Average

Why a simple average of averages goes wrong

Section A has 20 students with average 60 and section B has 80 students with average 40. The average of the two averages is 50, but the real class average is $\frac{20 \times 60 + 80 \times 40}{100} = 44$. The bigger section pulls the answer towards its own average. When groups have different sizes, each average must be weighted by its group size.

The weighted-average formula

$\bar{x} = \frac{n_1 \bar{x}_1 + n_2 \bar{x}_2 + \cdots}{n_1 + n_2 + \cdots}$ In words: turn every group's average into a total ($n \times$ average), add the totals, divide by the total count. Example: 30 boys average 42 kg and 20 girls average 37 kg → $\frac{1260 + 740}{50} = 40$ kg.

Where the combined average sits

• It always lies between the smallest and the largest group average.
• It is closer to the average of the bigger group.
• If the groups are equal in size, it is exactly the simple average of the group averages. These three facts let you reject options without any calculation.

Finding a missing group average

Class of 40 averages 65; the 25 boys average 62. Class total $= 2600$, boys' total $= 1550$, so the 15 girls total $1050$ → girls' average $70$. Always: missing total = overall total − known totals.

Finding group sizes: the balance (alligation) method

When both group averages and the combined average are known, the sizes are in the opposite ratio of the distances from the combined average: $\frac{n_1}{n_2} = \frac{\bar{x}_2 - \bar{x}}{\bar{x} - \bar{x}_1}$ Class average 58, boys 62, girls 52 → boys : girls $= (58 - 52) : (62 - 58) = 6 : 4 = 3 : 2$. Think of a see-saw: the heavier group sits closer to the balance point. Workers example: overall ₹8,000, 7 technicians at ₹12,000 (4000 above), the rest at ₹6,000 (2000 below) → technicians : rest $= 2000 : 4000 = 1 : 2$ → rest $= 14$, total $21$.

When sizes are given as a ratio or a fraction

Boys : girls $= 3 : 2$ with averages 150 cm and 140 cm → treat the ratio as the counts: $\frac{3 \times 150 + 2 \times 140}{5} = 146$ cm. "One-fourth of the class averages 72 and the class average is 60" → take the class as 4 parts: $4 \times 60 = 72 + 3x$ → $x = 56$ for the rest.

Common traps

• Reversing the ratio in the balance method (the group CLOSER to the mean is the BIGGER one).
• Taking the simple average of averages when sizes differ.
• Forgetting that the group averages must bracket the combined average — if they do not, re-read the question.

Quick revision

• Combined average $= \frac{\sum n_i \bar{x}_i}{\sum n_i}$.
• Missing group: overall total − known totals, then divide.
• Sizes: $n_1 : n_2 = (\bar{x}_2 - \bar{x}) : (\bar{x} - \bar{x}_1)$.
• Ratio or fraction of sizes → use them directly as counts.