Averages and ages for IBPS PO

Average questions look different every time: marks of a class, runs of a batsman, weights of a group, salaries of staff. But almost all of them collapse into one move: turn every average into a total. Totals can be added, subtracted and compared; averages cannot. Once you work in totals, most of these questions take a line or two.

Age questions sit next to averages because they use the same habit, plus one rule about time. Both appear as prelims word problems, as sides of a quantity comparison, and as data sufficiency stems in the mains.

The total method

Total = average × number of values. Convert first, then reason.

Why it works: an average is just a total shared out equally. Adding a member, removing one or replacing one changes the total by a known amount, and the new average is the new total divided by the new count.

The deviation shortcut

To average numbers that sit close together, guess a value near the middle and average the differences.

• Numbers: 47, 53, 58, 44, 50, 54. Guess 50.
• Differences: −3, +3, +8, −6, 0, +4. Sum = 6.
• Average = 50 + 6 ÷ 6 = 51.

Why it works: each number is 50 plus its difference, so the average is 50 plus the average of the differences.

Weighted averages

When two groups of different sizes are combined, you cannot average the two averages. Weight each by its size.

• Combined average = (n₁ × a₁ + n₂ × a₂) ÷ (n₁ + n₂).
• The combined average always lies between the two group averages, closer to the larger group.

Worked examples: averages

Example 1: The average of 11 results is 50. The first six average 49 and the last six average 52. Find the sixth result.

• The sixth result is in both groups, so it is counted twice.
• 6 × 49 + 6 × 52 − 11 × 50 = 294 + 312 − 550 = 56.

Example 2: A batsman scores 87 in the 17th innings, which raises their average by 3. Find the new average.

• Let the old average be a. Then 16a + 87 = 17(a + 3).
• 16a + 87 = 17a + 51, so a = 36. New average = 39.
• Shortcut: the 87 covers the new average plus 3 for each of the 16 earlier innings. 87 − 16 × 3 = 39.

Example 3: The average weight of 8 people rises by 2.5 kg when a new person replaces one who weighs 65 kg. Find the new person's weight.

• The total rises by 8 × 2.5 = 20 kg.
• New person = 65 + 20 = 85 kg.

Example 4: Class A has 30 students with an average of 60. Class B has 20 students with an average of 70. Find the combined average.

• (30 × 60 + 20 × 70) ÷ 50 = (1,800 + 1,400) ÷ 50 = 64.
• Not 65, because class A is larger.

Ages

• Use one letter for the ratio multiplier or for one person's present age.
• "n years ago" and "n years hence" change every person's age by n. The difference between two ages never changes.
• Write every statement at one point in time (usually the present, or the earliest time mentioned) before solving.

Worked examples: ages

Example 5: Six years ago, the ages of two people were in the ratio 6 : 5. Four years from now, the ratio will be 11 : 10. Find their present ages.

• Six years ago: 6x and 5x. Four years from now (10 years later): 6x + 10 and 5x + 10.
• (6x + 10) ÷ (5x + 10) = 11 ÷ 10, so 60x + 100 = 55x + 110 and x = 2.
• Present ages: 12 + 6 = 18 and 10 + 6 = 16. Check: in four years, 22 : 20 = 11 : 10.

Example 6: A parent is three times as old as their child. In 12 years, the parent will be twice as old as the child. Find their present ages.

• Child = c, parent = 3c. In 12 years: 3c + 12 = 2(c + 12).
• 3c + 12 = 2c + 24, so c = 12.
• Child 12, parent 36. Check: in 12 years, 48 = 2 × 24.

Example 7: The average age of a family of five is 24 years. The youngest member is 4 years old. What was the average age of the family just before the youngest was born?

• Present total = 5 × 24 = 120. The other four total 120 − 4 = 116.
• Four years ago, each of those four was 4 years younger: 116 − 16 = 100.
• Average then = 100 ÷ 4 = 25 years.

Common mistakes

• Averaging two averages without weighting by group size.
• Forgetting the double count in overlap questions. Adding two overlapping group totals counts the shared value twice.
• Adding years to only one person in age problems.
• Mixing time frames. Write every age at the same point in time before forming the equation.
• Forgetting that the count changes when someone joins or leaves.

Practice set

1. The average of 5 numbers is 30. If one number, 50, is removed, find the new average.
2. The average age of 20 students is 16. When the teacher is included, it becomes 17. Find the teacher's age.
3. Two ages are in the ratio 4 : 5. After 5 years, the ratio will be 5 : 6. Find the present ages.
4. A batsman scores 80 in the 11th innings, which raises their average by 2. Find the new average.
5. The average of 7 numbers is 40. The first four average 38 and the last four average 42. Find the fourth number.
6. The average of 10 numbers was found to be 25, but one number, 45, had been read as 54. Find the correct average.
7. The sum of the present ages of two people is 55. Five years ago, one was twice as old as the other. Find their present ages.
8. The average of five consecutive even numbers is 36. Find the largest.

Answers:

1. 25. (150 − 50) ÷ 4.
2. 37. 21 × 17 − 20 × 16 = 357 − 320.
3. 20 and 25. (4x + 5) ÷ (5x + 5) = 5 ÷ 6 gives x = 5.
4. 60. 10a + 80 = 11(a + 2), so a = 58.
5. 40. 4 × 38 + 4 × 42 − 7 × 40 = 152 + 168 − 280.
6. 24.1. The total was 9 too high; 25 − 9 ÷ 10.
7. 35 and 20. Five years ago: 30 and 15, and 30 = 2 × 15.
8. 40. The middle number is 36, so the numbers are 32, 34, 36, 38 and 40.

What to do next

• Solve ten average and five age questions a day for a week, converting every average to a total first.
• Use the deviation shortcut in simplification and DI averages.
• Revise the multiplier method in ratio and partnership; age ratios use it directly.
• Try age stems in data sufficiency, where the question is whether two statements fix the ages.