Average and problems on ages
Average = sum ÷ number of items. In age problems, everyone ages equally: the age gap between two people never changes.
Shortcuts for average and ages
| Situation | Shortcut | Example |
|---|---|---|
| Average | Sum ÷ number of items | 4, 6, 11 → 21 ÷ 3 = 7 |
| Evenly spaced numbers | (First + last) ÷ 2 | 11, 12, …, 29 → (11 + 29) ÷ 2 = 20 |
| A new member joins; average rises by d | New member = old average + (new count × d) | Worked example 2 |
| One member is replaced; average rises by d | New member = member replaced + (count × d) | Worked example 1 |
| Ages n years later or earlier | Add or subtract n for each person; a group of k changes by k × n in total | 3 people, 5 years later: total + 15 |
| Ages in a ratio | Take the ages as ak and bk; the age gap never changes | Worked example 3 |
Worked example 1: average after a replacement
The average weight of 8 students is 45 kg. A student weighing 41 kg leaves and a new student joins. The average becomes 46 kg. Find the weight of the new student.
- Standard method: old total = 8 × 45 = 360 kg; new total = 8 × 46 = 368 kg.
- New student = 368 − (360 − 41) = 368 − 319 = 49 kg.
- Shortcut: the average rose by 1 kg for 8 students, so the total rose by 8 kg. New student = 41 + 8 = 49 kg.
- Check: (360 − 41 + 49) ÷ 8 = 368 ÷ 8 = 46 kg.
49 kg.
Worked example 2: a new member raises the average
A batsman's average for 10 innings is 32 runs. How many runs must he score in the 11th innings to raise his average to 35?
- Standard method: runs so far = 10 × 32 = 320; runs needed after 11 innings = 11 × 35 = 385.
- Runs in the 11th innings = 385 − 320 = 65.
- Shortcut: new member = old average + (new count × rise) = 32 + 11 × 3 = 65.
- Check: (320 + 65) ÷ 11 = 385 ÷ 11 = 35.
65 runs.
Worked example 3: ages in a ratio
The present ages of a mother and her daughter are in the ratio 7 : 2. After 5 years the ratio will be 8 : 3. Find their present ages.
- Standard method: let the ages be 7k and 2k. Then (7k + 5) ÷ (2k + 5) = 8 ÷ 3.
- Cross-multiply: 21k + 15 = 16k + 40, so 5k = 25 and k = 5. Ages: 35 and 10.
- Shortcut (fixed gap): the gap is 7 − 2 = 5 parts now and 8 − 3 = 5 parts later, so the parts are the same size. The daughter goes from 2 parts to 3 parts in 5 years, so 1 part = 5 years. Ages: 7 × 5 = 35 and 2 × 5 = 10.
- Check: after 5 years, 40 : 15 = 8 : 3.
Mother 35 years, daughter 10 years.
Worked example 4: sum of ages and 'years ago'
The sum of the present ages of a father and his son is 60 years. Six years ago, the father was five times as old as the son. Find their present ages.
- Standard method: F + S = 60 and F − 6 = 5(S − 6), so F = 5S − 24.
- Substitute: 5S − 24 + S = 60, so 6S = 84 and S = 14. Then F = 46.
- Shortcut: six years ago, the sum was 60 − 2 × 6 = 48, split in the ratio 5 : 1. Son = 48 ÷ 6 = 8 and father = 40. Add 6 years: son 14, father 46.
- Check: 46 + 14 = 60, and six years ago 40 = 5 × 8.
Father 46 years, son 14 years.
Bank, SSC and railway papers ask replacement and new-member averages, cricket run averages and ratio-based age puzzles; placement tests add 'n years ago' twists. Use the change in total and the fixed age gap instead of long equations.
2 marks: “Find the average of the first 20 natural numbers.” Average of 1 to n = (n + 1) ÷ 2 = 21 ÷ 2 = 10.5.
5 marks: “The average age of 30 students is 15 years. When the teacher's age is included, the average rises by 1 year. Find the teacher's age.” Old total 450; new total 31 × 16 = 496; teacher = 46 years. Shortcut: 15 + 31 × 1 = 46.
10 marks: “Solve: (a) a replacement-average problem; (b) a ratio-of-ages problem; (c) the average of all odd numbers from 1 to 49.” (a) Use change in total, as in worked example 1 (49 kg). (b) Use ak and bk, as in worked example 3 (35 and 10). (c) 25 odd numbers; average (1 + 49) ÷ 2 = 25.
Change in average × count = change in total. In age sums, the gap never changes and every person adds the same years.