MathsCBSE Class 9 · How Quantities Combine: Understanding Data

Combined Mean and Weighted Mean: Class 9 Data

Q: When can I simply average two averages?

Only when the two groups contain the same number of items (or weights are equal).

Q: Can a weighted mean be outside the range of the values?

No. It always lies between the smallest and the largest value.

Q: Is the median always an actual data value?

Not always. With an even count it is the mean of the two middle values, which may not be in the data.

Updated 17 October 2026 · Fact-checked

The combined mean of two groups is $\frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2}$: add the totals, then divide by the total number of items. A weighted mean is $\frac{\sum w x}{\sum w}$, where each value counts as many times as its weight. A plain average of two averages is correct only when both groups have the same size. Always work with totals.

The bigger group sits closer to the balance point, so the combined mean 60.8 lies nearer 64 than 56, not at the plain average 60.

Illustration drawn for QuestionBuddy.

Understand Combined Mean and Weighted Mean

The mean of $n$ values is their total divided by $n$ , so $total = n \times mean$ . That one rearrangement solves almost every question here: convert each mean back to a total, combine totals, and divide again.

Average of averages. If 20 students average 56 marks and 30 students average 64, the class mean is not $\frac{56 + 64}{2} = 60$ . The correct value is $\frac{20 \times 56 + 30 \times 64}{50} = \frac{1120 + 1920}{50} = 60.8$ . The larger group pulls the mean towards its own value, so the answer lies nearer 64 than 56. A quick check: the combined mean must always lie between the two group means, and closer to the group with more members.

Weighted mean. When values matter unequally (marks in subjects with different credit, prices of goods bought in different quantities), multiply each value by its weight and divide by the sum of weights: $\overset{x}{ˉ}_{w} = \frac{\sum w _{i} x _{i}}{\sum w _{i}}$ . A student with scores $88, 76, 90, 70$ and weights $4, 3, 2, 1$ has weighted mean $\frac{352 + 228 + 180 + 70}{10} = 83$ . Weights can be scaled (multiplied by a common factor) without changing the answer.

Changing a mean. Adding a new value $v$ to $n$ values with mean $\overset{x}{ˉ}$ gives new mean $\frac{n x ˉ + v}{n + 1}$ . To lift the mean to a target $t$ the new value must be $(n + 1) t - n \overset{x}{ˉ}$ . If a value is mis-recorded, correct the total by the difference and divide by the same $n$ .

When the mean is the wrong tool. Rates need care. Travelling 60 km at 40 km/h and another 60 km at 60 km/h does not give an average speed of 50 km/h: the times are $1.5$ h and $1$ h, so the average speed is $\frac{120}{2.5} = 48$ km/h. Average speed is total distance over total time, a weighted mean of speeds with time as the weight.

Mean, median and mode. The median is the middle value of the ordered data (the mean of the middle two if the count is even), and the mode is the most frequent value. One extreme value can drag the mean a long way but moves the median very little, so skewed data such as salaries is better summarised by the median or mode. State which measure you choose and why.

Exam approach. Write every mean as "total = number x mean" first. Check that the answer lies in a sensible range. Label units. For decision questions, say what the data show and which measure represents "typical" best.

Combined Mean and Weighted Mean: formulas to remember

Mean and total: $\overset{x}{ˉ} = \frac{total}{n} \Rightarrow total = n \overset{x}{ˉ}$; Convert means back to totals first.
Combined mean: $\overset{x}{ˉ} = \frac{n _{1} x ˉ _{1} + n _{2} x ˉ _{2}}{n _{1} + n _{2}}$; Lies between $\overset{x}{ˉ}_{1}$ and $\overset{x}{ˉ}_{2}$ .
Weighted mean: $\overset{x}{ˉ}_{w} = \frac{\sum w _{i} x _{i}}{\sum w _{i}}$; Weights can be scaled by any common factor.
Adding one value: $new mean = \frac{n x ˉ + v}{n + 1}$; Value needed for a target $t$ : $(n + 1) t - n \overset{x}{ˉ}$ .
Average speed: $average speed = \frac{total distance}{total time}$; Not the mean of the speeds unless times are equal.
Median: $middle value (or mean of the two middle values)$; Order the data first.

How to solve Combined Mean and Weighted Mean questions

Turn means into totals, combine, and divide again.

1Write each mean as a total: total = (number of items) x (mean).
2For two groups add the totals and the counts; divide total by total count.
3For a weighted mean, multiply each value by its weight, add, and divide by the sum of weights.
4For a missing value, set up (known total + unknown) / (count) = target mean and solve.
5For rates such as speed, find total distance and total time, then divide.
6Check that the answer is in a sensible range and quote the unit.

Combined Mean and Weighted Mean in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Weighted mean

Suppose a grocer buys 3 kg of rice at 40 rupees per kg and 5 kg of another variety at 56 rupees per kg and mixes them.

At what price per kg should the mixture be sold to recover exactly the cost?

Try it first, then show the solutionSolution

Total cost $= 3 \times 40 + 5 \times 56 = 120 + 280 = 400$ rupees.
Total quantity $= 3 + 5 = 8$ kg.
Price per kg $= \frac{400}{8} = 50$ rupees.

Answer: The mixture must be sold at 50 rupees per kg to break even.

Memory hook: Weights are the quantities.

Tricks to remember Combined Mean and Weighted Mean

Short memory aids. Say each one aloud once, then use it on a question the same day.

Total first
Never average averages directly; multiply each by its count, add the totals, then divide by the combined count.
Use it when: Combined-mean questions.
Between and nearer the bigger
The combined mean sits between the two means and closer to the group with more items.
Use it when: Quick reasonableness check.
Speed needs time
Average speed is distance over time, never half the sum of speeds, unless the times are equal.
Use it when: Journey problems.

Combined Mean and Weighted Mean: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Combining two classes · 3 marks · Short answerBoard level

In a school test, 20 students of Section A have a mean of 56 marks and 30 students of Section B have a mean of 64 marks. Find the mean of all 50 students.

Try it first, then show the solutionSolution

Total of A $= 20 \times 56 = 1120$ . (1)
Total of B $= 30 \times 64 = 1920$ . (1)
Combined mean $= \frac{1120 + 1920}{50} = \frac{3040}{50} = 60.8$ . (1)

Answer: The combined mean is 60.8 marks.

Example 2: Adding a value · 3 marks · Short answerBoard level

The mean of five numbers is 18. A sixth number 24 is added. Find the new mean, and the sixth number that would make the mean of all six equal to 20.

Try it first, then show the solutionSolution

Total of five numbers $= 5 \times 18 = 90$ . (½)
With 24 added: $\frac{90 + 24}{6} = \frac{114}{6} = 19$ . (1)
For a mean of 20: required total $= 6 \times 20 = 120$ . (1)
Sixth number $= 120 - 90 = 30$ . (½)

Answer: The new mean is 19; the sixth number would have to be 30.

Example 3: Weighted marks target · 5 marks · Long answerHarder board level

A student's weighted score uses weights 4, 3, 2, 1 for Maths, Science, English and Hindi. Her marks are 88, 76, 90 and 70. (i) Find the weighted mean. (ii) What is the least whole number of marks in Maths that gives a weighted mean of at least 85, with the other marks unchanged?

Try it first, then show the solutionSolution

(i) Weighted sum $= 4 (88) + 3 (76) + 2 (90) + 1 (70) = 352 + 228 + 180 + 70 = 830$ . (1)
Sum of weights $= 10$ , so weighted mean $= \frac{830}{10} = 83$ . (1)
(ii) Let Maths marks be $m$ : $\frac{4 m + 228 + 180 + 70}{10} \geq 85$ . (1)
$4 m + 478 \geq 850$ , so $4 m \geq 372$ and $m \geq 93$ . (1)
Check: $m = 93$ gives $\frac{372 + 478}{10} = 85$ ; $m = 92$ gives $84.6 < 85$ . (1)

Answer: (i) 83; (ii) 93 marks.

Example 4: Average speed trap · 3 marks · Short answerHarder board level

A cyclist rides 60 km at 40 km/h and then 60 km at 60 km/h. Find the average speed for the whole journey and explain why it is not 50 km/h.

Try it first, then show the solutionSolution

Time for the first part $= \frac{60}{40} = 1.5$ h. (½)
Time for the second part $= \frac{60}{60} = 1$ h. (½)
Total distance $= 120$ km, total time $= 2.5$ h, so average speed $= \frac{120}{2.5} = 48$ km/h. (1)
The cyclist spends more time at the lower speed, so the lower speed carries more weight; hence the average is below 50. (1)

Answer: The average speed is 48 km/h, not 50 km/h.

Example 5: Correcting a recorded mean · 4 marks · Case-basedHOTS / competency

Case: a teacher records the marks of 40 students and finds a mean of 52. She then notices that one score of 82 was entered as 28. (i) Find the corrected mean of this class. (ii) Another section of 60 students has mean 50. Find the combined mean of the 100 students using the corrected value. (iii) Why is the combined mean not simply the average of 53.35 and 50?

Try it first, then show the solutionSolution

(i) Recorded total $= 40 \times 52 = 2080$ . Correction $= 82 - 28 = 54$ , so the true total is $2134$ and the mean is $\frac{2134}{40} = 53.35$ . (1½)
(ii) Second total $= 60 \times 50 = 3000$ . Combined mean $= \frac{2134 + 3000}{100} = 51.34$ . (1½)
(iii) The simple average is $51.675$ , which treats the groups as equal in size; the section of 60 has more weight, so the true mean is lower and nearer 50. (1)

Answer: (i) 53.35; (ii) 51.34; (iii) the groups have different sizes, so each mean must be weighted by its size.

Where students lose marks in Combined Mean and Weighted Mean

The slips examiners see most often, and what to write instead.

Mistake: Averaging the averages
Treating groups of different sizes as equal.
Wrong working
$\frac{56 + 64}{2} = 60$ .
Right working
$\frac{20 ( 56 ) + 30 ( 64 )}{50} = 60.8$ .
Do this: Multiply each mean by its count first.
Usually costs: 2 of 3 marks.
Mistake: Dividing by the wrong count after adding a value
Using the old $n$ .
Wrong working
$\frac{90 + 24}{5} = 22.8$ .
Right working
$\frac{90 + 24}{6} = 19$ .
Do this: Write the new count in brackets before dividing.
Usually costs: 1 to 2 marks.
Mistake: Mean of speeds as average speed
Forgetting that time differs for each part.
Wrong working
$\frac{40 + 60}{2} = 50$ km/h.
Right working
$\frac{120}{2.5} = 48$ km/h.
Do this: Find total distance and total time.
Usually costs: The full question.

Exam tips

Start every mean question with "total = n x mean".
Check the combined mean lies between the two group means.
For decision questions, give the measure and a one-line reason that mentions extreme values.
Give units and, if the question says so, the answer to the stated decimal places.

Practice questions on Combined Mean and Weighted Mean

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The mean of 8, 12, 16 and $x$ is 13. The value of $x$ is
(a) 12
(b) 14
(c) 16
(d) 18
Hint
The total of the four numbers is $4 \times 13$ .
Answer
(c) 16. $8 + 12 + 16 + x = 52$ , so $x = 16$ .
Q2 · 3 marks · Short answerBoard level
A shopkeeper mixes 3 kg of rice costing 40 rupees per kg with 5 kg costing 56 rupees per kg. Find the cost price per kg of the mixture.
Hint
Weighted mean with quantities as weights.
Answer
$\frac{3 \times 40 + 5 \times 56}{8} = \frac{120 + 280}{8} = \frac{400}{8} = 50$ rupees per kg.
Q3 · 3 marks · Short answerHarder board level
The mean of 11 numbers is 20. The mean of the first six is 18 and the mean of the last six is 22. Find the sixth number.
Hint
The sixth number is counted in both groups of six.
Answer
Sum of first six $= 108$ ; sum of last six $= 132$ ; together they count the sixth number twice, so $108 + 132 - 11 \times 20 = 240 - 220 = 20$ . The sixth number is 20.
Q4 · 5 marks · Long answerHarder board level
The mean marks of 50 students in a test is 40. The boys' mean is 38 and the girls' mean is 43. Find the number of boys and girls, and verify that the distances of the two means from 40 are in the inverse ratio of the numbers of students.
Hint
Let the number of boys be $b$ ; total marks of boys plus girls equals $50 \times 40$ .
Answer
$38 b + 43 (50 - b) = 2000$ , so $2150 - 5 b = 2000$ and $b = 30$ ; girls $= 20$ . Check: $38 \times 30 + 43 \times 20 = 1140 + 860 = 2000$ . Distances from 40 are $2$ (boys) and $3$ (girls), whose ratio $2 : 3$ is the inverse of $30 : 20 = 3 : 2$ .
Q5 · 4 marks · Case-basedHOTS / competency
Case: a company reports monthly salaries (in thousand rupees) of six staff: 18, 20, 20, 22, 25 and 95. (i) Find the mean. (ii) Find the median. (iii) Find the mode. (iv) Which measure best describes a typical salary here, and why?
Hint
Arrange the data in order; look at the effect of 95.
Answer
(i) Mean $= \frac{200}{6} \approx 33.3$ thousand rupees. (ii) Ordered data; the middle two are 20 and 22, so the median is 21. (iii) The mode is 20. (iv) The single large value 95 pulls the mean up to 33.3, higher than five of the six salaries, so the median (21) or mode (20) describes a typical salary better.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Combined mean of two groups?Show answer
$\frac{n _{1} x ˉ _{1} + n _{2} x ˉ _{2}}{n _{1} + n _{2}}$ .
2.Weighted mean formula?Show answer
$\frac{\sum w x}{\sum w}$ .
3.Total in terms of mean?Show answer
$n \times mean$ .
4.Average speed?Show answer
Total distance over total time.
5.Which measure resists outliers?Show answer
The median.

Combined Mean and Weighted Mean: frequently asked questions

When can I simply average two averages?

Only when the two groups contain the same number of items (or weights are equal).

Can a weighted mean be outside the range of the values?

No. It always lies between the smallest and the largest value.

Is the median always an actual data value?

Not always. With an even count it is the mean of the two middle values, which may not be in the data.