Averages

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Averages questions look simple and usually are — provided you resist the urge to find the individual values. Almost every one is solved faster by working with totals and deviations than by reconstructing the data.

Core Formulas

Average = Sum of observations / Number of observations

Sum = Average × Number  (the form you will actually use)

Weighted average = (n₁a₁ + n₂a₂) / (n₁ + n₂)

Average of first n natural numbers = (n + 1) / 2

Average speed over equal distances at x and y = 2xy / (x + y)

Think in totals, not averages. "The average of 600 people rose by 5" means nothing on its own, but "the total rose by 3,000" is immediately usable. Converting an average change into a total change is the key move in most correction questions.

Alligation

When two groups with averages A₁ and A₂ combine to give a mean Aₘ:

n₁ : n₂ = (A₂ − Aₘ) : (Aₘ − A₁)

In words: the quantities are in the inverse ratio of their distances from the mean.

Alligation turns a two-equation problem into a subtraction. Any question that gives you three averages and asks for a ratio is an alligation question.

Worked Examples from Real Papers

TS Police Constable 2022 — Q31

The average weight of a group of men is 77.5 kg and of a group of women is 70 kg. The average weight of all persons in both groups is 74 kg. Find the ratio of the number of men to women.

Apply alligation directly around the mean of 74:
men : women = (74 − 70) : (77.5 − 74) = 4 : 3.5 = 8 : 7.
Note the inversion: the men's distance from the mean (3.5) becomes the women's share of the ratio, and vice versa. Getting this backwards produces 7 : 8, which is sitting in the options.

TS Police SI 2022 — Q8

In a police battalion of 600, the average height is 150 cm. It is later discovered that for x policemen the heights were recorded as 160 cm instead of 190 cm. With the correct heights the average rises to 155 cm. Find x.

Each wrong entry understates the height by 190 − 160 = 30 cm.
The average rose by 5 cm across 600 people, so the total rose by 5 × 600 = 3,000 cm.
x = 3000 / 30 = 100.
Note: the actual heights never mattered — only the size of the error and the change in total.

TS Police SI 2022 — Q7

The averages of three different data sets D₁, D₂, D₃ are 30, 40 and 50. The average of D₁ and D₂ together is 38, and of D₂ and D₃ together is 42. Find the average of all three together.

Let the counts be n₁, n₂, n₃.
From D₁ & D₂: 30n₁ + 40n₂ = 38(n₁ + n₂) → 2n₂ = 8n₁ → n₂ = 4n₁.
From D₂ & D₃: 40n₂ + 50n₃ = 42(n₂ + n₃) → 8n₃ = 2n₂ → n₃ = n₂/4 = n₁.
So n₁ : n₂ : n₃ = 1 : 4 : 1.
Combined average = (30 + 160 + 50) / 6 = 240 / 6 = 40.
Trap: the plain average of 30, 40 and 50 is also 40 — but that is a coincidence here, not the method. Sets of unequal size require the weighted calculation.

Never average the averages unless the groups are equal in size. Doing so is the single most common error in this topic, and examiners rely on it when writing the distractors.

Also see: Ratio & Proportion · Time, Speed & Distance · Percentages

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