Ratio Problems

Learning goals

Write a ratio's parts as $as$ , $bs$ , $cs$ with one multiplier
Pin $s$ down with a total, a difference, or one known amount
Combine two ratios by matching the shared term, then pin down $s$ from a count
Adjust before-and-after amounts, then cross-multiply the new ratio
Check that the finished amounts reduce to the original ratio

The common multiplier

A ratio pins down how quantities compare, not how big they are. If two quantities stand in the ratio $2 : 3$ , they might be $2$ and $3$ , or $20$ and $30$ , or $2000$ and $3000$ . What every one of those pairs shares is a single hidden number, the size of one part. Call it $s$ . Then the first quantity is $2s$ and the second is $3s$ . That is because the ratio says the first quantity holds $2$ parts and the second holds $3$ parts, with each part the same size $s$ .

Write it the same way for any ratio. A two-term ratio $a : b$ describes quantities $as$ and $bs$ . A three-term ratio $a : b : c$ describes quantities $as$ , $bs$ , and $cs$ , all sharing the one multiplier $s$ . The multiplier is identical for every term, and that is the whole point: the parts must be equal in size for the ratio to mean what it says.

Why one multiplier captures a whole ratio#

Suppose the actual amounts are $x$ and $y$ , and they stand in the ratio $a : b$ . From the Ratios lesson, two ratios are equal exactly when one comes from the other by multiplying both parts by the same nonzero number. Since $x : y$ equals $a : b$ , the pair $x : y$ is an equivalent ratio to $a : b$ , so there is a single number $s$ with

x = as \qquad \text{and} \qquad y = bs.

The same $s$ scales both parts, never two different multipliers, because scaling the parts unequally would change the comparison and break the ratio.

Now separate what is fixed from what is free. The ratio nails down the proportion $a : b$ , but it says nothing about $s$ : every nonzero choice of $s$ rescales both amounts together and leaves the ratio $a : b$ unchanged. So a ratio, however many terms it carries, hides exactly one unknown, the multiplier $s$ . A fact that fixes the actual scale, such as a total or a difference, turns straight into one equation in that single unknown and solves it. A second ratio works differently: it links in another quantity and lets you write a longer ratio, but by itself it still leaves every part unscaled, so you still need one scale-setting fact before you can read off actual amounts. Once you know $s$ , multiply it back through $as$ , $bs$ , $cs$ and you have every amount. The whole method is this: turn the ratio into expressions in $s$ , use a scale-setting fact to solve for $s$ , then read off the quantities.

From a ratio and a total

The most common extra fact is a total. If the quantities $as$ and $bs$ add to a known total $N$ , then

as + bs = N, \qquad (a + b)s = N, \qquad s = \frac{N}{a + b}.

The sum $a + b$ is just the number of parts, so this is the part-counting rule from before, now written as an equation. The multiplier $s$ is then the value of one of those parts. The real payoff shows up with three terms, where sharing out by hand gets awkward but the algebra does not change at all.

Worked example 1 Three colors from a total

A necklace is strung with red, white, and blue beads in the ratio $3 : 5 : 7$ , and it uses $90$ beads in all. How many beads of each color does it use?

Write the three counts with one multiplier $s$ : there are $3s$ red, $5s$ white, and $7s$ blue beads. The extra fact is the total of $90$ beads, so add the three expressions and set the sum equal to $90$ :

3s + 5s + 7s = 90.

Combine the like terms and solve:

15s = 90, \qquad s = 6.

Each part is $6$ beads. Multiply back through each term:

3s = 18, \qquad 5s = 30, \qquad 7s = 42.

So the necklace has $18$ red, $30$ white, and $42$ blue beads. Check by adding them: $18 + 30 + 42 = 90$ , which matches the total, and the three counts still reduce to the ratio $3 : 5 : 7$ .

A ratio 3 : 5 : 7 splits the beads into 15 equal parts, each part worth s beads. The three colors take 3s, 5s, and 7s, so the whole bar is 15s. Fixing one fact (here the total of 90) fixes s, and every count follows.

Check your understanding

Two numbers are in the ratio $5 : 3$ and add to $64$ . What is the larger number?

40

24

8

32

From a ratio and a difference

Sometimes a problem gives a difference instead of a total. The setup is identical; only the equation changes. If the quantities are $as$ and $bs$ with $a > b$ , their difference is

as - bs = (a - b)s.

Set that equal to the known difference and solve for $s$ exactly as before.

Worked example 2 A difference instead of a total

An older sister and her younger brother collect baseball cards in the ratio $7 : 4$ . The sister has $15$ more cards than her brother. How many cards does each have?

Let one part be $s$ . The sister holds $7s$ cards and the brother holds $4s$ . The extra fact is that the sister has $15$ more, so their difference is $15$ :

7s - 4s = 15.

Combine and solve:

3s = 15, \qquad s = 5.

Multiply back: the sister has $7s = 35$ cards and the brother has $4s = 20$ . Check the difference: $35 - 20 = 15$ , as stated, and $35 : 20$ reduces to $7 : 4$ .

Check your understanding

Two numbers are in the ratio $7 : 4$ and differ by $21$ . What is the smaller number?

28

49

21

7

Combining two ratios

A problem sometimes links three quantities with two separate ratios that share a middle term, and asks you to merge them into a single three-term ratio. The two ratios agree on the shared quantity only if its number is the same in both. When it is not, scale each ratio so the shared term matches, using the same equivalent-ratio move as always, then read off all three terms at once.

Worked example 3 Combining two ratios

A workshop holds chairs, tables, and benches. The ratio of chairs to tables is $2 : 3$ , and the ratio of tables to benches is $5 : 4$ . Find the ratio of chairs to tables to benches, and then the actual counts if there are $60$ tables.

The shared quantity is tables. It appears as $3$ in the first ratio and as $5$ in the second, so the two ratios do not yet agree on it. Make them agree by scaling each so the tables term becomes the least common multiple of $3$ and $5$ , which is $15$ .

Scale the first ratio by $5$ so its tables term is $15$ :

\text{chairs} : \text{tables} = 2 : 3 = 10 : 15.

Scale the second ratio by $3$ so its tables term is also $15$ :

\text{tables} : \text{benches} = 5 : 4 = 15 : 12.

Both now agree that tables is $15$ , so line them up into one ratio:

\text{chairs} : \text{tables} : \text{benches} = 10 : 15 : 12.

For the counts, bring back the multiplier. There are $60$ tables, and tables is the $15$ term, so

15s = 60, \qquad s = 4.

Then chairs $= 10s = 40$ and benches $= 12s = 48$ . So the workshop has $40$ chairs, $60$ tables, and $48$ benches.

Check your understanding

If $A : B = 3 : 4$ and $B : C = 6 : 5$ , what is $A : B : C$ ?

9 : 12 : 10

3 : 4 : 6 : 5

9 : 12 : 5

6 : 8 : 5

Scaling a recipe or mixture

Recipes and mixtures are ratios you scale up or down. When a problem fixes the amount of one ingredient, that single amount determines $s$ , and every other ingredient follows. You do not need a total; one known part is enough.

Worked example 4 Scaling a recipe from one ingredient

A punch recipe mixes juice, soda, and sherbet in the ratio $4 : 3 : 1$ . You have exactly $6$ cups of soda and want to use all of it. How much juice and sherbet do you need, and how much punch will you make?

Write each ingredient with the multiplier $s$ : juice is $4s$ , soda is $3s$ , and sherbet is $1s$ cups. The fixed ingredient is soda, and it is the $3$ term, so

3s = 6, \qquad s = 2.

Each part is $2$ cups. Multiply back:

\text{juice} = 4s = 8, \qquad \text{sherbet} = 1s = 2.

So you need $8$ cups of juice and $2$ cups of sherbet. The total punch is $8 + 6 + 2 = 16$ cups, which also equals $(4 + 3 + 1)s = 8s = 16$ , a handy check.

Before and after: when the ratio changes

The richest ratio problems change one or both quantities partway through and then tell you the new ratio. Adding or removing an amount does not fit the equivalent-ratio picture, because the two quantities no longer scale together. The multiplier still rescues you: write the original quantities as $as$ and $bs$ . Then adjust those quantities by the stated change, and set the changed pair equal to the new ratio. That equation, cleared of its fraction, is a linear equation in $s$ .

Worked example 5 A before-and-after age ratio

The ratio of Anna’s age to Ben’s age is $4 : 3$ . In $6$ years, the ratio of their ages will be $6 : 5$ . How old is each now?

Write their current ages with the multiplier: Anna is $4s$ and Ben is $3s$ years old. In $6$ years each is $6$ years older, so Anna will be $4s + 6$ and Ben will be $3s + 6$ . Their new ratio is $6 : 5$ , which as a proportion says

\frac{4s + 6}{3s + 6} = \frac{6}{5}.

Clear the fraction the way Linear Equations in Disguise justified it, by cross-multiplying, which is really just multiplying both sides by both denominators:

5(4s + 6) = 6(3s + 6).

Expand both sides and solve:

20s + 30 = 18s + 36, \qquad 2s = 6, \qquad s = 3.

So Anna is $4s = 12$ and Ben is $3s = 9$ . Check against the words: right now $12 : 9$ is $4 : 3$ , and in $6$ years they are $18$ and $15$ , whose ratio $18 : 15$ reduces to $6 : 5$ . Both conditions hold, so Anna is $12$ and Ben is $9$ .

Check your understanding

A box holds red and green pens in the ratio $3 : 2$ . After $10$ more red pens are added, the ratio becomes $2 : 1$ . How many green pens are in the box?

30

20

10

40

Common mistakes

Treating the ratio numbers as the amounts. A ratio $3 : 5 : 7$ does not mean there are $3$ , $5$ , and $7$ of each thing. It means $3s$ , $5s$ , and $7s$ . Solve for $s$ first, then report the amounts.
Giving each term its own multiplier. Every term of a ratio shares the one multiplier $s$ . Writing one quantity as $3s$ and another as $5m$ lets the parts be different sizes, which breaks the comparison the ratio states.
Merging ratios without matching the shared term. To combine $A : B$ and $B : C$ you must first scale until $B$ is the same number in both, then line them up. You cannot staple $2 : 3$ and $5 : 4$ into $2 : 3 : 4$ .
Adding to the ratio number instead of the quantity. In a before-and-after problem the change ( $+10$ , $-8$ ) is added to the actual amount $as$ , not to the ratio number $a$ . The new ratio then applies to the changed amounts.
Confusing part-to-part with part-to-whole. A ratio $2 : 3$ makes the first quantity $\frac{2}{5}$ of the total, not $\frac{2}{3}$ , because the whole is the sum of the parts.

Key takeaways

Every quantity in a ratio $a : b : c$ is the same multiple of one part: write them $as$ , $bs$ , $cs$ , sharing a single multiplier $s$ .
A ratio hides exactly one unknown, $s$ . A scale-setting fact (a total, a difference, or one known amount) is one equation, and it pins $s$ down. A second ratio does not by itself; it only extends the comparison to a new quantity.
Total: $(a + b + c)s = N$ . Difference of two terms: $(a - b)s = D$ . Solve for $s$ , then multiply back through every term.
Combine two ratios by scaling until the shared term matches (their least common multiple keeps the numbers small), then read off all three terms and use a scale-setting fact to find $s$ .
Before and after: adjust $as$ and $bs$ by the change, set the result equal to the new ratio, cross-multiply, and solve the linear equation in $s$ .
Always check: the final amounts should add or differ as stated and reduce to the original ratio.

Ratio Problems

Learning goals

The common multiplier

From a ratio and a total

From a ratio and a difference

Combining two ratios

Scaling a recipe or mixture

Before and after: when the ratio changes

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice