MathsCBSE Class 9 · Chapter 10

How Quantities Combine: Understanding Data Class 9 Notes and Solved Questions

Updated 15 October 2026 · Fact-checked

How Quantities Combine: Understanding Data in Class 9 teaches the weighted mean, used to combine averages of groups of different sizes, find the concentration of mixtures and combine marks with set weights. It also teaches reading stacked and 100% stacked bar charts, and choosing between mean, median and mode.

What How Quantities Combine: Understanding Data is about

This chapter of *Ganita Manjari* Part II is CBSE's Statistics (Unit VI, Statistics and Probability). It extends the familiar average in one big direction: combining things fairly.

Average of averages: if 8 seniors average 165.5 cm and 3 juniors average 149.33 cm, the group's average is *not* the plain average of the two; each average must count according to how many people it stands for.
Mixtures: mixing 500 mL of 10% jaggery lemonade with 200 mL of 20% lemonade gives about 13%, not 15%.
Custom weights: a report card might combine internal tests, a project and a final exam in the ratio 3 : 2 : 5.

All three are the weighted mean $\overset{x}{ˉ} = \frac{w _{1} x _{1} + w _{2} x _{2} + \dots + w _{n} x _{n}}{w _{1} + w _{2} + \dots + w _{n}}$ , a formula Brahmagupta wrote down (c. 628 CE) for the mean depth of a pit; Śrīdharācārya used it for the purity of gold.

The second half is about seeing data: stacked bar charts show totals and their parts, and 100% stacked bar charts show only shares, like a pie chart but easier to compare. The CBSE syllabus also asks you to interpret mean, median and mode to make decisions.

Questions are mostly calculation plus a sentence of interpretation: say what your number means and which chart or average suits the question.

Weightage: This chapter is part of Unit VI, Statistics and Probability, which carries 10 of the 80 marks in the school annual exam, shared with Introduction to Probability.Source: CBSE Mathematics syllabus, Class IX 2026-27, course structure, Unit VI Statistics and Probability

Key concepts

Average of averages

NCERT section 10.1 Combining Things

If group 1 has n values with average a, its total is $an$ . If group 2 has m values with average b, its total is $bm$ . The average of everything is
$\frac{an + bm}{n + m}$
It is not $\frac{a + b}{2}$ unless the groups are the same size. The bigger group pulls the combined average towards its own average.

Worked example

In a class, 24 girls have an average score of 58 and 16 boys have an average of 63. Find the class average. Why is it not 60.5?

Try it first, then show the solutionSolution

Total for girls $= 24 \times 58 = 1392$ ; total for boys $= 16 \times 63 = 1008$ .
Class average $= \frac{1392 + 1008}{40} = \frac{2400}{40} = 60$ .
60.5 is the plain average of 58 and 63, which wrongly treats the groups as equal; there are more girls, so the class average is nearer 58.

Answer: 60

Mixtures and concentration

NCERT section 10.1 Combining Things

The concentration of something in a mixture is
$\frac{quantity of that thing}{total quantity of the mixture}$
When mixing, find each part's actual amount (volume × concentration), add, and divide by the total volume.

Equal volumes: the result is exactly midway between the two concentrations.
Unequal volumes: the result is nearer the concentration of the bigger volume.
Adding plain water adds volume but no extra substance, so the concentration falls.

Worked example

400 mL of a 5% salt solution is mixed with 600 mL of a 15% salt solution. Find the concentration of salt in the mixture.

Try it first, then show the solutionSolution

Salt: $400 \times 0.05 = 20$ mL-equivalent and $600 \times 0.15 = 90$ .
Total salt $= 110$ in 1000 mL.
Concentration $= \frac{110}{1000} = 11%$ , nearer 15% because there is more of the 15% solution.

Answer: 11%

The weighted mean

NCERT section 10.1 Combining Things

For values $x_{1}, x_{2}, \dots, x_{n}$ with weights $w_{1}, w_{2}, \dots, w_{n}$ :
$\overset{x}{ˉ} = \frac{w _{1} x _{1} + w _{2} x _{2} + \dots + w _{n} x _{n}}{w _{1} + w _{2} + \dots + w _{n}}$
The weights say how much each value counts: group sizes, volumes, lengths, or chosen importance.

History: Brahmagupta (Brāhmasphuṭasiddhānta, c. 628 CE) gave this formula for the mean depth of a pit dug to different depths along its length; Śrīdharācārya (c. 750 CE) used it to find the purity (varṇa) of melted-together gold. 16 varṇa meant pure gold, like today's 24 carat.

Multiplying all weights by the same number does not change the weighted mean, because it cancels from top and bottom.

Worked example

A goldsmith melts 6 units of gold of 14 varṇa with 10 units of 12 varṇa. Find the purity of the result in varṇa and in carats (16 varṇa = 24 carats).

Try it first, then show the solutionSolution

Weighted mean $= \frac{6 \times 14 + 10 \times 12}{6 + 10} = \frac{84 + 120}{16} = 12.75$ varṇa.
In carats: $12.75 \times \frac{24}{16} = 19.125$ carats.

Answer: 12.75 varṇa, about 19.1 carats.

Custom weights: marks and ratings

NCERT section 10.1 Combining Things

Weights can be chosen to show importance. If internal tests, a project and the final exam are combined in the ratio 3 : 2 : 5, the final exam counts most.

The ratio is the same as repeating the values: weights 3, 2, 5 on 60, 64, 73 is like averaging 60, 60, 60, 64, 64, 73, 73, 73, 73, 73.

The same idea is used in restaurant ratings (food, service, ambience weighted 4 : 3 : 2) and product ratings (recent reviews counted twice). Convert all marks to the same scale (for example, percentages) before weighting.

Worked example

Asha scores 70% in internal tests, 80% in a project and 64% in the final exam. Find her annual score if they are combined in the ratio 2 : 3 : 5, and compare with the plain average.

Try it first, then show the solutionSolution

Weighted: $\frac{2 \times 70 + 3 \times 80 + 5 \times 64}{2 + 3 + 5} = \frac{140 + 240 + 320}{10} = 70%$ .
Plain average: $\frac{70 + 80 + 64}{3} \approx 71.3%$ .
The weighted score is lower because her weakest score (the final) has the biggest weight.

Answer: 70% weighted (about 71.3% unweighted).

Updating an average when a value is added

NCERT section 10.1 Combining Things

Treat the old data as one group: its total is (old average) × (old count). Add the new value and divide by the new count:
$new average = \frac{old average \times n + new value}{n + 1}$
This is the weighted mean of the old average (weight n) and the new value (weight 1). The new value pulls the average only a little when n is large.

Cricket's run rate (runs per over) is updated this way after every over.

Worked example

Rohit's average over 15 tests is 68. He scores 84 in the 16th test. Find his new average. Guess first: will it rise by more or less than 2?

Try it first, then show the solutionSolution

Old total $= 15 \times 68 = 1020$ .
New average $= \frac{1020 + 84}{16} = \frac{1104}{16} = 69$ .
It rises by only 1, because one score of 84 is shared over 16 tests.

Answer: 69

Stacked bar charts

NCERT section 10.2 Visualising and Interpreting Data

A cluster (grouped) bar chart puts bars side by side; it is good for comparing one category across groups.

A stacked bar chart joins a group's bars end to end. The full length shows the total, and each piece shows a component. It is best when you want both the totals and what makes them up.

Drawback: only the first piece starts at 0, so the other pieces are harder to read exactly and harder to compare across bars.

Worked example

Weekly study hours: Arun: Maths 6, Science 4, Languages 2. Bina: 3, 3, 3. Charu: 5, 5, 10. In a stacked bar chart of this data, which student has the longest bar, and what does its length show? Which piece of Bina's bar starts at 0?

Try it first, then show the solutionSolution

Totals: Arun $6 + 4 + 2 = 12$ , Bina 9, Charu 20.
Charu's bar is longest; its full length is her total of 20 hours.
The first piece in the stacking order (Maths, if drawn first) starts at 0; the others start where the previous piece ends.

Answer: Charu (20 hours); the first-stacked piece (Maths) starts at 0.

100% stacked bar charts

NCERT section 10.2 Visualising and Interpreting Data

In a 100% stacked bar chart, every bar has the same length (100%), and each piece shows a share (percentage) of that bar's total. It does the job of a pie chart, but lengths along a line are easier to compare than angles.

To draw one: find each group's total, then each part as $\frac{part}{total} \times 100%$ .

Important: a 100% chart shows proportions, not amounts. You can make a 100% chart from a stacked chart (or a table), but not the other way round, because the totals are lost.

Worked example

A family's daily water use is 120 L for bathing, 60 L for washing clothes, 45 L for cooking and drinking, and 75 L for cleaning. Find the percentages for a 100% stacked bar.

Try it first, then show the solutionSolution

Total $= 120 + 60 + 45 + 75 = 300$ L.
Bathing $\frac{120}{300} = 40%$ , washing $20%$ , cooking and drinking $15%$ , cleaning $25%$ .
Check: $40 + 20 + 15 + 25 = 100$ .

Answer: 40%, 20%, 15% and 25%.

Mean, median and mode for decisions

The CBSE syllabus asks you to interpret the three measures of central tendency:

Mean: total ÷ number of values. Uses every value, but a few extreme values pull it.
Median: the middle value after arranging in order (for an even count, the average of the two middle values). Not affected by extreme values.
Mode: the most frequent value. Useful for 'most popular' questions, such as the shoe size to stock most.

Choose the measure that answers the question: median for a 'typical' salary or price when there are outliers, mode for the most common choice, mean when totals matter (sharing costs equally).

Worked example

Monthly salaries (in ₹ thousand) at a small office are 15, 18, 18, 20, 22, 25 and 90. Find the mean, median and mode, and say which best describes a typical salary.

Try it first, then show the solutionSolution

Mean $= \frac{15 + 18 + 18 + 20 + 22 + 25 + 90}{7} = \frac{208}{7} \approx 29.7$ .
Median: the 4th of 7 values in order $= 20$ .
Mode $= 18$ (appears twice).
The 90 (perhaps the owner) pulls the mean above six of the seven salaries, so the median, 20, best describes a typical salary.

Answer: Mean ≈ ₹29.7 thousand, median ₹20 thousand, mode ₹18 thousand; the median is the best summary.

Maths in real life: How Quantities Combine: Understanding Data

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 marks

Suppose your school combines internal tests, a project and the final exam in the ratio 3 : 2 : 5, and you score 80%, 90% and 70%.

What is your annual score?

Try it first, then show the solutionSolution

$\frac{3 \times 80 + 2 \times 90 + 5 \times 70}{10}$ .
$= \frac{240 + 180 + 350}{10} = 77%$ .

Answer: 77%

Memory hook: The heaviest weight decides most of the result.

Mixture concentration

Suppose you mix 300 mL of milk tea that is 20% milk with 100 mL of pure milk.

What percentage of the new drink is milk?

Try it first, then show the solutionSolution

Milk: $300 \times 0.2 + 100 = 160$ mL.
Total 400 mL: $\frac{160}{400} = 40%$ .

Answer: 40%

Memory hook: Count the actual milk, then divide by the whole cup.

Combined average

Suppose your cycling app shows you averaged 12 km a day for 5 school days and 20 km a day over the 2-day weekend.

What was your average for the whole week?

Try it first, then show the solutionSolution

Total $= 5 \times 12 + 2 \times 20 = 100$ km.
$\frac{100}{7} \approx 14.3$ km a day.

Answer: About 14.3 km a day (not 16).

Memory hook: Five weekdays outweigh two weekend days.

Ratings with weights

Suppose you rate a restaurant 4 for food, 3 for service and 5 for ambience, and the app weights them 4 : 3 : 2.

What overall rating does the app give?

Try it first, then show the solutionSolution

$\frac{4 \times 4 + 3 \times 3 + 2 \times 5}{9} = \frac{35}{9}$ .
$\approx 3.9$ , lower than the plain average of 4 because the best score (ambience) has the smallest weight.

Answer: About 3.9 out of 5

Memory hook: The item with the biggest weight moves the rating most.

100% stacked bars

Suppose your family's electricity use one month is 400 units for lighting, 250 for cooling, 150 for kitchen appliances and 200 for others.

Find the shares for a 100% stacked bar.

Try it first, then show the solutionSolution

Total $= 1000$ units.
Shares: 40%, 25%, 15%, 20%.

Answer: 40%, 25%, 15% and 20%.

Memory hook: Part over total, times 100, for every slice.

Topics in this chapter

How Quantities Combine: Understanding Data formula sheet

Mean: $\overset{x}{ˉ} = \frac{x _{1} + x _{2} + \dots + x _{n}}{n}$
Combined mean of two groups: $\overset{x}{ˉ} = \frac{an + bm}{n + m}$; Group averages a and b with sizes n and m.
Weighted mean: $\overset{x}{ˉ} = \frac{w _{1} x _{1} + w _{2} x _{2} + \dots + w _{n} x _{n}}{w _{1} + w _{2} + \dots + w _{n}}$; Brahmagupta's formula (c. 628 CE).
Concentration of a mixture: $c = \frac{V _{1} c _{1} + V _{2} c _{2} + \dots}{V _{1} + V _{2} + \dots}$; A weighted mean with volumes as weights.
Adding one value: $new mean = \frac{n x ˉ + x _{new}}{n + 1}$
Scaling weights: $\frac{\sum k w _{i} x _{i}}{\sum k w _{i}} = \frac{\sum w _{i} x _{i}}{\sum w _{i}}$; Multiplying every weight by k changes nothing.
Percentage share: $share = \frac{part}{total} \times 100%$; Used for 100% stacked bars and pie charts.
Median: $n odd: the \frac{n + 1}{2} -th value; n even: average of the \frac{n}{2} -th and (\frac{n}{2} + 1) -th$; Arrange the data in order first.
Mode: $the value with the highest frequency$
Purity of gold: $16 varṇa = 24 carats (pure gold)$

How the school exam asks How Quantities Combine: Understanding Data

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Combined average	Two or more groups with sizes and averages; find the overall average, or an unknown group size or average.	Turn each average into a total (average × size), add the totals, divide by the total size. For an unknown, set up one equation.
Mixture concentration	Mixing solutions or alloys; diluting with water; finding the amount of one batch for a given final concentration.	Work with the actual quantity of the substance in each part. Water adds volume but no substance. Solve a linear equation for an unknown quantity.
Weighted marks or ratings	Combine scores in a given ratio; compare two people; choose the right expression from options.	Put all scores on the same scale first. Multiply each by its weight, add, divide by the sum of weights.
Updating an average	Run rates, a migrating bird's daily distance, test averages, average share price after a purchase.	Old total = old average × old count; add the new values; divide by the new count.
Reading stacked and 100% stacked bars	Answer questions from a chart: totals, parts, shares, comparisons; or make a 100% chart from a table.	Use stacked bars for totals and amounts, 100% bars for shares. Say when a chart cannot answer a question (a 100% chart cannot give amounts).
Choosing mean, median or mode	Find all three and decide which to report for a purpose.	Outliers present: median. Most popular item: mode. Totals or equal sharing: mean. Give a one-line reason.

Tricks to remember How Quantities Combine: Understanding Data

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

Averages to totals and back
Never add averages; add totals. Average × count = total, total ÷ count = average.
Use it when: Every combined-average question.
The big group wins the tug of war
A combined average is pulled towards the group (or volume) that is bigger.
Use it when: Checking that your answer lies on the correct side of the midpoint.
Water dilutes, never adds
Plain water adds to the bottom of the fraction but not to the top.
Use it when: Dilution problems.
100% bars forget the totals
Stretch every bar to the same length and you lose how big each one was.
Use it when: Deciding which chart can answer a question.
Median ignores the millionaire
One huge value shifts the mean but not the middle.
Use it when: Choosing the median when there are outliers.

How Quantities Combine: Understanding Data: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Equal-sized groups · 1 mark · MCQExam level

Two sections of 35 students each have average scores of 70 and 80. The average of all 70 students is
(a) 72
(b) 75
(c) 78
(d) 150

Try it first, then show the solutionSolution

The groups are equal, so the simple average works.
$\frac{70 + 80}{2} = 75$ .

Answer: (b) 75

Example 2: Choosing a chart · 1 mark · MCQExam level

To compare what share of their monthly spending three families put on food, the most suitable chart is
(a) a cluster bar chart
(b) a stacked bar chart
(c) a 100% stacked bar chart
(d) a line graph

Try it first, then show the solutionSolution

The question is about shares, not amounts.
A 100% stacked bar shows each family's spending as parts of 100%.

Answer: (c)

Example 3: Unequal mixtures (assertion-reason) · 1 mark · Assertion-reasonExam level

Assertion (A): Mixing 500 mL of 10% lemonade with 200 mL of 20% lemonade does not give 15% lemonade.
Reason (R): In a mixture, each concentration counts in proportion to the quantity it contributes.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Try it first, then show the solutionSolution

$\frac{50 + 40}{700} \approx 12.9%$ , not 15%: A is true.
R is the reason: the 10% lemonade has more volume.

Answer: (a)

Example 4: Weights 1 : 1 : 2 · 2 marks · Very short answerExam level

Scores of 70, 76 and 82 are combined with weights 1, 1 and 2. Find the weighted mean.

Try it first, then show the solutionSolution

$\frac{70 + 76 + 2 \times 82}{1 + 1 + 2} = \frac{310}{4}$ .
$= 77.5$ .

Answer: 77.5

Example 5: Finding an unknown batch · 3 marks · Short answerHarder exam level

Batch A of an alloy weighs 150 kg and is 60% copper. Batch B is 80% copper. When they are melted together, the alloy is 72% copper. Find the weight of Batch B.

Try it first, then show the solutionSolution

Let Batch B weigh x kg. Copper: $0.6 \times 150 + 0.8 x = 90 + 0.8 x$ .
$\frac{90 + 0.8 x}{150 + x} = 0.72$ , so $90 + 0.8 x = 108 + 0.72 x$ .
$0.08 x = 18$ , so $x = 225$ kg.
Check: $\frac{90 + 180}{375} = 0.72$ .

Answer: 225 kg

Example 6: Diluting pani · 3 marks · Short answerHarder exam level

12 L of pani has 9% spice. How much water must be added so that the spice level becomes 6%?

Try it first, then show the solutionSolution

Spice $= 1.08$ L.
$\frac{1.08}{12 + w} = 0.06$ gives $12 + w = 18$ .
$w = 6$ L.

Answer: 6 L

Example 7: Average price of shares · 3 marks · Short answerHarder exam level

Meera holds 40 shares bought at an average of ₹120. She buys 20 more at ₹60 each. Find her new average price per share. How many shares at ₹60 would she have needed to bring the average down to ₹80?

Try it first, then show the solutionSolution

New average $= \frac{40 \times 120 + 20 \times 60}{60} = \frac{6000}{60} = ₹100$ .
For ₹80 with x new shares: $\frac{4800 + 60 x}{40 + x} = 80$ .
$4800 + 60 x = 3200 + 80 x$ , so $x = 80$ .

Answer: ₹100; she would need 80 shares at ₹60.

Example 8: Run rate in the last over · 3 marks · Short answerHarder exam level

After 19 overs a team's run rate is 7. What run rate will it have after 20 overs if it scores 19 runs in the last over? How many runs in the last over would make the final run rate 8?

Try it first, then show the solutionSolution

Runs after 19 overs $= 19 \times 7 = 133$ .
After 20 overs: $\frac{133 + 19}{20} = 7.6$ .
For a run rate of 8: $20 \times 8 = 160$ runs needed, so $160 - 133 = 27$ in the last over.

Answer: 7.6; 27 runs.

Example 9: From table to 100% stacked bars · 5 marks · Long answerHarder exam level

A family's daily water use (litres) was: Year 1: bathing 120, washing 60, cooking 45, cleaning 75. Year 2: bathing 100, washing 50, cooking 50, cleaning 50. (i) Find the percentage shares for both years. (ii) Did bathing use fall? Did its share fall? (iii) Which chart would show both answers?

Try it first, then show the solutionSolution

Year 1 total 300: 40%, 20%, 15%, 25%.
Year 2 total 250: bathing $\frac{100}{250} = 40%$ , others 20% each.
(ii) Bathing use fell from 120 L to 100 L, but its share stayed 40%.
Cooking's share rose (15% to 20%) though its amount rose only by 5 L, because the total fell.
(iii) A stacked bar chart shows the amounts and totals; a 100% stacked bar shows the shares. To see both, use both charts (or the table).

Answer: Shares 40/20/15/25 and 40/20/20/20; bathing amount fell but its share did not; use a stacked chart for amounts and a 100% chart for shares.

Example 10: Which average to report? · 3 marks · Short answerHOTS / competency

Monthly salaries (₹ thousand) in a small office are 15, 18, 18, 20, 22, 25 and 90. A job advert says 'Average salary nearly ₹30,000!' Is this honest? Which measure should a job-seeker look at, and why?

Try it first, then show the solutionSolution

Mean $= \frac{208}{7} \approx 29.7$ thousand, so the advert's number is the mean.
Six of the seven people earn ₹25,000 or less; the one salary of 90 pulls the mean up.
The median (₹20,000) shows what a typical employee earns; the mode (₹18,000) is the most common salary.

Answer: Technically true but misleading; a job-seeker should look at the median, ₹20,000.

Example 11: Doubling the weights · 3 marks · Short answerHOTS / competency

Show that if every weight in a weighted mean is multiplied by the same number k, the weighted mean does not change. Check with values 60, 80 and weights 1, 3, and then weights 2, 6.

Try it first, then show the solutionSolution

$\frac{k w _{1} x _{1} + k w _{2} x _{2}}{k w _{1} + k w _{2}} = \frac{k ( w _{1} x _{1} + w _{2} x _{2} )}{k ( w _{1} + w _{2} )}$ , and k cancels.
Weights 1, 3: $\frac{60 + 240}{4} = 75$ .
Weights 2, 6: $\frac{120 + 480}{8} = 75$ . Same.

Answer: Unchanged (75 in both cases): only the ratio of the weights matters.

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 2 marks.

Run rate in a T20 match

A cricket team's run rate is its runs divided by the overs it has faced. In a 20-over match, a side makes 6 runs off the first over and 12 off the second. After 10 overs its run rate is 7.5, and after 19 overs it is 7.

(i) · 1 mark
What is the run rate after 2 overs?
Answer
Answer: 9
$\frac{6 + 12}{2} = 9$ .
(ii) · 1 mark
The team scores 45 runs in overs 11 to 15. What is the run rate after 15 overs?
Answer
Answer: 8
Runs after 10 overs $= 75$ ; $\frac{75 + 45}{15} = 8$ .
(iii) · 2 marks
How many runs must the team score in the 20th over to finish with a run rate of 7.6? Explain using totals.
Answer
Answer: 19 runs
Runs after 19 overs $= 19 \times 7 = 133$ ; for a run rate of 7.6 after 20 overs it needs $20 \times 7.6 = 152$ (1 mark). So it must score $152 - 133 = 19$ in the last over (1 mark).

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

A class has 24 girls with an average score of 58 and 16 boys with an average score of 63. Find the class average. (2 marks)

Step	Marks
Totals: $24 \times 58 = 1392$ and $16 \times 63 = 1008$	1
Average $= \frac{1392 + 1008}{40} = 60$	1

A pani-puri seller has 12 L of pani with 9% spice. How much plain water must be added to bring the spice level down to 6%? (3 marks)

Step	Marks
Spice in the pani $= 12 \times 0.09 = 1.08$ L; water adds none	1
With w L of water: $\frac{1.08}{12 + w} = 0.06$	1
$12 + w = 18$ , so $w = 6$ L	1

In a fitness test, strength, flexibility and agility are combined in the ratio 4 : 5 : 6. Rashi scores 60, 65 and 70; Keerthi scores 55, 65 and 75. Without calculating, predict who scores more; then find both final scores. (5 marks)

Step	Marks
Prediction: Keerthi is 5 less in strength (weight 4) but 5 more in agility (weight 6), so she should come out ahead	1
Rashi: $\frac{4 \times 60 + 5 \times 65 + 6 \times 70}{15}$	1
$= \frac{240 + 325 + 420}{15} = \frac{985}{15} \approx 65.67$	1
Keerthi: $\frac{220 + 325 + 450}{15} = \frac{995}{15} \approx 66.33$	1
Keerthi scores more, as predicted	1

Where students lose marks in How Quantities Combine: Understanding Data

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

Mistake: Averaging the averages of unequal groups
It looks natural to add and halve.
Wrong working
30 students average 72, 25 students average 76: combined $= \frac{72 + 76}{2} = 74$
Right working
$\frac{30 \times 72 + 25 \times 76}{55} = \frac{4060}{55} \approx 73.8$
Do this: Weight each average by its group size.
Usually costs: Most of the marks.
Mistake: Averaging concentrations without volumes
The quantities mixed are ignored.
Wrong working
500 mL at 10% with 200 mL at 20%: 15%
Right working
$\frac{500 \times 0.1 + 200 \times 0.2}{700} = \frac{90}{700} \approx 12.9%$
Do this: Find the actual amount of substance from each part first.
Usually costs: The answer and the method marks.
Mistake: Weighting marks on different scales
Raw marks out of 50, 60 and 100 are weighted directly.
Wrong working
$\frac{35 \times 3 + 44 \times 4 + 80 \times 5}{12}$ for marks out of 50, 60 and 100
Right working
Convert to percentages first: 70%, 73.3%, 80%, then weight 3 : 4 : 5.
Do this: Put every score on the same scale before weighting.
Usually costs: 1 to 2 marks.
Mistake: Reading amounts from a 100% stacked bar
The bar pieces look like amounts.
Wrong working
Family A's housing piece is longer than Family C's, so Family A spent more on housing.
Right working
A 100% chart only shows that housing is a bigger share of A's spending; the actual amounts need the totals.
Do this: Say 'share' for 100% charts and 'amount' only for ordinary stacked charts.
Usually costs: Interpretation marks.
Mistake: Finding the median without ordering
The middle of the list as given is taken.
Wrong working
Data 25, 15, 90, 18, 20: median $= 90$
Right working
Ordered: 15, 18, 20, 25, 90; median $= 20$
Do this: Always arrange the data first.
Usually costs: 1 mark.

NCERT exercise guide: How Quantities Combine: Understanding Data

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercise Set 10.1

What it tests: Combined averages of sections, mixing fertilisers of different nitrogen content, Śrīdharācārya's gold purity problem, combined rainfall averages, and mixtures of spice drinks.

How to approach it: Turn each part into an amount (size × average, or volume × concentration), add the amounts, divide by the total size.

Like the gold-purity question in Exercise Set 10.1

A goldsmith melts 4 units of gold at 15 varṇa, 8 units at 12 varṇa and 4 units at 10 varṇa. Find the purity of the combined gold in varṇa and carats.

Show the solutionSolution

$\frac{4 \times 15 + 8 \times 12 + 4 \times 10}{4 + 8 + 4} = \frac{60 + 96 + 40}{16} = \frac{196}{16} = 12.25$ varṇa.
Carats: $12.25 \times \frac{24}{16} = 18.375$ .

Answer: 12.25 varṇa, about 18.4 carats

Exercise Set 10.2

What it tests: Choosing the correct expression for an annual score built from marks on different scales combined in a ratio.

How to approach it: Convert each mark to a fraction or percentage of its maximum, multiply by its weight, add, and divide by the sum of the weights.

Like the annual-score question in Exercise Set 10.2

Dev scores 42 out of 60 in internals, 18 out of 20 in a project and 75 out of 100 in the final, combined in the ratio 2 : 1 : 3. Find his annual percentage.

Show the solutionSolution

Percentages: $\frac{42}{60} = 70%$ , $\frac{18}{20} = 90%$ , $75%$ .
$\frac{2 \times 70 + 1 \times 90 + 3 \times 75}{6} = \frac{140 + 90 + 225}{6} = \frac{455}{6} \approx 75.8%$ .

Answer: About 75.8%

Exercise Set 10.3

What it tests: Profit percentage on combined sales, updating a daily average, mixing a salt solution with a sugar solution, diluting pani, fitness scores in a ratio, and restaurant ratings.

How to approach it: Identify what the weights are (amounts of money, days, volumes or chosen importance) and apply the weighted mean.

Like the profit-percentage question in Exercise Set 10.3

A shop sells ₹6,000 of books at 25% profit and ₹4,000 of stationery at 40% profit. What is the profit percentage on total sales?

Show the solutionSolution

Profit: $6000 \times 0.25 = 1500$ and $4000 \times 0.40 = 1600$ .
Total profit $= 3100$ on sales of 10000.
$\frac{3100}{10000} = 31%$ .

Answer: 31%

Exercise Set 10.4

What it tests: Reading a stacked column chart of threatened animal species by class over time.

How to approach it: Read totals from the top of each column and components from the lengths of the pieces; describe trends in both.

Like the stacked-chart questions in Exercise Set 10.4

A stacked column shows a school's library loans: in 2024, 300 fiction, 150 non-fiction and 50 comics (total 500); in 2025, 320 fiction, 240 non-fiction and 40 comics (total 600). Which total rose, by how much, and which category grew most?

Show the solutionSolution

Totals: 500 to 600, a rise of 100 loans.
Changes: fiction +20, non-fiction +90, comics −10.
Non-fiction grew most.

Answer: Total up by 100; non-fiction grew most (+90).

Exercise Set 10.5

What it tests: Completing a 100% stacked bar chart of time spent lying down, sitting and standing by people in different jobs, and inferring an unknown job.

How to approach it: Each bar is 24 hours = 100%; each part is its hours ÷ 24. Compare shapes of bars to infer the type of work.

Like the body-state chart in Exercise Set 10.5

A person spends 8 hours lying down, 12 hours sitting and 4 hours standing or moving each day. Find the shares for a 100% stacked bar, and suggest a job that might fit.

Show the solutionSolution

Total 24 hours.
Lying $\frac{8}{24} \approx 33%$ , sitting $\frac{12}{24} = 50%$ , standing $\frac{4}{24} \approx 17%$ .
Half the day sitting suggests desk work, such as an office worker or a computer programmer.

Answer: About 33%, 50% and 17%; a desk job fits.

Practice questions on How Quantities Combine: Understanding Data

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 · MCQExam level
The mean of 5 numbers is 12. A sixth number, 18, is added. The new mean is
(a) 12
(b) 13
(c) 15
(d) 30
Hint
Old total $= 60$ .
Answer
(b) 13
Q2 · 1 mark · MCQExam level
A 100% stacked bar chart is used to compare
(a) the actual amounts in each category
(b) the totals of each group
(c) the shares (proportions) of each category
(d) changes over time only
Hint
Every bar has the same length.
Answer
(c)
Q3 · 1 mark · Assertion-reasonExam level
Assertion (A): A stacked bar chart can always be drawn from a 100% stacked bar chart of the same data.
Reason (R): A 100% stacked bar chart shows only the shares of each part.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Can you recover the totals?
Answer
(d). Since only shares are shown, the totals are lost, so the stacked chart cannot be rebuilt.
Q4 · 2 marks · Very short answerExam level
20 students have an average weight of 45 kg and 30 students have an average weight of 50 kg. Find the average weight of all 50.
Hint
Add the totals.
Answer
$\frac{900 + 1500}{50} = 48$ kg
Q5 · 2 marks · Very short answerHarder exam level
300 mL of a 5% salt solution and 200 mL of a 10% salt solution are mixed, and then 500 mL of plain water is added. Find the final salt concentration.
Hint
Water adds volume only.
Answer
Salt $= 15 + 20 = 35$ in 1000 mL: 3.5%.
Q6 · 2 marks · Very short answerHarder exam level
A shop sells ₹6,000 of books at 25% profit and ₹4,000 of stationery at 40% profit. Find the overall profit percentage.
Hint
Add actual profits.
Answer
$\frac{1500 + 1600}{10000} = 31%$
Q7 · 3 marks · Short answerHarder exam level
A pool is dug in four sections of lengths 3, 5, 4 and 8 hastas to depths of 6, 4, 5 and 2 hastas. Use Brahmagupta's weighted mean to find its mean depth.
Hint
Lengths are the weights.
Answer
$\frac{3 \times 6 + 5 \times 4 + 4 \times 5 + 8 \times 2}{20} = \frac{74}{20} = 3.7$ hastas
Q8 · 2 marks · Very short answerHarder exam level
Three families have 4, 5 and 3 members with average ages 30, 26 and 38 years. Find the average age of all 12 people.
Hint
Total ages first.
Answer
$\frac{120 + 130 + 114}{12} = \frac{364}{12} \approx 30.3$ years
Q9 · 3 marks · Short answerHarder exam level
Find the mean, median and mode of: 7, 3, 9, 3, 5, 8, 3, 10. Which best represents the data?
Hint
Order first; n is even.
Answer
Ordered: 3, 3, 3, 5, 7, 8, 9, 10. Mean $= \frac{48}{8} = 6$ ; median $= \frac{5 + 7}{2} = 6$ ; mode $= 3$ . Mean and median agree at 6, so 6 is a good summary; the mode only shows the most frequent value.
Q10 · 5 marks · Long answerHarder exam level
Weekly study hours: Arun: Maths 6, Science 4, Languages 2; Bina: 3, 3, 3; Charu: 5, 5, 10. (i) Who studies most in total? (ii) Who gives the largest share of their time to Maths? (iii) Which chart answers (i) at a glance, and which answers (ii)?
Hint
Totals 12, 9, 20.
Answer
(i) Charu (20 hours). (ii) Maths shares: Arun 50%, Bina 33%, Charu 25%, so Arun. (iii) A stacked bar chart for totals; a 100% stacked bar chart for shares.
Q11 · 3 marks · Short answerHOTS / competency
Shop X has an average rating of 4.6 from 50 reviews; Shop Y has 4.8 from 5 reviews. Find the combined average of all 55 reviews. Which shop's rating would you trust more, and why?
Hint
Weights are the numbers of reviews.
Answer
$\frac{50 \times 4.6 + 5 \times 4.8}{55} = \frac{254}{55} \approx 4.62$ . Shop X's rating is based on ten times as many reviews, so it is more reliable; Shop Y's high average could change a lot with a few more reviews.
Q12 · 3 marks · Short answerHOTS / competency
Rashi scores 60, 65, 70 and Keerthi 55, 65, 75 in strength, flexibility and agility, combined in the ratio 4 : 5 : 6. Without calculating, decide who scores more. Then change the ratio to 6 : 5 : 4. Who scores more now? Explain.
Hint
Look at where each girl is stronger and how heavily it is weighted.
Answer
With 4 : 5 : 6, agility counts most and Keerthi leads there, so Keerthi scores more (66.33 against 65.67). With 6 : 5 : 4, strength counts most and Rashi leads there, so Rashi scores more: $\frac{360 + 325 + 280}{15} \approx 64.33$ against $\frac{330 + 325 + 300}{15} \approx 63.67$ .

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.When is the combined average equal to the average of the group averages?Show answer
When the groups are the same size.
2.Write the weighted mean formula.Show answer
$\frac{w _{1} x _{1} + \dots + w _{n} x _{n}}{w _{1} + \dots + w _{n}}$
3.Who wrote the weighted mean formula, and for what?Show answer
Brahmagupta (c. 628 CE), for the mean depth of an irregular pit.
4.What did Śrīdharācārya use the weighted mean for?Show answer
Finding the purity (varṇa) of gold melted together.
5.What happens to a weighted mean if all weights are doubled?Show answer
Nothing; the 2 cancels from numerator and denominator.
6.What does the full length of a stacked bar show?Show answer
The total of that group.
7.What does a 100% stacked bar chart compare?Show answer
Proportions (shares), not actual amounts.
8.Can you make a stacked bar chart from a 100% stacked bar chart?Show answer
No; the totals are lost.
9.Which average is not affected by a few extreme values?Show answer
The median.
10.What is the mode?Show answer
The most frequent value.

Best order to study it

Work through the badminton-height example to see why averages cannot simply be averaged.
Practise mixture problems, including dilution with water (Exercise Set 10.1).
Learn the weighted mean formula and use it with marks and ratings (Exercise Sets 10.2 and 10.3).
Practise updating an average when values are added (run rates, test scores).
Read stacked bar charts and 100% stacked bar charts, and convert a table into a 100% chart (Exercise Sets 10.4 and 10.5).
Revise mean, median and mode and when to use each, then try the end-of-chapter questions.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you find the combined average of 30 values averaging 72 and 25 values averaging 76?
Can you find the concentration when 500 mL at 10% is mixed with 200 mL at 20%?
Can you find how much water dilutes 10 L at 8% to 6%?
Can you combine three marks in the ratio 3 : 2 : 5?
Can you update a run rate after one more over?
Can you explain why doubling all weights does not change the weighted mean?
Can you say which chart shows totals and which shows shares?
Can you make a 100% stacked bar from a small table?
Can you find the median of an even number of values?
Can you choose between mean, median and mode for a given purpose?

How Quantities Combine: Understanding Data: frequently asked questions

Is this chapter CBSE's 'Statistics' for Class 9?

Yes. CBSE's 2026-27 syllabus lists Statistics (graphical representation of data and measures of central tendency) in Unit VI, with competencies on weighted averages, stacked and 100% stacked bar graphs, and using mean, median and mode for decisions. NCERT teaches it as 'How Quantities Combine: Understanding Data' in Ganita Manjari Part II.

Are histograms and frequency polygons in this chapter?

No. The new chapter's graphs are cluster, stacked and 100% stacked bar charts (with pie charts as a comparison).

Do I need to remember Brahmagupta and Śrīdharācārya?

Know that Brahmagupta wrote the weighted mean formula (for the mean depth of a pit) and that Śrīdharācārya used it for gold purity in varṇa. Exam questions mainly ask you to use the formula.

What is the difference between a pie chart and a 100% stacked bar chart?

Both show shares of a whole. A pie chart splits 360° by angle; a 100% stacked bar splits a fixed length. Comparing lengths across several bars is usually easier than comparing angles in several pies.

Are the projects at the end of the chapter assessed?

They suit internal assessment (portfolio and activities). In written exams, expect calculations and chart-reading like the solved questions on this page.

Based on: CBSE Mathematics syllabus, Class IX 2026-27; NCERT Ganita Manjari, Part II, Mathematics Textbook for Grade 9 (First Edition 2026)