Recap

Label the original as $100\%$ .
Divide by $1+p/100$ to reverse an increase of $p\%$ .
Divide by $1-p/100$ to reverse a decrease below $100\%$ .
Check by applying the stated change to your recovered original.

The opening ticket originally cost $\pounds60$ : $67.20\div1.12=60$ .

Last Accessed

Number

Algebra

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Handling data

1.6 - Reverse percentages

Recovering an amount before an increase

Recovering an amount before a decrease

Checking the base and the inverse

Quick Recap

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