Find the quantity that did not change
A before-and-after problem gives a ratio at the start, then something happens, and it gives the ratio at the end. Beads are added to a jar, money is spent, or some years go by. The first job is to find the quantity that did not change. The question tells you which one it is, in words such as "only red beads are added" or "nothing was taken out".
A jar holds red beads and blue beads in the ratio 2 : 5. Then 12 red beads are added, and the ratio becomes 4 : 5. No blue beads were added, so the blue beads are the same amount before and after. Blue is 5 parts in both ratios, so a part is the same amount of beads in both, and the two bars can be drawn on one scale.
The colored parts are the red beads and the plain parts are the blue beads. Every part is the same width in both rows, so the 5 blue parts are the same length before and after, while the red share grows from 2 parts to 4.
Count the change in parts
Now compare the red share. It went from 2 parts to 4 parts, a gain of 4 − 2 = 2 parts. Those 2 parts are the 12 red beads that were added, so 1 part is 12 ÷ 2 = 6 beads.
A part is 6 beads in both ratios. At the start there were 2 × 6 = 12 red beads and 5 × 6 = 30 blue beads. At the end there are 4 × 6 = 24 red beads and still 30 blue beads.
Check the answer against the question. 24 : 30 with both numbers divided by 6 is 4 : 5, and 24 − 12 = 12 red beads were added.
Three things that can stay the same
The quantity that stays the same is not always one of the shares. There are three common cases, and each one is solved by making that quantity the same number of parts in both ratios.
One share does not change. Something is added to, or taken from, the other share only, like the red beads. Make the unchanged share the same number of parts in both ratios.
The total does not change. Some of one share moves to the other, and nothing enters or leaves. Make the two totals the same number of parts.
The difference does not change. Both shares grow, or both shrink, by the same amount. A father and his son both get 10 years older, so the difference between their ages stays the same. Make the two differences the same number of parts.
Both amounts gain 2: 3 and 7 become 5 and 9. The difference between them is 4 before and 4 after.
The ratios alone do not say
The two ratios on their own never tell you which quantity stayed the same. In 2 : 5 and 4 : 5 the second share is 5 parts both times, but that does not mean its amount is unchanged. If some of the second share had moved to the first instead, the total would be the one that stayed the same, and a part after the change would be smaller than a part before it.
In the drawing below, the top bar is the ratio 2 : 5 as 18 and 45, which is 63 in all. The lower bar is 4 : 5, and you choose how much one of its parts is worth.
nothing is held here — 2 : 5 → 4 : 5 alone does not fix any amount
Make the total the quantity that did not move.
Drag the size of a part in the lower bar. At 9 the second share stays at 45, as when only the first share has something added. At 7 the total stays at 63, and the second share falls from 45 to 35: still 5 parts, but less of it.
When the difference is fixed
In the next problem, a father and his son both grow older by the same number of years, so the difference between their ages is the quantity that does not change. The two ratios give that difference different numbers of parts, so the first step is to rewrite one ratio until the differences match.
Worked example: Constant Difference (Equal Changes / Age Progression)
Question Five years ago, the ratio of Mr. Wong's age to his son's age was 7 : 2. In 10 years' time from now, the ratio of Mr. Wong's age to his son's age will be 2 : 1. (a) How old is Mr. Wong's son now? (b) What is the ratio of Mr. Wong's age to his son's age now in simplest form?
1.Past Model: Father (7 units), Son (2 units). Difference = 5 units.
Five years ago: father 7u, son 2u. The gap is 5 units. 2.Future Model: Father (2 parts), Son (1 part). Difference = 1 part.
In ten years: 2p and p. The gap is 1 part. Two people age by the same amount, so the gap in years never changes. 3.To align difference, cut each part into 5 units: Father = 10 units, Son = 5 units.
Make the gaps match: cut each part into 5, so the future is 10u and 5u. 4.Each person gains: 10 − 7 = 3 units.
Father from 7u to 10u, son from 2u to 5u: each gained 3 units. 5.Years passed = 5 + 10 = 15 years.
From 5 years ago to 10 years ahead is 15 years. 6.3 units = 15 ⟹ 1 unit = 5 years.
3u = 15, so u = 5 years. 7.Son's age 5 years ago = 2 × 5 = 10.
The son was 2u = 10 five years ago. 8.(a) Son's age now = 10 + 5 = 15 years old.
Slide the years: at 5 years after the past model the son is 15 and the gap stays 25. 9.(b) Father now = 35 + 5 = 40. Ratio = 40 : 15 = 8 : 3.
Now (5 years on): father 40, son 15, ratio 40:15 = 8:3.
Answer: (a) 15 years old; (b) 8 : 3
Common mistakes
- Calculating the time interval as 10 − 5 = 5 years instead of 5 + 10 = 15 years.
- Forgetting to add 5 years back to find the present age, reporting the age from 5 years ago as the final answer.
The usual mistakes
Subtracting ratio numbers when the parts are different sizes. Blue and white paint are mixed 3 : 4, and white is added until the ratio is 2 : 5. The white did not gain 5 − 4 = 1 part, because the blue is 3 parts before and 2 parts after, so a part is a different amount of paint in each ratio. First rewrite both ratios with the blue as 6 parts: 3 : 4 = 6 : 8 and 2 : 5 = 6 : 15. Now the white gained 15 − 8 = 7 parts.
Answering for the wrong moment. A question may ask how many there were at the start, and the working may have found how many there are at the end. In the jar, the answer "at first" is 12 red beads, not 24.