This Year 9 percentage change and simple interest lesson is written for pupils aged 13-14 and follows the National Curriculum. In short, it shows you how to solve percentage increase, decrease, original value and simple interest problems. Before you start, it helps to revise number: approximation, error intervals and calculators, because this lesson builds on it.
Year 9 percentage change and simple interest lesson: what you will learn
This lesson covers multipliers, percentage increase and decrease, finding a percentage change, reverse percentages, repeated change and simple interest.
Starter activity
A jacket costs £50. Shop A takes 20% off. Shop B takes £10 off. Work out both prices. Now the jacket costs £200: compare again. Which shop is better now? Then try this: a price of £100 goes up by 10% and then down by 10%. Predict the final price, then work it out. Were you right?
Percentages as multipliers
Definition: multiplier
A percentage is a number out of 100. A multiplier is the number you multiply by to change an amount by a percentage. To increase by 15% you end up with \(100\% + 15\% = 115\%\), so you multiply by \(1.15\). To decrease by 15% you keep \(100\% - 15\% = 85\%\), so you multiply by \(0.85\).
| Change | Calculation | Multiplier |
|---|---|---|
| Increase by 8% | 100% + 8% = 108% | 1.08 |
| Decrease by 8% | 100% − 8% = 92% | 0.92 |
| Increase by 2.5% | 100% + 2.5% = 102.5% | 1.025 |
| Decrease by 35% | 100% − 35% = 65% | 0.65 |
| Increase by 100% | 200% | 2 |
Percentage increase and decrease
Worked example 1: increase
Increase £80 by 15%.
Multiplier \(1.15\). \(80 \times 1.15 = £92\). By finding the percentage: 10% is £8, 5% is £4, so 15% is £12 and \(80 + 12 = £92\). The multiplier method is quicker and is essential for harder problems.
Worked example 2: decrease
A coat costs £60. In a sale it is reduced by 35%. Find the sale price.
Multiplier \(0.65\). \(60 \times 0.65 = £39\).
Finding a percentage change
Rule: percentage change
\(\text{percentage change} = \dfrac{\text{actual change}}{\text{original}} \times 100\%\)
Always divide by the original value, which is the value before the change happened.
Worked example 3: increase
A console costs £250 and later costs £290. Find the percentage increase.
The change is \(290 - 250 = 40\). \(\dfrac{40}{250} \times 100 = 16\%\) increase.
Worked example 4: decrease
A phone costs £240 and is reduced to £204. Find the percentage decrease.
The change is £36. \(\dfrac{36}{240} \times 100 = 15\%\) decrease.
Original value problems (reverse percentages)
Method: reverse percentages
- Write the multiplier for the change.
- Divide the after-value by the multiplier to get the original.
- Check by multiplying forwards.
Worked example 5: after an increase
After a 20% increase a price is £36. Find the original price.
The multiplier is 1.2. Original \(= 36 \div 1.2 = £30\).
Worked example 6: after a sale
In a sale all prices are cut by 15%. A jacket costs £51 in the sale. Find the original price.
The multiplier is 0.85. Original \(= 51 \div 0.85 = £60\).
Watch out: finding 15% of the new price
A very common mistake is to find 15% of £51 and add it on. That is wrong, because the 15% was taken off the original price, not off £51.
Repeated percentage change
Rule: repeated change
If a quantity changes twice, multiply the multipliers. A £400 television goes up by 10% and then comes down by 10%: \(400 \times 1.1 \times 0.9 = 400 \times 0.99 = £396\). It is not back to £400, because the second 10% is taken from a bigger number.
Simple interest
Definition: simple interest
Interest is the extra money paid when you save or borrow. With simple interest the same amount is paid every year, calculated on the original amount only (the principal).
Rule: simple interest formula
\(I = \dfrac{P \times R \times T}{100}\)
\(P\) is the principal in pounds, \(R\) is the yearly percentage rate and \(T\) is the time in years.
Worked example 7: interest earned
£500 is saved at 3% simple interest for 4 years. Find the interest and the total.
Interest each year is 3% of £500 = £15, so over 4 years \(4 \times 15 = £60\). Using the formula: \(\dfrac{500 \times 3 \times 4}{100} = 60\). The total is \(500 + 60 = £560\).
Worked example 8: finding the rate
Interest of £84 is earned on £700 over 3 years. Find the rate.
Rearrange: \(R = \dfrac{100 \times I}{P \times T} = \dfrac{100 \times 84}{700 \times 3} = \dfrac{8400}{2100} = 4\%\).
Note: choosing a method
Do I know the original and want the new value? Multiply by the multiplier. Do I know the new value and want the original? Divide by the multiplier. Do I know both values and want the percentage change? Use change divided by original. Writing the question type down first prevents most errors.
Watch out: common mistakes
- Dividing by the new value instead of the original value when finding percentage change.
- Using \(0.15\) instead of \(0.85\) as the multiplier for a 15% decrease.
- Adding the percentages for two changes (10% up then 10% down is not 0%).
- In simple interest, forgetting to multiply by the number of years, or forgetting to add the interest to the principal when the total is asked for.
Worked example 9: a decimal percentage
Increase £480 by 7.5%.
The multiplier is \(1+0.075=1.075\). So \(480\times1.075=516\). The new amount is £516.
Worked example 10: compound growth with a power
£1 000 earns 5% compound interest each year. Find the value after 10 years.
Each year the value is multiplied by \(1.05\). After 10 years: \(1000\times1.05^{10}=1000\times1.628894\ldots=\) £1 628.89 to the nearest penny. Simple interest would give only \(1000+10\times50=\) £1 500, so compound interest earns £128.89 more.
Worked example 11: percentage profit
A shop buys a bike for £160 and sells it for £212. Find the percentage profit.
Profit \(=212-160=52\). Percentage profit \(=52\div160\times100=32.5\%\). Always divide by the original amount, which here is the cost price.
Worked example 12: finding the time
How long does it take £400 to earn £60 simple interest at 3% per year?
\(60=\dfrac{400\times3\times T}{100}=12T\), so \(T=60\div12=5\). It takes 5 years.
Worked example 13: two-step reverse percentage
A pair of boots is reduced by 30% in a sale, and then costs £52.50. Find the original price and the amount saved.
The sale price is 70% of the original, so the multiplier is \(0.7\). Original \(=52.50\div0.7=75\). The boots cost £75 before the sale. The saving is \(75-52.50=22.50\), which is £22.50. Check: 30% of 75 is 22.50, which agrees.
Method: any percentage problem
- Decide whether the quantity is going up, going down or being compared.
- Write the multiplier, or the fraction change over original.
- Multiply to find the new value, divide to find the original value.
- Check that the answer is sensible: an increase must give a bigger number.
Note: which method?
Use multipliers for increases and decreases, for repeated change and for reverse percentages (divide). Use change over original to find a percentage change. Use \(I=PRT/100\) only for simple interest. If the problem says "per year" and the interest is added each year, think compound; if it says "simple", the interest is always worked out on the original sum.
Watch out: percentages of different amounts
- An increase of 10% followed by a decrease of 10% does not return to the start, because the second percentage is of a different amount.
- To undo a percentage change, divide by the multiplier; do not subtract or add the same percentage.
- Percentage change is always compared with the original amount, not the new one.
Did you know?
Interest is thousands of years old: clay tablets from ancient Babylon, about 4 000 years ago, record loans of silver and grain with interest. Today the Bank of England sets a base rate that affects the interest on savings and loans across the whole of the UK.
Percentages in everyday money
Worked example 14: comparing two savings accounts
Account A pays 4% simple interest per year. Account B pays 3.8% compound interest per year. Which gives more on £2 000 after 5 years?
For account A the interest is the same every year: \(2000\times4\times5\div100=400\), so the total is £2 400. For account B we multiply by 1.038 once for every year: \(2000\times1.038^5=2409.97\) to the nearest penny. Even though the rate is lower, compound interest earns interest on the interest, so account B gives about £10 more after 5 years. Over a shorter time, simple interest at the higher rate would win, so always check the number of years in the question.
Worked example 15: finding the percentage of a percentage
In a school, 60% of pupils have a bus pass, and 25% of those with a pass also have a bike. What percentage of all pupils have both?
Find 25% of 60%. Using decimals, \(0.25\times0.60=0.15\), so 15% of all pupils have both a pass and a bike. If the school has 840 pupils, that is \(840\times0.15=126\) pupils. Multiplying the decimals works for the same reason as using repeated multipliers.
Note: percentage points
When a rate changes, newspapers sometimes talk about percentage points. If an interest rate rises from 4% to 5%, it has gone up by 1 percentage point, but the percentage increase in the rate is \(1\div4\times100=25\%\). Read questions carefully to see which one is wanted.
Otto’s tip
Write the multiplier before you press any buttons. If you want to go back to the original, divide by the same multiplier. Hoot hoot!
How to use this Year 9 percentage change and simple interest lesson
First, read the key points so that you know where the Year 9 percentage change and simple interest lesson is going. Then work through each definition and worked example, and cover the solution before you read it. After that, practise with the percentage change and simple interest worksheet and check every question with the worksheet answers. When you feel ready, test yourself with the multiple choice questions or the quick quiz. Finally, sit the percentage change and simple interest test and mark it with the test answers.
Also, the starters are a good warm-up, so you can keep the key facts fresh. Moreover, all the pages of this chapter work together, so you can move between them at any time. For example, many pupils read the lesson on Monday, practise on Wednesday and test themselves on Friday. In short, this Year 9 percentage change and simple interest lesson gives you everything you need for the chapter.
To sum up, regular practice works better than one long session. For instance, ten minutes a day is often enough. Besides, mistakes are useful, because they show what to revise next. Then, once a method feels easy, try a harder question. As a result, you build confidence step by step.
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