1c Percentages, growth and surds | IGCSE Maths 0580

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🎯What you need to be able to do

Find a percentage of a quantity, and express one quantity as a percentage of another.
Calculate percentage increase and decrease, profit and loss, discount and deposit.
Calculate simple and compound interest (no formulas are given).
Use reverse percentages to find an original amount EXTENDED.
Use exponential growth and decay, such as depreciation and population change EXTENDED.
Simplify surds and rationalise denominators EXTENDED.

📚The mathematics

Multipliers

Almost every percentage question is quickest with a multiplier: increasing by 15% is multiplying by 1.15; decreasing by 15% is multiplying by 0.85. Then

new $ = $ original $ \times $ multiplier

percentage change $ = \dfrac{\text{change}}{\text{original}} \times 100 $

original $ = $ new $ \div $ multiplier EXTENDED

Percentage change is always measured against the original amount. For a reverse percentage, never take the percentage of the new amount and add it back: if a price after a 15% discount is $68, the original is $ 68 \div 0.85 = $80 $, not $ 68 \times 1.15 = $78.20 $.

Simple and compound interest

Simple interest is the same amount every year: $ \text{interest} = \text{principal} \times \text{rate} \times \text{years} $. Compound interest adds interest to interest: after $n$ years the amount is $ P \times \left(1 + \tfrac{r}{100}\right)^{n} $. Questions sometimes ask for the interest (amount minus principal) and sometimes for the amount; read which.

A line chart of the value of 3000 dollars over 10 years at 3.5 percent per year. Simple interest grows in a straight line to 4050 dollars; compound interest curves slightly upwards to about 4232 dollars, pulling further ahead each year. — $3000 at 3.5% a year: simple interest adds $105 every year; compound interest adds a little more each year.

Exponential growth and decay EXTENDED

Anything that changes by the same percentage each period follows value $ = $ start $ \times $ multiplier$^{n}$. For “after how many years” questions, try whole numbers of years on the calculator until the value first crosses the target, and write down the values either side to show why. (Logarithms are not required, and nor is $e$.)

A bar chart of a car's value as a fraction of its new price, falling by 12 percent each year: 1, 0.88, 0.774, 0.681, 0.600, 0.528 and 0.464 after 6 years. A dashed line marks one half; year 6 is the first bar below it. — Depreciation of 12% a year: $ 0.88^{5} = 0.528 $ is still above half, $ 0.88^{6} = 0.464 $ is the first value below.

Surds EXTENDED

$ \sqrt{ab} = \sqrt{a}\sqrt{b} $

$ \sqrt{\tfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}} $

$ \sqrt{a} \times \sqrt{a} = a $

$ (a + \sqrt{b})(a - \sqrt{b}) = a^{2} - b $

Simplify by taking out the largest square factor: $ \sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3} $. To rationalise $ \dfrac{k}{\sqrt{a}} $, multiply top and bottom by $ \sqrt{a} $; for $ \dfrac{k}{a - \sqrt{b}} $, multiply by $ a + \sqrt{b} $, which removes the surd from the denominator using the difference of two squares.

✏️Worked example

(a) (Calculator.) $3000 is invested at 3.5% per year compound interest. Find the value after 4 years. [2] (b) (Calculator.) EXTENDED A car loses 12% of its value each year. After how many complete years is it first worth less than half its value when new? [2] (c) (Non-calculator.) EXTENDED Simplify $ \sqrt{75} - \sqrt{12} $, and rationalise $ \dfrac{6}{3 - \sqrt{3}} $. [4]

(a) $ 3000 \times 1.035^{4} = 3442.569\ldots $, so $3442.57.

(b) $ 0.88^{5} = 0.528 $ (more than half) and $ 0.88^{6} = 0.464 $ (less than half), so after 6 years.

(c) $ 5\sqrt{3} - 2\sqrt{3} = 3\sqrt{3} $. Then

\[ \frac{6}{3 - \sqrt{3}} \times \frac{3 + \sqrt{3}}{3 + \sqrt{3}} = \frac{6(3 + \sqrt{3})}{9 - 3} = 3 + \sqrt{3} \]

Check it. (c) $ 6 \div (3 - 1.732) = 4.732 $, and $ 3 + \sqrt{3} = 4.732 $ ✓. (a) Simple interest would give $ 3000 + 4 \times 105 = 3420 $, and compound must be a little more ✓.

Using $ 1 - 0.12 \times n $. Losing 12% a year for 6 years is not losing 72%: each year’s 12% is of a smaller amount. Always multiply by 0.88 once per year.

📝Practise

Questions in the style of the current papers. EXTENDED marks Extended-only content.

1. (Calculator.) Write 45 as a percentage of 360. [1]

$ \tfrac{45}{360} \times 100 = 12.5\% $.

2. (Calculator.) The price of a bicycle falls from $250 to $215. Find the percentage decrease. [2]

Change $35; $ \tfrac{35}{250} \times 100 = 14\% $.

3. (Calculator.) $5000 is invested at 2.8% per year simple interest. Find the interest earned in 6 years. [2]

$ 5000 \times 0.028 \times 6 = $840 $.

4. (Calculator.) A shop buys a jacket for $40 and sells it for $52. Find the percentage profit. [2]

Profit $12 on a cost of $40: $ \tfrac{12}{40} \times 100 = 30\% $.

5. (Calculator.) EXTENDED After a 20% pay rise, Rina earns $33 600 a year. Find her salary before the rise. [2]

$ 33\,600 \div 1.2 = $28\,000 $.

6. (Calculator.) EXTENDED A town’s population of 12 000 grows by 4% each year. Find the population after 5 years, to the nearest hundred. [2]

$ 12\,000 \times 1.04^{5} = 14\,599.8\ldots \approx 14\,600 $.

7. (Non-calculator.) EXTENDED (a) Simplify $ \sqrt{50} + \sqrt{18} $. (b) Rationalise the denominator of $ \dfrac{10}{\sqrt{5}} $. [3]

(a) $ 5\sqrt{2} + 3\sqrt{2} = 8\sqrt{2} $. (b) $ \dfrac{10\sqrt{5}}{5} = 2\sqrt{5} $.

8. (Non-calculator.) EXTENDED Expand and simplify $ (2 + \sqrt{3})(5 - \sqrt{3}) $. [2]

$ 10 - 2\sqrt{3} + 5\sqrt{3} - 3 = 7 + 3\sqrt{3} $.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

Corbettmaths — reverse percentages and compound interest
Khan Academy — simplifying radicals and rationalising denominators