1.1 Fractions, ratios and percentages

Fractions, decimals, percentages, ratios, and proportions are different ways to compare quantities. They are taught early because they form a base for much of the mathematics that follows. Everyone should be comfortable using them: they appear again in probability, averages, and model evaluation. The calculation is often short once you know what the whole is and which two quantities the question is comparing.

If you need to learn or review these ideas, Khan Academy's Arithmetic course covers fractions and decimals, while its Pre-algebra course covers percentages, ratios, and proportions.

A quick tour

Suppose three of eight objects are blue. The blue part is $3/8$ of the whole, or $0.375$, or $37.5\%$. If the other five objects are not blue, the ratio of blue to not blue is $3:5$. The fraction uses all eight as its denominator; the ratio compares two groups. That distinction will matter in both examples below.

Fractions and decimals

A fraction $a/b$ means $a$ divided by $b$, where $b\ne 0$. Multiplying or dividing its numerator and denominator by the same nonzero number leaves its value unchanged: $6/8=3/4$. To write a fraction as a decimal, divide the numerator by the denominator. To turn a terminating decimal into a fraction, write it over a power of ten and simplify: $0.375=375/1000=3/8$.

Use a common denominator to add fractions. Multiply numerators and denominators to multiply them; to divide, multiply by the reciprocal:

$$\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd},\qquad \frac{a}{b}\cdot\frac{c}{d}=\frac{ac}{bd},\qquad \frac{a}{b}\div\frac{c}{d}=\frac{ad}{bc}.$$

Here $b$ and $d$ must be nonzero, and division by $c/d$ also requires $c\ne 0$. Keeping an exact fraction through the calculation can prevent a rounded decimal from changing the final answer.

Percentages

A percentage is a fraction per hundred: $37.5\%=37.5/100=0.375$. If $p$ is the percentage number, such as $25$ for $25\%$, then

$$\text{part}=\frac{p}{100}\times\text{whole},\qquad p=100\times\frac{\text{part}}{\text{whole}}.$$

For a percentage change, the original value is the base:

$$\text{percentage change}= \frac{\text{new}-\text{original}}{\text{original}}\times100\%.$$

The original value must be nonzero. When a question says “25% more water,” the base is the amount of water already there, unless the question defines another base. Always identify that base before multiplying.

Ratios and proportions

The ratio $a:b$ says that for every $a$ parts of the first quantity there are $b$ parts of the second. If these are the only two parts of a whole, the first quantity's fraction of that whole is $a/(a+b)$. For example, a syrup-to-water ratio of $2:3$ means that syrup is $2/5$ of the mixture, not $2/3$.

A proportion says that two ratios have the same value. If $b,d\ne0$, then

$$\frac{a}{b}=\frac{c}{d}\quad\Longleftrightarrow\quad ad=bc.$$

You can use this to scale a recipe or a diagram while keeping its ratio fixed. A ratio alone tells you the relative sizes; you need a total or one actual amount to find the quantities themselves.

Application 1: a mixture that changes

A drink contains syrup and water in the ratio $2:3$, with $40$ litres in total. You add water equal to $25\%$ of the water already in the drink. What fraction and percentage of the final mixture is syrup?

1. Turn the ratio into amounts. There are $2+3=5$ equal parts, so each part is $40/5=8$ litres. Syrup is $2\times8=16$ litres, and water is $3\times8=24$ litres. Equivalently, syrup is $2/5$ of $40$ litres.
2. Apply the percentage to its stated base. The added water is $25\%$ of $24$, which is $(25/100)\times24=6$ litres. Water becomes $30$ litres. Syrup stays at $16$ litres, so the new total is $46$ litres.
3. Use the new whole. The syrup fraction is $16/46=8/23\approx0.348$, or about $34.8\%$. The new syrup-to-water ratio is $16:30=8:15$.

The result passes a quick check: adding only water should lower syrup's share from its original $2/5=40\%$. Using $40$ as the final denominator, or taking $25\%$ of the entire original drink, would answer a different question.

Application 2: reading a classifier's results

The 2025 Stage 1 paper includes a longer problem about classifier results. Here is a smaller, invented example. A spam filter checks $240$ messages:

There are several sensible percentages you could calculate. The wording decides which one is wanted:

• What percentage of all messages are spam? The whole is all $240$ messages. The answer is $60/240=1/4=25\%$; the spam-to-ordinary ratio is $60:180=1:3$.
• What percentage of flagged messages are actually spam? The whole is the $50$ flagged messages. The answer is $45/50=9/10=90\%$. In classification, this is the filter's precision for spam.
• What percentage of actual spam messages did the filter flag? The whole is the $60$ spam messages. The answer is $45/60=3/4=75\%$. This is its recall for spam.

The last two questions share the same numerator, $45$, but use different denominators. Before you reach for a formula, say in words which group is the whole. That one step is what connects the arithmetic in this table to the arithmetic in the mixture problem.