Ratios as Fractions

Suppose red counters and blue counters are in the ratio

$\text{red}:\text{blue}=2:3.$

The whole collection contains both colours, so it has

$2+3=5$

equal parts altogether. Red occupies two of those five parts, so red is $\frac25$ of the whole. Blue occupies three of them, so blue is $\frac35$ of the whole.

The denominator is therefore not simply the other number in the ratio. For example, the first part of $2:3$ is not $\frac23$ of the whole: the whole contains five ratio parts, not three.

The general rule

For a two-part ratio

$A:B=a:b,$

the total number of equal parts is $a+b$. Therefore

$A=\frac{a}{a+b}\text{ of the whole}$

and

$B=\frac{b}{a+b}\text{ of the whole}.$

This works even when the ratio numbers are not the actual quantities. If the true amounts are $ak$ and $bk$ for some scale factor $k$, their total is $(a+b)k$. The fraction belonging to $A$ is

$\frac{ak}{(a+b)k}=\frac{a}{a+b}.$

The unknown scale factor cancels, so the ratio alone is enough to determine each fraction of the whole.

Keep the labels in order

The order of a ratio carries meaning. If

$\text{cats}:\text{dogs}=3:7,$

then there are $3+7=10$ parts in total. Cats correspond to the first number, so cats form $\frac3{10}$ of all the animals. Dogs correspond to the second number, so dogs form $\frac7{10}$.

Reversing the labels would reverse the fractions. Always write the labels above or beside the ratio before calculating.

Using the fraction with a known total

OCR gives the example of £9 divided in the ratio $2:1$. There are $2+1=3$ parts, so the two shares are

$\frac23\times £9=£6$

and

$\frac13\times £9=£3.$

The key interpretation comes first: the ratio tells us the fractions $\frac23$ and $\frac13$. Multiplying by £9 then turns those fractions into amounts.

Check the whole

The fractions belonging to all parts must combine to make one whole. For the ratio $a:b$,

$\frac{a}{a+b}+\frac{b}{a+b}=\frac{a+b}{a+b}=1.$

For example, if adults and children are in the ratio $4:5$, their fractions are $\frac49$ and $\frac59$. These add to $\frac99=1$, confirming that every member of the group has been included.