Proportions and percentages: exercises

Late September 2026: in Year 1 at technical institutes (istituti tecnici), the entrance test is already behind you and in class you are now working on natural numbers, integers and rational numbers, powers — and, within the same September-October block, proportions and percentages. The first maths test typically arrives between mid-October and the start of November, so three to six weeks from today. Anyone on the economics track meets them again identical, the following week or in parallel, inside Business economics (Economia aziendale), where they are called “calculations: proportions, allocations” — and that is usually where the first real difficulty arrives: invoices with VAT and discounts, above- and below-100 calculations. Here I don’t go into the invoice: let’s look at proportions, percentages and above- and below-100 calculations as a topic in its own right, because they need to be properly understood before seeing them applied to a commercial document.

Proportions and percentages seem like middle school topics — and in part they are, you have already seen them — but at the technical institute they come back harder: with more steps, with less “round” numbers, and above all with a specific mistake that almost always costs the exercise: applying the percentage to the wrong number.

Proportions: the method

A proportion is an equality between two ratios:

a : b = c : d

read ” $a$ is to $b$ as $c$ is to $d$ ”. The four numbers are called terms: $a$ and $d$ are the extremes, $b$ and $c$ the means.

Fundamental property. In every proportion, the product of the extremes equals the product of the means:

a \cdot d = b \cdot c

It is the property from which you find the unknown term, wherever it is. If an extreme is missing:

a : b = c : x \Rightarrow x = \frac{b \cdot c}{a}

If a mean is missing:

a : x = c : d \Rightarrow x = \frac{a \cdot d}{c}

A frequent special case is the mean proportional: when the two means are equal, $a : x = x : d$ , and then

x = a \cdot d

(with $a$ and $d$ of the same sign, because under a root you need a non-negative number).

The properties of compounding and dividing. From $a : b = c : d$ you get two other equivalent proportions, useful when a problem gives sums or differences instead of the individual terms:

compounding: (a + b) : b = (c + d) : d

dividing: (a - b) : b = (c - d) : d

Direct and inverse proportionality. Two quantities are directly proportional if their ratio is constant ( $y = k x$ ): double one, and the other doubles too. They are inversely proportional if their product is constant ( $y = k / x$ ): double one, and the other halves. In problems the signal is in the text: “the time to do a job” grows if the number of workers falls — inverse proportionality; “the total cost” grows if more items are bought — direct proportionality. The first step, always, is to decide which of the two situations you have in front of you, before writing any proportion.

Percentages: what they really mean

“20%” is just a way of writing the fraction $\frac{20}{100}$ , that is $0.20$ . Calculating the percentage of a number is therefore a multiplication:

p % of N = N \cdot \frac{p}{100}

15% of 80 is $80 \cdot \frac{15}{100} = 12$ .

Finding the whole, knowing the part and the percentage, is the inverse operation: if I know that 12 is 15% of a number, that number is

N = \frac{12}{15} \cdot 100 = 80

Percentage change. If a value goes from $V_{i}$ to $V_{f}$ , the percentage change is

change % = \frac{V _{f} - V _{i}}{V _{i}} \cdot 100

positive if the value has increased, negative if it has decreased.

Increases, discounts and why you don’t get back to the starting price

The fastest way — and the one you will always use from here on — is to think of the increase or discount as a multiplier, not as two separate operations (calculate the percentage, then add or subtract it).

increase of p % : N_{new} = N \cdot (1 + \frac{p}{100})

discount of p % : N_{new} = N \cdot (1 - \frac{p}{100})

A price of €250 with a 20% discount becomes $250 \cdot 0.80 = € 200$ , in a single step.

From here comes a fact that always surprises, the first time: a 20% discount followed by a 20% increase doesn’t bring you back to the starting price. Starting from €100:

100 \cdot 0.80 = 8080 \cdot 1.20 = 96

It doesn’t go back to 100, it goes back to 96. The reason is that the second 20% is applied to a different, smaller base (80, not 100): increasing an already reduced number by 20% recovers less than the reduction took away. It is the same principle as successive discounts: a 10% discount followed by another 10% is not equivalent to a single 20% discount, because the second 10% is calculated on the already discounted price:

100 \cdot 0.90 \cdot 0.90 = 81

A 19% discount, not 20%: the difference between “applying two discounts in sequence” and “adding the two percentages” is a mistake that in problems with successive discounts almost always costs the whole mark for the exercise.

Above-100 and below-100 calculations

“Above 100” (sopra cento) and “below 100” (sotto cento) are two ways of referring to the same idea of percentage increase and discount, but with a different language, typical of commercial problems: you speak of how many euros correspond to a starting €100 (above 100, when from the net price you want the price with VAT or a mark-up) and, conversely, of how much the net price is worth starting from a total that already includes the mark-up (below 100).

A classic example is extracting the VAT from a final price that already includes it. If a product costs €122 including VAT, with VAT at 22%, the net price is not found by taking 22% off 122 — that would be right only if 122 were the price without VAT. The price with VAT is 122% of the net price (100% + 22%), so:

net price = \frac{122}{122} \cdot 100 = € 100 (in general: \frac{final price}{100 + p} \cdot 100)

The reasoning doesn’t change if the numbers aren’t as convenient as in this example: you always divide by $100 + p$ , not by $100$ , and multiply by $100$ — never the other way round.

Simple direct allocation

Simple direct allocation (riparto semplice diretto) divides a total quantity $S$ among several parts, in proportion to given numbers $a_{1}, a_{2}, \dots, a_{n}$ (for example hours worked, shares of capital, weights). First you calculate the unit allocation value, dividing the total by the sum of the numbers:

k = \frac{S}{a _{1} + a _{2} + \dots + a _{n}}

and then each part is assigned $a_{i} \cdot k$ . The sum of all the parts must come back to equal $S$ : it is a check that costs a few seconds and immediately exposes a calculation error.

The mistake that costs the most

Marking Year 1 papers at the technical institute, the mistake that recurs most often — by far — is applying the percentage to the wrong number. The typical case is exactly the extraction of VAT: whoever has a final price of €122 and, to find the net price, calculates “122 minus 22% of 122” gets $122 - 26.84 = € 95.16$ , a wrong number — because the 22% to take off is not 22% of the final price, it is 22% of the net price (which is smaller). The practical rule: if the number you hold already includes the percentage, divide by $100 + p$ ; if it doesn’t include it yet, multiply by $100 + p$ over $100$ (or, for a discount, by $100 - p$ over $100$ ). Before writing any calculation, the first question is always: does this number already have the percentage inside it or not?

Three worked exercises

1. Unknown term in a proportion

Solve the proportion $15 : 6 = x : 10$ .

I apply the fundamental property: the product of the extremes equals the product of the means.

15 \cdot 10 = 6 \cdot x \Rightarrow 150 = 6 x \Rightarrow x = \frac{150}{6} = 25

Quick check: $15 : 6 = 2.5$ and $25 : 10 = 2.5$ — the two ratios coincide, the proportion is correct.

To practise on the same numbers and the same block as the first test you’ll find the practice test on numbers, fractions and powers: the correct answer to every question is free, and the guided step-by-step solution unlocks with your first lesson.

2. Successive discounts

A garment costs €90. The shop first applies a 15% discount, then — on the already discounted price — a further 10% discount for those who pay cash. How much do you pay in the end, and what single discount is it equivalent to?

First discount, as a multiplier:

90 \cdot (1 - \frac{15}{100}) = 90 \cdot 0.85 = € 76.50

Second discount, applied to the new price of €76.50, not to the starting €90:

76.50 \cdot (1 - \frac{10}{100}) = 76.50 \cdot 0.90 = € 68.85

You pay €68.85. The equivalent single discount comes from the ratio between final price and initial price:

\frac{68.85}{90} = 0.765 \Rightarrow single discount = (1 - 0.765) \cdot 100 = 23.5%

Not 25% (which would be the naive sum of 15 and 10): the two discounts in sequence are worth, overall, 23.5%. The check calculation: $90 \cdot 0.765 = 68.85$ , it matches the result found step by step.

3. Extracting VAT and the below-100 calculation

An article is sold for €244 including VAT, with the VAT rate at 22%. What is the net price, and how much is the VAT in euros?

The final price represents 122% of the net price (100% of the net plus 22% VAT), so the net price is found by dividing by $122$ and multiplying by $100$ :

net price = \frac{244}{122} \cdot 100 = 2 \cdot 100 = € 200

The VAT, by difference, is

244 - 200 = € 44

Check: 22% of €200 is $200 \cdot 0.22 = € 44$ , it matches. Note the mistake that is easiest to make here: calculating “22% of 244” instead of “22% of 200” — it would give $€ 53.68$ , a wrong number because applied to the price that already contains the VAT, not to the net price.

What to do now

On proportions and percentages we don’t yet have a dedicated practice test; the one available, the practice test on numbers, fractions and powers, covers the rest of the same first test — natural numbers, integers and rational numbers, powers — and is in any case a good test to see whether the rest of the September-October syllabus is solid. The correct answer to every question is free; the guided solution unlocks with your first lesson.

If your school’s syllabus follows a different pattern — more oriented towards sets and logic, as at the science and classical high schools — you’ll find the same first test told from that angle in the first maths test in Year 1 of high school and, for the art high school, in numbers, fractions and powers in Year 1 at art high school. If you are on the economics track and in a few weeks you will find yourself facing invoices with VAT and discounts inside Business economics, these same calculations — multiplier for the increase or discount, division by $100 + p$ to extract — are exactly the base you start from: when you get to double-entry bookkeeping, in Year 3, I wrote about it here: double-entry bookkeeping made easy. The full programme of maths tutoring is here, year by year.

If proportions and percentages remain a weak point, or if it is extracting VAT in Business economics that doesn’t work out for you, the free introductory call is for exactly this: you tell me what you have done in class and I tell you honestly where to start before the test arrives.

Ratios, proportions and percentages in Year 1 of high school