When both change by the same amount
Two amounts are 3 and 7, so the difference between them is 7 − 3 = 4. Now add 2 to both: they become 5 and 9, and the difference is 9 − 5 = 4 again. Adding the same amount to both numbers moves both ends of the gap by the same distance, so the gap does not change. Taking the same amount away from both works the same way.
This happens in many problems. Two people's ages both grow by the same number of years, because the same time passes for both of them. Two friends who each spend the same amount of money both go down by that amount. In every case the difference between the two stays the same.
Both bars grow by 2, so the gap between them is 4 before and 4 after.
Make the differences match first
In a ratio problem, the difference is the anchor, because it is the same real amount before and after. To use it, both ratios must give the difference the same number of parts.
The ages of two brothers are in the ratio 3 : 5. In 6 years' time, their ages will be in the ratio 3 : 4. In 3 : 5 the difference is 5 − 3 = 2 parts, but in 3 : 4 it is 4 − 3 = 1 part. These parts are different sizes, so they cannot be compared yet. Multiply both numbers of the second ratio by 2: 3 : 4 = 6 : 8, and now the difference is 8 − 6 = 2 parts in both ratios. One part is the same number of years in both.
Now compare. The younger brother goes from 3 parts to 6 parts, a gain of 3 parts, and so does the older brother, from 5 parts to 8 parts. Those 3 parts are the 6 years that pass, so one part is 6 ÷ 3 = 2 years. Now the brothers are 3 × 2 = 6 and 5 × 2 = 10 years old. Check: in 6 years they will be 12 and 16, and 12 : 16 = 3 : 4.
The two ratios with the difference matched, measured in parts of 2 years each. Each brother gains 3 parts, and the gap stays 2 parts.
The usual mistakes
Comparing parts before the differences match. From 3 : 5 to 3 : 4, the younger brother seems to stay at 3 parts, which would mean he did not get older. The parts in the two ratios are different sizes until the difference has the same number of parts in both.
Adding the change to the gap. If 3 and 7 both gain 2, the gap is still 4, not 4 + 2 = 6. Each number gained its own 2, and the two gains cancel.
Changing only one of the amounts. When both people spend $18, take $18 from both of them, not only from one.
Counting the time wrongly. From 5 years ago to 10 years from now is 5 + 10 = 15 years, not 10 − 5 = 5.
When the differences already match
Sometimes the two ratios already give the difference the same number of parts, and you can compare them straight away. In the next problem, 2 : 5 and 1 : 4 both have a difference of 3 parts.
Worked example: Constant Difference (Equal Reductions/Additions)
Question Ethan had 25 as much pocket money as Fiona. After both of them spent $18 each on stationery, Ethan had 14 as much money as Fiona. How much money did Fiona have at first?
1.Before: Ethan : Fiona = 2 : 5 ⟹ Difference = 5 − 2 = 3 units.
Ethan : Fiona = 2:5. The gap is 3 units. 2.After: Ethan : Fiona = 1 : 4 ⟹ Difference = 4 − 1 = 3 units.
Both spend the same $18. Slide it: the gap between the bars never changes. 3.Since the difference is already identical (3 units in both states), compare before and after directly.
After: 1:4, and the gap is again 3 units. Same gap, same unit, so compare directly. 4.Ethan dropped from 2u to u: Change = 2u − u = u.
Ethan went from 2u to u. That u is what he spent. 5.Therefore: u = $18.
u = $18. 6.Fiona at first = 5u = 5 × $18 = $90.
Fiona at first: 5u = $90.
Answer: $90
Common mistakes
- Equating (2 − 1) and (5 − 4) when the difference units are not aligned to a common multiple.
- Subtracting 18 from only one of the parties in the model.
Ages, when the differences must be matched
In the next problem, the difference is 7 − 2 = 5 parts in the first ratio and 2 − 1 = 1 part in the second, so the second ratio is multiplied by 5 before the two are compared.
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.