Approach
Write present ages with one variable (or as ratio terms 3x and 4x). Move to the past by subtracting years and to the future by adding years - for every person. Then form one equation. Target: 40 seconds.
Standard forms
| Statement | Equation |
|---|---|
| Ratio of present ages is a : b | ages = ax, bx |
| n years ago the ratio was c : d | (ax - n)/(bx - n) = c/d |
| n years hence | (ax + n)/(bx + n) = c/d |
| Father is k times son | F = k S |
| Children born at intervals of t years | y, y + t, y + 2t, ... |
| Average age of a group after n years | old average + n |
Key facts
- The difference between two people's ages never changes.
- The sum of ages of k people increases by k x n after n years.
- If the ratio is given at two different times, the difference of ratio terms must represent the same age gap - use this for quick solving.
Ratio-difference shortcut
Present ratio 3 : 4, after 5 years 4 : 5. Gap in the first ratio is 1 part, in the second also 1 part, so the parts are equal in size. Change from 3 to 4 parts = 5 years, so 1 part = 5. Present ages = 15 and 20.
(Check by equation: (3x + 5)/(4x + 5) = 4/5 gives 15x + 25 = 16x + 20, x = 5.)
Worked example
A father is three times as old as his son. After 15 years he will be twice as old. Present ages?
3s + 15 = 2(s + 15), s = 15. Father = 45 years.
Worked example 2
Three years ago, the average age of a husband, wife and child was 27. Five years ago, the average of wife and child was 20. Husband now?
All three now = 81 + 9 = 90. Wife + child now = 40 + 10 = 50. Husband = 40 years.
Common traps
- Subtracting years from only one person.
- "n years ago" for one person and "n years hence" for another in the same sentence - read twice.
- Average age of a group: after n years, add n to the average, not n times the count.
- Twins: same age, count them as two people in sums.