Average

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high importance~1 Q in Tier 117 formulas⚡ 12 shortcuts4 subtopics

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Weighted average & two-group problems

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⏱ 4 min read🧩 5 question types🎯 13 practice Q

The idea in one minute

When groups of different sizes are combined, the combined average is the weighted mean:

total of all values ÷ total count. Turn each group's average into a total (size × average) first.

The combined average always sits closer to the bigger group's average. With equal group sizes it is the plain middle of the two averages.

With two group averages and the combined average known, the group sizes sit in the reverse ratio of their distances from the combined average.

Why averaging averages fails

Section A has 20 students with average 60. Section B has 80 students with average 40. The middle of 60 and 40 is 50, but the real class average is $\dfrac{20 \times 60 + 80 \times 40}{100} = 44$ .

The bigger section pulls the answer towards its own average. So each group must be weighted by its size.

Rule: Combined average = (sum of all group totals) ÷ (sum of all sizes). A group total = size × its average.

The weighted-average method

Work in three moves, every time.

Total of each group = size × average.
Add all the totals. Add all the sizes.
Divide.

Example: 30 boys average 42 kg and 20 girls average 37 kg. Combined = $\dfrac{1260 + 740}{50} = 40$ kg.

Where the combined average sits

Three facts that kill wrong options without calculation:

It always lies between the smallest and the largest group average.
It lies closer to the average of the bigger group.
Equal group sizes → it is the plain middle of the group averages.

Tip: If an option sits outside the two group averages, cross it out at once.

One group's average missing

Class of 40 averages 65 marks. The 25 boys average 62. Find the girls' average.

Class total = $40 \times 65 = 2600$ . Boys' total = $25 \times 62 = 1550$ . Girls' total = $2600 - 1550 = 1050$ over 15 girls → 70.

Rule: Missing total = overall total − known totals. Then divide by that group's own size.

Finding the sizes: the balance method

Both group averages and the combined average are known; a size is asked. The sizes sit in the reverse 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. Distances: boys 4, girls 6. So boys : girls = $6 : 4 = 3 : 2$ .

Picture a see-saw: the heavier group sits closer to the balance point. Scale the ratio up to the given total when a count is asked.

Watch: The group closer to the combined average is the bigger one. Reversing this ratio is the classic wrong option.

Sizes given as a ratio or a fraction

No real counts? Use the ratio parts as the counts.

Boys : girls = 3 : 2, averages 150 cm and 140 cm → $\dfrac{3 \times 150 + 2 \times 140}{5} = 146$ cm.

"One-fourth of a class averages 72 and the whole class averages 60." Take the class as 4 parts: $4 \times 60 = 240 = 72 + 3x$ , so the other three parts average $x = 56$ .

Tip: The weighted average depends only on the proportions, never on the actual headcount.

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common3 practice Q

Combined average of two or more groups

How to spot it:

Two or three groups (boys/girls, sections, batches) with sizes and averages; the overall average is asked.

\bar{x} = \frac{n_1 \bar{x}_1 + n_2 \bar{x}_2 + n_3 \bar{x}_3}{n_1 + n_2 + n_3}

Method

Multiply each group's size by its average to get its total.
Add all the totals; add all the sizes.
Divide. The answer must lie closer to the bigger group's average.

Why it works:

An average is a total shared by a count, so only totals can be added.

Try this

In a school, 30 boys average 42 kg and 20 girls average 37 kg. Find the average weight of all 50 students.

Show solution

Boys: $30 \times 42 = 1260$ ; girls: $20 \times 37 = 740$ .
Total = $2000$ kg over 50 students.
Average = $2000 \div 50 = 40$ kg.

Answer

40 kg

Type 2very common2 practice Q

Missing average of one group

How to spot it:

The overall average and one group's average are given; the other group's average is asked.

\bar{x}_2 = \frac{(n_1 + n_2)\bar{x} - n_1 \bar{x}_1}{n_2}

Method

Overall total = total count × overall average.
Subtract the known group's total.
Divide the rest by the other group's size.

Why it works:

The two group totals must add up to the overall total.

Try this

The average marks of 40 students is 65. The 25 boys average 62 marks. Find the girls' average.

Show solution

Class total = $40 \times 65 = 2600$ .
Boys = $25 \times 62 = 1550$ , so girls have $1050$ marks.
Girls: $1050 \div 15 = 70$ .

Answer

Type 3common3 practice Q

Group ratio from the averages

How to spot it:

All three averages are known (two groups plus combined) and the ratio of sizes is asked.

n_1 : n_2 = (\bar{x}_2 - \bar{x}) : (\bar{x} - \bar{x}_1)

Method

Find how far each group average is from the combined average.
The sizes are in the reverse ratio of these distances.
Scale the ratio to the given total if a count is asked.

Why it works:

The surplus above the mean from one group must balance the shortfall from the other.

Try this

A class averages 58 marks. The boys average 62 and the girls 52. Find boys : girls.

Show solution

Distances: boys $62 - 58 = 4$ ; girls $58 - 52 = 6$ .
Sizes are the reverse: boys : girls = $6 : 4 = 3 : 2$ .

Answer

3 : 2

Type 4common3 practice Q

Sizes given as a ratio or fraction

How to spot it:

'Boys and girls are in the ratio 3 : 2' or 'one-fourth of the students average …' instead of real counts.

\bar{x} = \frac{a\bar{x}_1 + b\bar{x}_2}{a + b}

Method

Use the ratio parts (or fraction parts) as if they were the counts.
Apply the weighted-average formula.
For a missing average, solve the one equation that is left.

Why it works:

The weighted average depends only on the proportions of the groups.

Try this

Boys and girls in a class are in the ratio 3 : 2. Boys average 150 cm in height and girls 140 cm. Find the average height of the class.

Show solution

Take 3 parts of boys and 2 of girls.
Total height = $3 \times 150 + 2 \times 140 = 730$ per 5 parts.
Average = $730 \div 5 = 146$ cm.

Answer

146 cm

Type 5common

A group count from the averages

How to spot it:

The overall average and both group averages are known, one group's count is known, and the other count (or total) is asked.

n_1 : n_2 = (\bar{x} - \bar{x}_2) : (\bar{x}_1 - \bar{x})

Method

Find the two distances from the overall average.
The sizes follow the reverse ratio; pair each size with the group on the far side.
Scale the ratio by the known group's count, then add for the total.

Why it works:

Once the size ratio is fixed, the known count prices one part of it.

Try this

A firm pays an average salary of ₹8,000. Its 7 technicians earn ₹12,000 each and the rest earn ₹6,000 each. How many employees does the firm have in total?

Show solution

Distances from 8000: technicians $+4000$ , others $-2000$ .
Size ratio technicians : others = $2000 : 4000 = 1 : 2$ .
Others = $7 \times 2 = 14$ ; total = $7 + 14 = 21$ .

Answer

21 employees

Formula sheet

Weighted mean

\bar{x} = \frac{\sum n_i \bar{x}_i}{\sum n_i}

Missing group average

\bar{x}_2 = \frac{N\bar{x} - n_1\bar{x}_1}{n_2}

N = total count, overall average known.

Sizes from distances

\frac{n_1}{n_2} = \frac{\bar{x}_2 - \bar{x}}{\bar{x} - \bar{x}_1}

Reverse ratio of the distances.

Shortcuts that save time

⚡ Totals, not averages

Turn each group into a total, add, divide by the combined count.

Example

In a class of 60 students, the 20 girls average 40 marks. The class average is 50. Find the boys' average.

Show solution

Class total = $60 \times 50 = 3000$ ; girls = $20 \times 40 = 800$ .
Boys' total = $3000 - 800 = 2200$ over 40 boys.
Boys' average = $2200 \div 40 = 55$ .

Answer

⚡ Feel the balance point

The combined average splits the gap between the group averages in the ratio n₂ : n₁, the reverse of the sizes.

Example

20 boys average 12 years and 30 girls average 11 years. Find the combined average age.

Show solution

Totals: $240 + 330 = 570$ .
Combined = $570 \div 50 = 11.4$ years.
The answer sits 0.4 above 11 and 0.6 below 12 — closer to the girls, the bigger group.

Answer

11.4 years

⚡ One newcomer is a tiny second group

A single new member changing a group average is just a weighted average with k = 1. Use value = B + n(B − A).

Example

15 workers average ₹250 in daily wages. A manager joins and the average becomes ₹300. Find the manager's wage.

Show solution

Jump = $300 - 250 = 50$ .
Manager = $300 + 15 \times 50$ .
= ₹1050.

Answer

₹1,050

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Averaging the two averages when the groups differ in size.

Weight each average by its group size: (n₁x̄₁ + n₂x̄₂) ÷ (n₁ + n₂).

Mistake 02

Dividing by the wrong count (students + teacher, workers + manager).

List every member of the combined group before dividing.

Mistake 03

Reading the balance ratio forwards (closer group = smaller).

The group closer to the combined average is the bigger one.

Mistake 04

Rounding a total halfway through.

These questions come out exact. A messy decimal mid-way means a slip, so recheck the totals.

Mistake 05

Group averages that do not bracket the combined average.

The combined average must sit between the group averages. If not, re-read the question.

Quick revision

Read this the night before the exam.

Combined average = sum of (size × average) ÷ sum of sizes.
Missing group: overall total − known totals, then ÷ its own size.
Sizes = reverse ratio of distances from the combined average.
Ratio or fraction given → use the parts as counts.
Combined average sits closer to the bigger group.

Practice: 13 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 13 questions

Suggested time 8 min · wrong answers go to your mistake notebook automatically.

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