Percentage Increase and Decrease
Every question with a fully worked solution.
- Percentage Increase and Decrease - Video - Percentage Increase and Decrease 2 Watch
Theory
An increase or a decrease by a percentage is one multiplication. An increase of \(p\%\) ends at \((100 + p)\%\) and a decrease ends at \((100 - p)\%\): a \(15\%\) decrease on \(\$120\) is \(0.85 \times 120 = \$102\). This Year 8 maths page shows how.
The whole amount is \(100\%\). An increase of \(p\%\) ends at \((100 + p)\%\); a decrease of \(p\%\) ends at \((100 - p)\%\). Divide by \(100\) for the multiplier: a \(12\%\) rise is \(\times\,1.12\), and a \(30\%\) fall is \(\times\,0.7\).
Two steps or one. Two steps: find the change, then add it on or take it away. One step: multiply by the multiplier. They agree, and the two-step way also gives the size of the change.
Read a multiplier backwards. Above \(1\) is an increase and below \(1\) is a decrease: \(\times\,1.04\) is a \(4\%\) rise, and \(\times\,0.55\) is a \(45\%\) fall.
Decimals and big percentages work the same way. A \(4.5\%\) rise is \(\times\,1.045\). A \(150\%\) rise ends at \(250\%\), so the amount more than doubles.
Percentage change from a before and an after amount: \(\dfrac{\text{change}}{\text{original amount}} \times 100\%\).
Profit and loss. A profit is an increase on the cost price and a loss is a decrease on it. Selling price \(=\) cost price \(\times\) the multiplier.
Two changes in a row. Multiply the two multipliers. A \(10\%\) fall then another \(10\%\) fall is \(0.9 \times 0.9 = 0.81\), a \(19\%\) fall, not \(20\%\). A value that drops by the same percentage each year (depreciation) is multiplied again each year.
Working backwards. New amount \(=\) multiplier \(\times\) old amount, so divide to go back: \(\$96\) after a \(20\%\) rise came from \(96 \div 1.2 = \$80\).
Multipliers:
| Change | Ends at | Multiplier |
|---|---|---|
| increase by \(12\%\) | \(112\%\) | \(1.12\) |
| decrease by \(30\%\) | \(70\%\) | \(0.7\) |
| increase by \(4.5\%\) | \(104.5\%\) | \(1.045\) |
| increase by \(150\%\) | \(250\%\) | \(2.5\) |
| decrease by \(8\%\) | \(92\%\) | \(0.92\) |
Increasing or decreasing by a percentage
- Write the new amount as a percentage of the old: \((100 \pm p)\%\).
- Change it to a decimal multiplier.
- Multiply the original amount (change units first if needed).
- Round money to the cent and check: above \(1\) gives more, below \(1\) gives less.
Finding the percentage change
- Subtract to find the change.
- Divide by the original amount and multiply by \(100\%\).
- Say whether it is an increase or a decrease.
(ii) A winter shirt priced at \(\$120\) was decreased by \(15\%\). What was the reduced price?
| \(2400 \times 4\%\) | \(=\) | \(2400 \times \dfrac{4}{100}\) |
| \(\) | \(=\) | \(24 \times 4 = 96\) |
| \(2400 + 96\) | \(=\) | \(2496\) |
| \((100 - 15)\%\) | \(=\) | \(85\%\) |
| \(120 \times 0.85\) | \(=\) | \(102\) |
(i) The increase is \(\$96\), so her new pay is \(\$2496\).
(ii) \(\$102\)
| \(0.12 \times 245\) | \(=\) | \(29.40\) |
| \(245 + 29.40\) | \(=\) | \(274.40\) |
| \(1.12 \times 245\) | \(=\) | \(274.40\) |
\(\$274.40\) either way.
| \(1.25 \times 1000\) | \(=\) | \(1250~\text{mL}\) |
| \(0.76 \times 1250\) | \(=\) | \(950\) |
\(950\) mL. Check: \(24\%\) of \(1250\) is \(300\), and \(1250 - 300 = 950\).
(ii) A roll rose from \(\$6.00\) to \(\$7.50\). Find the percentage increase.
(iii) A shop multiplies every price by \(0.92\). What change is this?
| \(\dfrac{18}{90} \times 100\%\) | \(=\) | \(20\%\) |
| \(\dfrac{1.50}{6.00} \times 100\%\) | \(=\) | \(25\%\) |
| \(100\% - 92\%\) | \(=\) | \(8\%\) |
(i) a \(20\%\) decrease
(ii) a \(25\%\) increase
(iii) an \(8\%\) decrease
(ii) Increase \(40\) g by \(150\%\).
(iii) A car worth \(\$20\,000\) loses \(10\%\) of its value each year. Find its value after \(2\) years.
| \(1.2 \times 250\) | \(=\) | \(300\) |
| \(2.5 \times 40\) | \(=\) | \(100\) |
| \(0.9 \times 20\,000\) | \(=\) | \(18\,000\) |
| \(0.9 \times 18\,000\) | \(=\) | \(16\,200\) |
(i) \(\$300\)
(ii) \(100\) g
(iii) \(\$16\,200\), a \(19\%\) fall over two years.
(i) Find the single multiplier and the overall change.
(ii) The dam now holds \(4400\) kL. How much did it hold before the rain?
| \(1.25 \times 0.88\) | \(=\) | \(1.1\) |
| \(4400 \div 1.1\) | \(=\) | \(4000\) |
(i) \(1.1\), a \(10\%\) rise overall
(ii) \(4000\) kL. Check: \(1.25 \times 4000 = 5000\) and \(0.88 \times 5000 = 4400\).
Common pitfalls
Frequently asked questions
How do I increase an amount by a percentage?
Multiply by \(1 + \dfrac{p}{100}\): increasing \(\$80\) by \(15\%\) is \(1.15 \times 80 = \$92\).
How do I decrease an amount by a percentage?
Multiply by \(1 - \dfrac{p}{100}\): decreasing \(\$80\) by \(15\%\) is \(0.85 \times 80 = \$68\).
How do I find the percentage change?
Divide the change by the original amount and multiply by \(100\%\).
Does a 10% rise then a 10% fall get back to the start?
No. \(1.1 \times 0.9 = 0.99\), so the result is \(1\%\) less than the start.
How do I find the original amount after a change?
Divide the new amount by the multiplier.