HookThe VAT trap that costs shopkeepers thousands
On 4 January 2011 the Chancellor, George Osborne, raised the standard rate of VAT from \(17.5\%\) to \(20\%\), and overnight every shop in Britain faced one deceptively simple question: given a price that already includes VAT, how much of it is tax? The instinct — knock \(20\%\) off the ticket — is wrong, and the mistake costs real money. VAT is added to the smaller pre-tax price, so it is removed by dividing by \(1.20\), not by subtracting a fifth. A £150 ticket splits into £125 of goods (from \(150 \div 1.20\)) and £25 of VAT; the wrong method, \(150 \times 0.80 = 120\), leaves the shopkeeper £5 short on every single item sold.
The fix is to treat every percentage as a multiplier — a single number you multiply by — and that one idea runs through this whole section, from reverse-percentage problems like the VAT above to compound growth and decay. Around it sit two more Higher skills that reward precision: converting between fractions and decimals in both directions, including turning a recurring decimal such as \(0.1\dot{6}\) into the exact fraction \(\frac{1}{6}\) by algebra; and reading fractions inside ratio problems, where 'Amir gets \(\frac{3}{8}\) of the money' and 'the ratio is \(3:5\)' turn out to be two descriptions of one situation. Fractions, decimals and percentages are not three topics — they are three costumes worn by the same number, and the marks go to whoever changes costume fastest.
ModelTerminating decimals and fractions — one number, two costumes
A terminating decimal is just a fraction with a power of ten on the bottom: \(0.35 = \frac{35}{100} = \frac{7}{20}\), and \(0.6 = \frac{6}{10} = \frac{3}{5}\). Going the other way, a fraction terminates exactly when its denominator, in lowest terms, is built only from \(2\)s and \(5\)s — the primes of ten. So \(\frac{7}{20}\) terminates because \(20 = 2^2 \times 5\), while \(\frac{1}{6}\) cannot, because the \(3\) in \(6 = 2 \times 3\) has no partner in ten, and the decimal is forced to recur.
Working interchangeably — the exact word AQA uses — means choosing whichever form is easier for the job. To compare or order a mix of fractions and decimals, decimals usually win; to multiply or cancel, fractions usually win. The skill being tested is the fluency to switch without friction, because a later step often needs the other costume.
Put \(\frac{5}{8}\), \(0.6\) and \(\frac{3}{5}\) in order, smallest first. Convert to decimals with the same number of places: \(\frac{5}{8} = 0.625\), \(0.6 = 0.600\) and \(\frac{3}{5} = 0.600\). So \(0.6 = \frac{3}{5}\), and both are smaller than \(0.625\): the order is \(0.6 = \frac{3}{5} \lt \frac{5}{8}\). Padding the decimals to three places is what stops the classic slip of thinking \(0.6\) beats \(0.625\) — line up the columns and the third-place \(5\) decides it.
MechanismTurning a recurring decimal into an exact fraction
Every recurring decimal is secretly a fraction, and Higher expects you to find it — exactly, not as a rounded guess. The method is a small piece of algebra that makes the repeating tail cancel. Call the decimal \(x\). Multiply by a power of \(10\) that shifts it exactly one full repeat along, subtract a version that shares the same tail to annihilate the recurring part, and solve the linear equation that remains. The power of ten has as many zeros as there are recurring digits: one recurring digit means \(\times 10\), a two-digit repeat means \(\times 100\), and so on.
The reverse direction — fraction to recurring decimal — is just division: \(\frac{1}{7} = 0.\dot{1}4285\dot{7}\), the famous six-digit cycle, because dividing \(1\) by \(7\) can only ever throw up the remainders \(1\) to \(6\) before it must repeat. The dots sit over the first and last digits of the block that repeats.
Convert \(0.1\dot{6}\), meaning \(0.16666\ldots\), to a fraction. Only the \(6\) recurs, so first shift the non-recurring part clear: let \(x = 0.16666\ldots\), then \(10x = 1.6666\ldots\) and \(100x = 16.6666\ldots\). Subtract the two lines that share the same tail: \(100x - 10x = 16.6666\ldots - 1.6666\ldots\), giving \(90x = 15\), so \(x = \frac{15}{90} = \frac{1}{6}\). For a pure repeat such as \(0.\dot{4}\dot{5} = 0.4545\ldots\), one step does it: \(100x - x = 45\), so \(99x = 45\) and \(x = \frac{45}{99} = \frac{5}{11}\). The denominator is always a string of \(9\)s (then \(0\)s), one \(9\) per recurring digit — a check worth knowing.
ModelFractions inside ratio problems
A ratio and a fraction describe the same split from two angles. If a bag of red and blue counters is in the ratio \(2:3\), then reds are \(\frac{2}{5}\) of the whole — the \(5\) being the total number of parts, \(2 + 3\). Reading the other way: if \(\frac{5}{8}\) of a class are girls, the ratio of girls to boys is \(5:3\), because the missing \(\frac{3}{8}\) are boys. Getting fluent both ways is exactly what this leaf tests, and it is the hinge of most multi-step ratio questions.
The trap is confusing a part-to-part ratio (\(2:3\)) with a part-to-whole fraction (\(\frac{2}{5}\)). The ratio's numbers add up to the whole; the fraction's denominator is the whole. Mix them up and every later step inherits the error.
Amir and Beth share money in the ratio \(3:5\), and Amir receives £18 less than Beth. Find the total. The difference in shares is \(5 - 3 = 2\) parts, and that gap is the £18, so one part is \(18 \div 2 = 9\) pounds. There are \(3 + 5 = 8\) parts altogether, so the total is \(8 \times 9 = 72\) — £72. Check with fractions: Amir's \(\frac{3}{8}\) of £72 is \(\frac{3}{8} \times 72 = 27\), Beth's \(\frac{5}{8}\) is \(45\), and \(45 - 27 = 18\) as required. Reading '£18 less' as the difference of parts, not as one person's share, is the whole battle.
MechanismPercentages as multipliers — forwards and backwards
A percentage is a fraction of \(100\), and its real power is as an operator: 'find \(30\%\) of £80' is simply \(0.30 \times 80 = 24\). To increase by a percentage, multiply by \(1\) plus the decimal — a \(15\%\) rise is \(\times 1.15\); to decrease, multiply by \(1\) minus it — a \(15\%\) fall is \(\times 0.85\). This is faster and far less error-prone than finding the percentage and then adding or subtracting it separately, and it is the only sensible way to handle the problems that follow.
Reverse percentage problems undo a multiplier by dividing. If a price after a \(20\%\) increase is £150, the original was \(150 \div 1.20 = 125\), because the £150 is already \(120\%\) of the original — exactly the VAT logic from the introduction. The classic error is to take \(20\%\) off the new figure, which finds \(20\%\) of the wrong number. Always ask: is the amount I have the value before or after the change? The multiplier goes forward; division comes back.
A coat is reduced by \(30\%\) in a sale to £59.50. What was the original price? The sale price is \(70\%\) of the original — the original multiplied by \(0.70\) — so the original is \(59.50 \div 0.70 = 85\), that is £85. Check forwards: \(85 \times 0.70 = 59.50\). Now a 'percentage of' the other way: express £27 as a percentage of £45. Divide and scale: \(\frac{27}{45} \times 100 = 60\%\). The two moves — multiply to go forward, divide to come back — cover almost every percentage mark on the paper.
CaseCompound change — why 20% then 10% is not 30%
Percentage changes multiply, they do not add. A \(20\%\) discount followed by a \(10\%\) discount is \(\times 0.80\) then \(\times 0.90\), which is \(\times 0.72\) overall — a \(28\%\) reduction, not \(30\%\), because the second cut is taken off an already-smaller price. The same machinery drives compound interest and depreciation: a quantity that changes by the same percentage each year is multiplied by the same factor each year, so \(n\) years means that factor raised to the power \(n\).
This is why savings and car values move the way they do, and why the multiplier method scales so cleanly — a single power replaces a table of year-by-year sums. It also sets up the growth-and-decay work later in the ratio strand, where the very same \(\times (1 + r)^n\) reappears.
A car bought for £20000 loses \(15\%\) of its value each year. What is it worth after \(3\) years? Each year multiplies by \(0.85\), so after three years the value is \(20000 \times 0.85^3\). Now \(0.85^3 = 0.85 \times 0.85 \times 0.85 = 0.614125\), so the car is worth \(20000 \times 0.614125 = 12282.50\) — £12,282.50. Note it is not a flat \(45\%\) loss, which would give \(20000 \times 0.55 = 11000\); compounding three \(15\%\) falls leaves more than that, because each year's loss is a slice of a smaller amount.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
The multiplier is the single most examined idea here, so make it a reflex: an increase is \(\times \left(1 + \frac{p}{100}\right)\), a decrease is \(\times \left(1 - \frac{p}{100}\right)\), and 'find the original' means divide by that multiplier. Reverse-percentage questions are where marks are won and lost — before you calculate, write down whether the figure you have is the 'before' or the 'after', because taking a percentage off the wrong one is the single most common error AQA reports.
On recurring-decimal conversions, show the algebra in full: define \(x\), state the multiplied line, subtract to kill the tail, then simplify the fraction to lowest terms. The subtraction line is the method mark; a bare fraction with no working scores little even when it is right. Match the power of ten to the number of recurring digits, and use the tidy check that a pure recurrence over \(n\) digits has \(n\) nines on the denominator.
In ratio-and-fraction problems, label your parts. Decide early whether a given number is one share, a difference of shares, or the whole, and convert freely between the ratio \(a:b\) and the fraction \(\frac{a}{a+b}\) — examiners award method marks for a correct part-value even when the final arithmetic slips.