Choose your next learning step
Solve comparisons involving unequal groups, combined means, percentage changes and fixed baselines, with worked checks and transfer practice.
1. Understand · 2. Calculate · 3. Reason · 4. Compare · 5. Practise
Use the 2027 K310 syllabus for G3 Mathematics. The worked numbers and practice tasks below are original teaching examples; this workshop does not supply an official mark scheme.
Chapter index
Chapters 1–3
A mean represents total divided by number of observations. When two groups have different sizes, their means do not contribute equally to the combined mean. Rebuild each total first, add the totals, then divide by the combined number.
For group sizes n and m with means a and b, the combined mean is (na + mb)/(n + m). The quantities n and m are the weights because they count how many observations each group represents. This formula assumes you are combining comparable observations measured on the same basis.
The simple average (a + b)/2 works when the two groups have equal size. It also happens to give the same result when a = b. It is not a general rule for combining means. Explain what the denominator counts before using it.
A fictional revision centre records scores from Group A: 12 students, mean 68; and Group B: 18 students, mean 78. Find the combined mean.
Group A total = 12 × 68 = 816. Group B total = 18 × 78 = 1404. Combined total = 2220, and number of students = 30. Combined mean = 2220/30 = 74.
Check the result against the structure. It lies between 68 and 78 and is nearer 78 because the higher-scoring group has more students. The simple average 73 treats the two groups as equal-sized and answers a different question.
Reverse-check: a mean of 74 across 30 observations gives total 2220, which agrees with the sum of the group totals. This check tests the original data relationship; pressing the same calculator sequence again merely repeats the arithmetic route.
Group A has 10 students with mean score 60. Group B has mean score 75. The combined mean is 69. Find the number of students in Group B.
Let that number be n. The total-score equation is 600 + 75n = 69(10 + n). Hence 600 + 75n = 690 + 69n, so 6n = 90 and n = 15.
Verify: total = 600 + 1125 = 1725; number = 25; 1725/25 = 69. The positive integer answer is appropriate because the unknown counts students. An algebraic answer such as 15.6 would require rechecking the model and data; you cannot silently round a count that is constrained by an exact mean relationship.
The crucial first decision is not an equation-solving trick. It is recognising that mean × count recovers the total. An equation based on 60 + 75 = 2 × 69 incorrectly gives each group equal weight.
A price rises from $80 to $100. The increase is $20, so the percentage increase is 20/80 × 100% = 25%. The percentage decrease needed to return from $100 to $80 is 20/100 × 100% = 20%. The absolute change is the same; the starting amount is different.
Likewise, a success rate rising from 40% to 50% rises by 10 percentage points. Relative to the original 40%, that is a 25% increase in the rate. Write which measure the question asks for; “up 10%” is ambiguous when you mean percentage points.
For unequal groups, reconstruct the counts. If 20 of 25 students in one group and 15 of 50 in another succeed, the combined success rate is 35/75 × 100% = 46.7% to three significant figures. Averaging 80% and 30% gives 55%, which ignores the unequal denominators.
Keep exact fractions during calculation and round the final requested quantity. Rounding a group rate early can change a combined result.
Attempt three tasks. A: 8 readings have mean 15 and 12 readings have mean 20. Find the combined mean. B: 6 observations have mean 10; another group has mean 16; together their mean is 14. Find the second group’s size. C: A value falls from 250 to 200. Find the percentage decrease, then the percentage increase required to return to 250.
Answers with working: A = (8 × 15 + 12 × 20)/20 = 18. B: 60 + 16n = 14(6 + n), so 2n = 24 and n = 12. C: decrease = 50/250 × 100% = 20%; return increase = 50/200 × 100% = 25%.
If A is wrong, inspect whether you reconstructed totals. If B is wrong, inspect whether the combined count was 6 + n. If C has identical percentages, inspect the chosen base. These are different errors and deserve different repairs.
Later, retest using a mean price, a mean journey time or a combined attendance rate. Ask whether the observations are comparable. This reasoning does not imply that every average can be pooled in the same way: average speed, for example, must be rebuilt from total distance and total time.
Continue learning and official sources
Official K310 Mathematics syllabus · Complete EMS Learner’s Guide
