How Quantities Combine: Understanding Data: Class 9 Ganita Manjari · Chapter 10

Ganita Manjari Part 2 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 13 Oct 2026 · Updated 13 Oct 2026

60 min
33 practice questions
19 MCQs

What you will learn in How Quantities Combine: Understanding Data

Explain why the mean of group means can mislead when groups differ in size, and find the correct combined mean
Use mean × number of values = total to combine groups and to add or remove values
Find the concentration of a mixture, and the amount needed to reach a target concentration
Calculate a weighted mean whose weights stand for size or for importance, such as marks combined in a ratio
Choose between clustered, stacked and 100% stacked bar charts to suit the question being asked
Convert amounts to percentages to draw a 100% stacked bar chart, and judge which conclusions it supports

Original companion text aligned to NCERT Mathematics Class 9, Chapter 10, pp. 8–36. Not affiliated with or endorsed by NCERT.

10.1 Combining Things

Section 9A scored a mean of 68 in a class test and Section 9B scored 74. Is the mean for the whole of Class 9 simply 71, halfway between? The two averages cannot tell you. The missing fact is how many students sat in each section.

Recall that the mean (माध्य), or average, of a set of values is their sum divided by how many values there are. In this chapter we first learn to combine data correctly, which leads to the weighted mean. Then we turn to charts that show how a whole splits into parts.

10.1.1 Average of Averages

Our school's Eco Club collects plastic waste for recycling. It has two groups: 7 members from Class 10 and 3 members from Class 6. At the end of the term, each member's collection was weighed.

Example 10.1NCERT Example 1 · p. 8

The Eco Club's average

The 7 Class 10 members collected 32, 28, 35, 30, 26, 33 and 26 kg. The 3 Class 6 members collected 12, 15 and 18 kg. What is the mean amount collected per member for the whole club?

Try it yourself first. Then reveal the solution one step at a time.

3/3Pooling all 10 members gives 25.5 kg, pulled towards the 7 Class 10 members.

Fig. 10.1The true mean sits nearer the bigger group.

A number line runs from 10 to 40 kilograms. Seven circles mark the Class 10 members between 26 and 35, and three squares mark the Class 6 members at 12, 15 and 18. In the first step dashed lines mark the group means at 15 and 30. In the second step a diamond marks 22.5, halfway between the two means, labelled mean of means. In the third step a solid line and a filled triangle mark the true mean at 25.5, closer to the larger Class 10 group, while the halfway marker fades.

The repair uses a fact you already know. If a group of $n$ values has mean $a$ , the group's total is $a \times n$ . So the Class 10 total is $30 \times 7 = 210$ kg and the Class 6 total is $15 \times 3 = 45$ kg. Add the totals and divide by the number of members: $255 \div 10 = 25.5$ kg. We never needed a single member's value.

In general, suppose one collection has $n$ values with mean $a$ , and another has $m$ values with mean $b$ . Their totals are $a n$ and $b m$ , so the mean of all $n + m$ values is

Mean of two combined groups

\overset{ˉ}{x} = \frac{a n + b m}{n + m}

(10.1)

where

$a, n$: mean and size of the first group
$b, m$: mean and size of the second group

Notice how the bigger group pulls the combined mean towards its own mean. The Class 10 group has 7 of the 10 members, so 25.5 kg lies much nearer 30 kg than 15 kg.

Example 10.2NCERT Example 2 · p. 9

When the shortcut does work

Bhola runs a samosa stall six days a week. The table shows how many plates he sold each day for three weeks. Find the mean number of plates sold per day.

Table 10.1Plates sold at Bhola's stall

Week	Mon	Tue	Wed	Thu	Fri	Sat	Weekly mean
Week 1	42	45	48	50	55	60	50
Week 2	48	50	52	55	60	65	55
Week 3	50	52	54	58	62	66	57

Try it yourself first. Then reveal the solution one step at a time.

Why did the shortcut work for Bhola but not for the Eco Club? Take three collections with means $a$ , $b$ and $c$ and sizes $p$ , $q$ and $r$ . Their combined mean is

Mean of three combined groups

\overset{ˉ}{x} = \frac{a p + b q + c r}{p + q + r}

(10.2)

where

$a, b, c$: the means of the three collections
$p, q, r$: their sizes

So when every collection has the same size, the mean of the means is the true mean. When the sizes differ, as in the Eco Club, each mean must be multiplied by its group size before you add.

Quick check

1 of 3

Section 9A has 40 students with a mean score of 62. Section 9B has 20 students with a mean score of 74. What is the mean score of all 60 students?

68666470

Show answer

Answer (B) 66

The totals are

40 \times 62 = 2480

and

20 \times 74 = 1480

, so the mean is

3960 \div 60 = 66

. The tempting 68 is the halfway point, which ignores that 9A has twice as many students.

10.1.2 Mixtures

Pour two drinks together and ask how sweet the mixture is. It is the same question in a new costume. The concentration (सांद्रता) of sugar in a drink is the fraction, or percentage, of the drink that is sugar. For any mixture,

c = \frac{s}{q}

where

$c$: concentration of the mixture
$s$: total amount of sugar (or salt, or fat …)
$q$: total amount of the mixture

Example 10.3NCERT Example 3 · p. 11

Two equal jugs

Jug A holds 1 litre of aam panna that is 8% sugar. Jug B holds 1 litre that is 14% sugar. Both are poured into one big jar. What percentage of the mixture is sugar?

Try it yourself first. Then reveal the solution one step at a time.

Equal jugs behave like Bhola's equal weeks: each jug gets an equal say. Now change the amounts.

Example 10.4NCERT Example 4 · p. 11

Unequal jugs

This time Jug A holds 1.5 litres (8% sugar) and Jug B holds 0.5 litre (14% sugar). Before calculating, guess: is the mixture 11% sugar, more than 11% or less?

Try it yourself first. Then reveal the solution one step at a time.

2/2Unequal amounts: the mixture is 9.5%, nearer the bigger glass's 8%.

Fig. 10.2The bigger glass drags the mixture's concentration towards its own.

Two glasses of drink stand on the left, the first 8% sugar and the second 14% sugar, with the sugar shown as a band at the bottom of each. A tall jar on the right holds the 2 litre mixture. Below, a scale runs from 8% to 14% with a pointer. In the first step both glasses hold 1 litre and the pointer is at 11%, halfway. In the second step the first glass grows to 1.5 litres and the second shrinks to 0.5 litre; the mixture's sugar band gets thinner and the pointer slides to 9.5%, nearer 8%.

Think of each jug as tugging the mixture towards its own sweetness, and tugging harder the more of it you pour. Group sizes did the same job in Example 10.1. We can even run the idea backwards to find a missing amount.

Example 10.5NCERT Example 5 · p. 12

Finding a missing amount

A dairy blends three lots of milk: 40 L with 6% fat, 60 L with 3% fat, and some more milk with 5% fat. The blend must be 4.5% fat. How many litres of the 5% milk are needed?

Try it yourself first. Then reveal the solution one step at a time.

Every calculation so far has the same shape: multiply each value by how much of it there is, add up, and divide by the total amount. Call the values $x_{1}, x_{2}, \dots, x_{n}$ and the amounts $w_{1}, w_{2}, \dots, w_{n}$ . Then the weighted mean, written $\overset{ˉ}{x}$ , is

Weighted mean

\frac{w_{1} x_{1} + \dots + w_{n} x_{n}}{w_{1} + \dots + w_{n}}

where

$x_{1}, \dots, x_{n}$: the values: marks, concentrations, group means …
$w_{1}, \dots, w_{n}$: their weights: how much of each, or how important each is

Definition

Weighted mean

Multiply each value by its weight, add the products, and divide by the sum of the weights. It is also called the weighted average.

ExampleMarks of 90 and 60 with weights 1 and 2: $(90 + 120) \div 3 = 70$ .

A weight (भार) need not be a mass in kilograms. It is any number that says how much a value should count: a group's size, a jug's volume, or how important a test is. If every weight is the same, the weighted-mean formula becomes the ordinary mean.

Fig. 10.3Change a value or its weight and watch the balance point, the weighted mean, move. The dashed line is the simple mean.

Quick check

1 of 3

300 mL of a 10% salt solution is mixed with 100 mL of a 30% salt solution. What is the concentration of the mixture?

20%25%15%40%

Show answer

Answer (C) 15%

Salt:

30 + 30 = 60

mL in 400 mL, and

60 \div 400 = 15 %

. 20% would be right only for equal volumes; 40% wrongly adds the two percentages.

Practice · 5 questions

Practice 10.1: questions with answers

NumericalEasyNCERT Exercise Set 10.1, Q1
Coaching batch P has 36 students whose mean mock-test score is 68. Batch Q has 24 students whose mean is 78. Find the mean score of all 60 students.
Show answer
Answer 72
$(36 \times 68 + 24 \times 78) \div 60 = (2448 + 1872) \div 60 = 4320 \div 60 = 72$ .
NumericalEasyNCERT Exercise Set 10.1, Q2
A gardener mixes equal amounts of three potting mixes. The fraction of compost in them is $\frac{1}{6}$ , $\frac{1}{4}$ and $\frac{1}{12}$ . What fraction of the final mix is compost?
Show answer
Answer 0.1667
Equal amounts, so the plain mean works: $(\frac{2}{12} + \frac{3}{12} + \frac{1}{12}) \div 3 = \frac{6}{12} \div 3 = \frac{1}{6}$ .
3
NumericalMediumNCERT Exercise Set 10.1, Q3
(In the style of Śrīdharācārya, c. 750 CE.) A goldsmith melts together 4 units of gold of 15 varṇa, 10 units of 12 varṇa and 6 units of 13 varṇa. Pure gold is 16 varṇa.
1. What is the purity of the melted gold, in varṇa?
  varṇa
  Show answer
  Answer 12.9 varṇa
  $(4 \times 15 + 10 \times 12 + 6 \times 13) \div 20 = 258 \div 20 = 12.9$ varṇa.
2. Today purity is measured in karats, and pure gold is 24 karat. Convert your answer to karats.
  karat
  Show answer
  Answer 19.35 karat
  16 varṇa = 24 karat, so multiply by $\frac{24}{16} = 1.5$ : $12.9 \times 1.5 = 19.35$ karat.
4
Short answerMediumNCERT Exercise Set 10.1, Q4
A village dairy sold a mean of 42 L of milk a day in February 2026, 38 L a day in March and 35 L a day in April.
1. Write an expression for the mean amount sold per day over the three months.
  Show solution
  February 2026 has 28 days, March 31 and April 30. Weight each monthly mean by its number of days:
  $\overset{ˉ}{x} = (42 \times 28 + 38 \times 31 + 35 \times 30) \div (28 + 31 + 30)$
2. Evaluate it, to one decimal place.
  L
  Show answer
  Answer 38.2 L
  $(1176 + 1178 + 1050) \div 89 = 3404 \div 89 \approx 38.2$ L a day.
5
NumericalMediumNCERT Exercise Set 10.1, Q5
A sweet shop combines three pans of sugar syrup.
1. 1 L at 20% sugar, 2 L at 30% and 3 L at 40%. Find the concentration of the mixture.
  %
  Show answer
  Answer 33.33 %
  Sugar: $0.2 + 0.6 + 1.2 = 2$ L in 6 L, so $2 \div 6 \approx 33.3 %$ .
2. Now the volumes are swapped: 3 L at 20%, 2 L at 30% and 1 L at 40%.
  %
  Show answer
  Answer 26.67 %
  Sugar: $0.6 + 0.6 + 0.4 = 1.6$ L in 6 L, so $1.6 \div 6 \approx 26.7 %$ .
3. Why do the answers differ, although the same three concentrations were used?
  Show solution
  In (a) most of the syrup is the 40% kind, so the mixture is pulled up; in (b) most is the 20% kind, so it is pulled down. Only equal volumes would give the simple mean, 30%.

10.1.3 Custom Weights

So far the weights were sizes: members, litres, days. Weights can also express importance. A school may decide that the final exam should count for more than a class test, because it covers the whole year.

Example 10.6NCERT Example 6 · p. 14

Marks that matter differently

Ananya scored 90% in her Science practicals, 70% in the periodic tests and 62% in the final exam. The school combines these in the ratio $1 : 3 : 6$ . What is her final percentage?

Try it yourself first. Then reveal the solution one step at a time.

2/2Copies in the ratio 1 : 3 : 6: the mean of the ten tiles is 67.2.

Fig. 10.4A ratio of weights works like making copies of each value.

A row of ten tile spaces holds Ananya's scores: a 90 for the practical, 70 for tests and 62 for the final exam. In the first step there is one tile of each and the others are faint outlines; a pointer on the number line below marks the plain mean, 74. In the second step the outlines fill in: one 90, three 70s and six 62s, labelled times 1, times 3 and times 6, and the pointer slides left to 67.2, the mean of all ten tiles.

Practice · 1 question

Practice 10.2: questions with answers

1
Quick answerMediumNCERT Exercise Set 10.2, Q1
Karan's Sanskrit marks are 18 out of 25 in tests, 36 out of 40 in a project and 54 out of 80 in the final exam. They are combined in the ratio $2 : 3 : 5$ . Which of the expressions below give his final score as a percentage? Choose all that apply.
- (i) $(72 \times 2 + 90 \times 3 + 67.5 \times 5) \div 10$
- (ii) $(18 \times 2 + 36 \times 3 + 54 \times 5) \div 10$
- (iii) $(\frac{18}{25} \times 2 + \frac{36}{40} \times 3 + \frac{54}{80} \times 5) \div 10 \times 100$
- (iv) $(\frac{18}{100} \times 2 + \frac{36}{100} \times 3 + \frac{54}{100} \times 5) \div 10 \times 100$
(i)(ii)(iii)(iv)
Select all that apply
Show answer
Answer
(A) (i)
(C) (iii)
Each mark must first become a percentage (i) or a fraction of its total (iii); both give 75.15%. Expression (ii) mixes raw marks out of 25, 40 and 80, and (iv) treats every mark as if it were out of 100. Both give 41.4, which is not his score.

Group averages, mixtures, marks and ratings all lead to the same tool. Whenever some parts deserve more say than others, give them bigger weights.

Quick check

1 of 3

If every weight is equal, the weighted mean equals the ordinary mean.

TrueFalse

Show answer

Answer True

With every weight equal to

w

, the top is

w (x_{1} + \dots + x_{n})

and the bottom is

n w

. The

w

cancels, leaving the ordinary mean.

Practice · 8 questions

Practice 10.3: questions with answers

NumericalEasyNCERT Exercise Set 10.3, Q1
A juice stall's sales in May were ₹12,000 from juices, with a 25% profit, and ₹3,000 from snacks, with a 40% profit. What was the profit as a percentage of total sales?
%
Show answer
Answer 28 %
Profit $= 0.25 \times 12000 + 0.40 \times 3000 = 3000 + 1200 = 4200$ rupees out of 15,000 rupees of sales, which is 28%. It is a weighted mean of 25% and 40%, weighted by sales.
NumericalEasyNCERT Exercise Set 10.3, Q2
A delivery rider averaged 38.5 deliveries a day over 24 days. On day 25 she made 76 deliveries. Guess first: will her mean rise a lot or a little? Then find her mean over the 25 days.
Show answer
Answer 40
Total for 24 days $= 38.5 \times 24 = 924$ . New mean $= (924 + 76) \div 25 = 1000 \div 25 = 40$ . One big day among 25 moves the mean only a little.
Quick answerMediumNCERT Exercise Set 10.3, Q3
800 mL of a solution with 6% salt is mixed with 400 mL of a solution with 9% sugar. Find the percentage of salt and the percentage of sugar in the mixed liquid.
salt 6%, sugar 9%salt 7.5%, sugar 7.5%salt 4%, sugar 3%salt 5%, sugar 4.5%
Show answer
Answer (C) salt 4%, sugar 3%
Salt: 48 mL in 1200 mL is 4%. Sugar: 36 mL in 1200 mL is 3%. Averaging 6% and 9% makes no sense here: they measure different substances.
NumericalMediumNCERT Exercise Set 10.3, Q4
A 12 L can of rose sherbet is 15% sugar. How many litres of water must be added to bring the sugar down to two-thirds of its present concentration?
L
Show answer
Answer 6 L
Target: $\frac{2}{3} \times 15 % = 10 %$ . The sugar stays $0.15 \times 12 = 1.8$ L, so $1.8 \div (12 + w) = 0.10$ . Then $12 + w = 18$ and $w = 6$ L.
5
Short answerMediumNCERT Exercise Set 10.3, Q5
In a sports trial the sprint, long jump and endurance run are combined in the ratio $2 : 3 : 5$ . Meera scored 80, 70 and 60. Nikhil scored 60, 70 and 80.
1. Without calculating, who has the higher final score? Why?
  Show solution
  Nikhil. They have the same three scores, but Nikhil's best score, 80, is in the endurance run, which carries the largest weight.
2. Find both final scores.
  Show solution
  AnswerMeera 67, Nikhil 73.
  Meera: $(160 + 210 + 300) \div 10 = 67$ .
  Nikhil: $(120 + 210 + 400) \div 10 = 73$ .
NumericalHardNCERT Exercise Set 10.3, Q6
Twelve guests at a homestay rated three things from 1 to 5 stars. The table shows how many guests gave each rating. The owner weights cleanliness : food : location in the ratio $5 : 3 : 2$ . Find the overall rating.
Table 10.2Number of guests giving each rating
5★ 4★ 3★ 2★ 1★
Cleanliness 6 4 2 0 0
Food 3 5 3 1 0
Location 2 2 4 3 1
stars
Show answer
Answer 3.93 stars
Row totals of stars: cleanliness $30 + 16 + 6 = 52$ , food $15 + 20 + 9 + 2 = 46$ , location $10 + 8 + 12 + 6 + 1 = 37$ , each from 12 guests. Overall $= (5 \times 52 + 3 \times 46 + 2 \times 37) \div (12 \times 10) = 472 \div 120 \approx 3.93$ stars.
7
NumericalHardNCERT Exercise Set 10.3, Q7
A sanctuary has 50 spotted deer. The stags (males) have a mean weight of 52 kg, the does (females) 40 kg, and the whole herd 44.8 kg.
1. Without calculating, are there more stags or more does? Explain.
  Show solution
  More does. The herd's mean, 44.8 kg, is nearer the does' mean (40 kg) than the stags' (52 kg); equal numbers would give the midpoint, 46 kg.
2. How many stags are there?
  Show answer
  Answer 20
  $52 s + 40 (50 - s) = 44.8 \times 50 = 2240$ , so $12 s + 2000 = 2240$ and $s = 20$ .
3. A doe weighing 46 kg joins the herd. What is the new mean weight of the does, to two decimal places?
  kg
  Show answer
  Answer 40.19 kg
  There are 30 does with total $30 \times 40 = 1200$ kg. New mean $= 1246 \div 31 \approx 40.19$ kg.
4. Instead, two stags weighing 58 kg and 50 kg move to another sanctuary. What is the new mean weight of the stags, to two decimal places?
  kg
  Show answer
  Answer 51.78 kg
  The stags' total was $20 \times 52 = 1040$ kg. Now $(1040 - 58 - 50) \div 18 = 932 \div 18 \approx 51.78$ kg: the two who left were heavier than average, so the mean drops.
8
NumericalHardNCERT Exercise Set 10.3, Q8
A pickle maker has strong brine that is 25% salt and weak brine that is 5% salt.
1. What is the salt concentration if she mixes 1 L of strong brine with 3 L of weak brine?
  %
  Show answer
  Answer 10 %
  $(0.25 + 0.15) \div 4 = 0.10$ , which is 10%.
2. Can any mixture of these two brines be 30% salt? Why or why not?
  Show solution
  No. The mixture's concentration is a weighted mean of 25% and 5%, so it must lie between them.
3. How much weak brine should she mix with 2 L of strong brine to get 10% salt?
  L
  Show answer
  Answer 6 L
  $0.5 + 0.05 w = 0.10 (2 + w)$ , so $0.3 = 0.05 w$ and $w = 6$ L.
4. Plain water has 0% salt. How much water should she add to 1 L of strong brine to make 5% brine?
  L
  Show answer
  Answer 4 L
  $0.25 \div (1 + w) = 0.05$ , so $1 + w = 5$ and $w = 4$ L.

	5★	4★	3★	2★	1★
Cleanliness	6	4	2	0	0
Food	3	5	3	1	0
Location	2	2	4	3	1

10.2 Visualising and Interpreting Data

A table is exact but slow to read. A chart trades a little precision for speed, because the eye compares lengths much faster than it compares digits. Every chart, though, makes some comparisons easy and others hard. The skill is to pick the chart that suits your question.

10.2.1 Stacking Columns

A residents' association asked three households to measure how many litres of water they use on a typical day, sorted by purpose.

Table 10.3Daily water use (litres) of three households

	Bathing	Toilet	Laundry	Kitchen	Garden
Desai	180	120	90	60	50
Joseph	120	100	60	100	20
Rao	70	60	50	45	25

A clustered bar chart draws one bar per number and stands related bars side by side, every bar starting at 0. This table can be clustered in two ways: by purpose, comparing the households within each use, or by household, comparing one home's uses. Fig. 10.5 shows both.

1/2Clusters by purpose: compare the three households within each use.

Fig. 10.5The same fifteen numbers, clustered two ways.NCERT Fig. 10.1 · p. 19

A column chart of daily litres from 0 to 200. In the first step there are five clusters, one per purpose (bathing, toilet, laundry, kitchen, garden), each with three columns marked D, J and R for the Desai, Joseph and Rao households. In the second step the same fifteen columns slide into three clusters, one per household, and each column is now marked B, T, L, K or G for its purpose and coloured by purpose.

Now try some harder questions. Which household uses the least water altogether? About how many litres does the Joseph household use in a day? Roughly what percentage of the Desai household's water goes on bathing? Neither cluster chart answers these quickly, because neither shows a household's total.

The fix is to lay each household's bars end to end. One long bar then shows the total, and its pieces show the parts. This is a stacked bar chart.

2/2End to end, the five bars make one bar for the household's total.

Fig. 10.6Joining a household's bars end to end shows its total.NCERT Fig. 10.2 · p. 20

A horizontal chart with a litre scale from 0 to 500. In the first step each household (Desai, Joseph, Rao) has five thin bars, one per purpose, all starting at 0. In the second step the thin bars slide end to end and thicken into one stacked bar per household, in the order bathing, toilet, laundry, kitchen, garden, labelled B, T, L, K and G, with the totals 500, 400 and 250 litres at the ends.

Definition

Stacked bar chart

A bar chart in which the parts of each whole are joined end to end, always in the same order, so that the full bar shows the total.

ExampleOne bar per household, split into bathing, toilet, laundry, kitchen and garden water.

The totals now jump out: Desai 500 L, Joseph 400 L, Rao 250 L. Bathing takes 180 L of the Desai household's 500 L, which is $\frac{180}{500} = 36 %$ . The price is that only the first piece of each bar starts at 0. To read the Joseph household's kitchen use, you must subtract where its piece starts (280 L) from where it ends (380 L). Pieces that start in different places are harder to compare than columns on a common base.

In short: if your question is about the wholes, stack the bars; if it is about individual parts, cluster them.

Quick check

1 of 3

In Fig. 10.6, the Rao household's laundry piece runs from 130 L to 180 L. How many litres of water does it stand for?

Show answer

Answer 50 L

Read the end minus the start:

180 - 130 = 50

Fig. 10.7For Practice 10.4, Question 1: vehicles registered in a town (thousands).

Four stacked columns for 2010, 2015, 2020 and 2025 on a scale of 0 to 70 thousand vehicles. Each column is split, from the bottom, into two-wheelers, cars, autos, and buses and trucks. The totals printed above the columns are 28, 38, 50 and 63 thousand. Two-wheelers and cars grow a lot; autos and buses barely change.

Practice · 2 questions

Practice 10.4: questions with answers

1
Short answerEasyNCERT Exercise Set 10.4, Q1
The stacked column chart in Fig. 10.7 shows how many vehicles (in thousands) were registered in a town, by type.
1. What does the number 63 above the 2025 column stand for?
  Show solution
  The total number of vehicles of all four types registered in 2025: about 63 thousand.
2. About how many cars were registered in 2020, in thousands?
  thousand
  Show answer
  Answer 14 thousand
  The car piece of the 2020 column runs from 30 to 44 thousand, so about $44 - 30 = 14$ thousand.
3. Which types of vehicle changed least between 2010 and 2025?
  Show solution
  Autos (about 3 → 5 thousand) and buses and trucks (about 1 → 2 thousand). Two-wheelers grew by about 18 thousand and cars by about 14 thousand.
2
DiagramMediumNCERT Exercise Set 10.4, Q2
Athletes from one state won these medals at the national school games.
Table 10.4Medals won
Gold Silver Bronze
Boys 24 31 35
Girls 28 22 30
1. Sketch a stacked bar chart with one bar for boys and one for girls, split by medal type.
  Show solution
  Boys: total 90, pieces gold 0–24, silver 24–55, bronze 55–90. Girls: total 80, pieces gold 0–28, silver 28–50, bronze 50–80. Use one scale from 0 to 100 for both bars.
2. Sketch a second stacked bar chart with one bar per medal type, split into boys and girls.
  Show solution
  Gold: total 52 (boys 0–24, girls 24–52). Silver: total 53 (boys 0–31, girls 31–53). Bronze: total 65 (boys 0–35, girls 35–65).
3. Which chart shows at a glance who won more medals in all? Which shows the most common type of medal?
  Show solution
  Chart (a): the boys' bar is longer (90 against 80). Chart (b): the bronze bar is longest (65).

	Gold	Silver	Bronze
Boys	24	31	35
Girls	28	22	30

10.2.2 An Alternative to Pie Chart

Here is a question of a different kind: which household uses the smallest share of its water in the kitchen? The stacked chart shows litres, not shares. The Desai household uses more kitchen water (60 L) than the Rao household (45 L), but its total is twice as large, so its kitchen share might still be smaller.

In Class 8 you met the pie chart, which shows each part as a slice of a whole circle. Fig. 10.8 draws one pie per household.

2/2Kitchen shares: Desai 12%, Joseph 25%, Rao 18%.

Fig. 10.8One pie per household: each slice is a share of that household's water.NCERT Fig. 10.3 · p. 22

Three pie charts side by side for the Desai, Joseph and Rao households, each split into bathing, toilet, laundry, kitchen and garden slices in the same order, starting at the top and going clockwise. In the second step every slice except the kitchen slice fades, and labels under the pies give the kitchen shares: 12% for Desai, 25% for Joseph and 18% for Rao.

Pies do show shares. But try ranking the three kitchen slices: they sit at different angles in different circles, and our eyes judge angles poorly. We judge lengths much better. So stretch every household's stacked bar to the same length; then the length of each piece is its share of the whole.

2/2Each bar stretched to 100%: the pieces now show percentages.

Fig. 10.9Stretch every bar to the same length and each piece becomes a share.NCERT Fig. 10.4 · p. 23

Three horizontal stacked bars for the Desai, Joseph and Rao households. In the first step they are drawn in litres on a scale from 0 to 500, so their lengths differ (500, 400 and 250 litres). In the second step each bar stretches to the same full length and the scale changes to 0% to 100%; the pieces now show shares, with the kitchen pieces labelled 12%, 25% and 18%.

Definition

100% stacked bar chart

A stacked bar chart in which every bar has the same length, standing for 100% of its own total, so each piece shows a percentage share rather than an amount.

ExampleDesai household: bathing 36%, toilet 24%, laundry 18%, kitchen 12%, garden 10%.

A 100% stacked bar chart is like a pie chart unrolled into a strip: where the pie shares out 360°, the bar shares out one fixed length. The answer to our question is now visible. The Desai household's kitchen piece is the shortest (12%), even though it uses more kitchen litres than the Rao household (18%).

Some comparisons stay awkward, though. The Desai and Rao toilet pieces are both exactly 24%, but one starts at 36% and the other at 28%. Pieces of equal length in different positions are hard to match at a glance.

From amounts to shares

To draw a 100% stacked bar chart from amounts, turn every amount into a percentage of its own bar's total.

Example 10.7NCERT Example 7 · p. 23

A family's travel, then and now

The table shows how many kilometres the Sharma family travelled in one month, by each mode, in 2016 and in 2026. Make a 100% stacked bar chart.

Table 10.5Kilometres travelled in a month

	Bus	Metro	Auto	Two-wheeler
2016	450	50	150	350
2026	250	500	150	350

Try it yourself first. Then reveal the solution one step at a time.

2/2Shares: the two-wheeler piece shrinks from 35% to 28%.

Fig. 10.10Kilometres become shares when each bar is stretched to 100%.NCERT p. 24

Two horizontal stacked bars for 2016 and 2026, split into bus, metro, auto and two-wheeler. In the first step they are drawn in kilometres on a scale from 0 to 1250, with the 2026 bar longer (1250 km against 1000 km) and the kilometres written in the pieces. In the second step both bars stretch to the same length, the scale becomes 0% to 100%, and the pieces show percentages: 45, 15 and 35 for 2016 (metro 5% is too thin to label) and 20, 40, 12 and 28 for 2026.

Look closely at the two-wheeler. The family rode it 350 km in both years, yet its share fell from 35% to 28%. A share can shrink while the amount stays the same, simply because the total grew. Metro travel, almost nothing in 2016, became the biggest piece by 2026.

What a 100% chart can and cannot tell

Because every bar in a 100% chart has the same length, the chart hides the totals. That limits what you can conclude from it.

Example 10.8NCERT Example 8 · p. 24

Two stalls at a mela

Asha and Vikram run snack stalls at the same mela. A 100% stacked bar chart shows what share of each stall's plates were pani puri, chaat and kulfi: Asha 45%, 35% and 20%; Vikram 30%, 30% and 40%. Which of these follow from the chart alone?

(i) At Vikram's stall, kulfi sold more plates than chaat.
(ii) Asha sold more plates of pani puri than Vikram did.
(iii) The best-selling snack is different at the two stalls.
(iv) Vikram's stall sold more plates altogether.

Try it yourself first. Then reveal the solution one step at a time.

3/3If Asha sold 120 and Vikram 300, Vikram sold more pani puri: 90 against 54.

Fig. 10.11The same shares can come from very different numbers of plates.NCERT p. 25

Two horizontal bars for Asha's and Vikram's stalls, split into pani puri, chaat and kulfi. In the first step both bars have the same length and show shares: 45, 35 and 20 percent for Asha and 30, 30 and 40 percent for Vikram. In the second step the bars shrink to a plate scale with both stalls selling 200 plates: Asha 90, 70 and 40, Vikram 60, 60 and 80. In the third step Asha sells 120 plates (54, 42, 24) and Vikram 300 (90, 90, 120), so now Vikram sold more pani puri.

Whose typical day?

Charts of how people spend their 24 hours are popular with survey planners. India's National Statistical Office runs a Time Use Survey in which people record what they did through a whole day. Ours is a made-up survey of 120 people from one school community: 30 students, 60 working-age adults and 30 elders.

2/2Students, adults and elders: three very different days.

Fig. 10.12An average day, then the same people split into groups (made-up survey).NCERT Fig. 10.5, 10.6 · p. 26

Horizontal bars of 24 hours, also marked 0% to 100%, split into sleep, personal care and meals, learning, paid work, home and care work, and leisure and travel, with the hours written in the larger pieces. In the first step only the Everyone bar shows: 8, 2.5, 2, 4, 2.5 and 5 hours. In the second step three more bars appear. Students: 9 hours sleep, 2 personal care, 8 learning, no paid work, 1 home work and 4 leisure. Adults: 7, 2.5, no learning, 7.5 paid work, 3.5 and 3.5. Elders: 9, 3, no learning, 1, 2 and 9.

Example 10.9NCERT Example 9 · p. 26

One day, read two ways

Use Fig. 10.12. (a) Does anyone's day really look like the Everyone bar? (b) Show that the Everyone sleep time is a weighted mean of the three groups' sleep times.

Try it yourself first. Then reveal the solution one step at a time.

Breaking the 120 people into groups told us far more than the single bar did. Other splits, such as village and city, income, or one decade against another, could tell us more still.

Practice · 1 question

Practice 10.5: questions with answers

1
ActivityMediumNCERT Exercise Set 10.5, Q1
A fitness band recorded the mean number of hours per weekday that some people spent lying down, sitting, and on their feet (standing, walking or working).
Table 10.6Hours per weekday
Lying down Sitting On their feet
Gurpreet (farmer) 7 3 14
Meenal (bank clerk) 7 12 5
Dadi (retired, 70) 9 8 7
Mystery person 7 14 3
You
Your entries are saved automatically.
1. Fill in your own row for a typical school day. Your three numbers must add up to 24.
  Show solution
  Answers vary. A typical student might spend about 8.5 h lying down, 9 h sitting (school, homework, meals, travel) and 6.5 h on their feet.
2. Guess what kind of work the mystery person might do. Compare your guess with a classmate's.
  Show solution
  14 h sitting and only 3 h on their feet suggests desk work or driving: perhaps a software developer, a call-centre worker or a long-distance bus driver.
3. Turn each person's hours into percentages of the day and sketch a 100% stacked bar chart, one bar per person.
  Show solution
  Each hour is $\frac{100}{24} \approx 4.17 %$ of the day. Gurpreet: 29.2%, 12.5%, 58.3%. Meenal: 29.2%, 50%, 20.8%. Dadi: 37.5%, 33.3%, 29.2%. Mystery person: 29.2%, 58.3%, 12.5%.
4. Would Meenal's bar change if Sundays were included? Would another bank clerk's bar look exactly like hers?
  Show solution
  Probably yes: on a holiday she may sit less and move more, so a 7-day average would shift towards time on her feet. Another clerk's bar would be similar but not identical, because people in the same job still live different days.

	Lying down	Sitting	On their feet
Gurpreet (farmer)	7	3	14
Meenal (bank clerk)	7	12	5
Dadi (retired, 70)	9	8	7
Mystery person	7	14	3
You

End-of-chapter exercises

Fig. 10.13For Question 7: library members by year.

Six stacked columns for 2016, 2018, 2020, 2022, 2024 and 2026 on a scale of 0 to 3500 members. The bottom piece of each column is adult members, rising slowly from 900 to 1400; the top piece is student members, rising fast from 300 to 2100. The column totals are about 1200, 1450, 1700, 2200, 2800 and 3500.

1/2Each eatery's ratings as shares of 100%.

Fig. 10.14For Question 9: ratings as shares, then as counts.

Three horizontal bars for eateries P, Q and R, split into 5-star, 4-star, 3-star, 2-star and 1-star ratings. In the first step all bars are the same length and show shares, with the 5-star shares labelled 60% for P, 45% for Q and 33% for R; R has a large 1-star piece. In the second step the bars shrink to a count scale from 0 to 600: P has only 20 ratings and becomes a tiny sliver, Q has 600 and fills the scale, and R has 150.

Practice · 16 questions

Practice questions with answers (NCERT “End-of-chapter exercises”)

1
NumericalEasyNCERT End-of-chapter Q1
In cricket, the run rate is the mean number of runs scored per over.
1. After 10 overs a team's run rate is 7.2. It scores 15 runs in the 11th over. What is the run rate now, to two decimal places?
  Show answer
  Answer 7.91
  Runs so far $= 7.2 \times 10 = 72$ . New run rate $= (72 + 15) \div 11 = 87 \div 11 \approx 7.91$ .
2. In a 50-over match the run rate after 49 overs is 5.4. The last over yields 20 runs. What is the final run rate?
  Show answer
  Answer 5.692
  $(5.4 \times 49 + 20) \div 50 = 284.6 \div 50 = 5.692$ .
2
Short answerMediumNCERT End-of-chapter Q2
Four friends noted the price of rice (₹ per kg) at shops near them, marking each price B for basmati or N for non-basmati.
- Priya: 62 (N), 140 (B), 58 (N)
- Rahul: 150 (B), 65 (N), 160 (B), 55 (N)
- Sana: 75 (N), 135 (B)
- Tenzin: 145 (B), 60 (N), 155 (B)
1. Each friend found the mean of their own prices: $a_{P}$ , $a_{R}$ , $a_{S}$ and $a_{T}$ . Write an expression for the mean of all 12 prices.
  Show solution
  Weight each friend's mean by the number of prices they noted (3, 4, 2 and 3):
  $\overset{ˉ}{x} = (3 a_{P} + 4 a_{R} + 2 a_{S} + 3 a_{T}) \div 12$
2. Instead, they group the prices by type, with means $a_{B}$ and $a_{N}$ . Write the expression now.
  Show solution
  There are 6 prices of each type, so $\overset{ˉ}{x} = (6 a_{B} + 6 a_{N}) \div 12$ . Because the groups are equal in size, this is simply $(a_{B} + a_{N}) \div 2$ .
3. Calculate the mean price both ways. Do the two ways agree?
  rupees per kg
  Show answer
  Answer 105 rupees per kg
  By friend: $(260 + 430 + 210 + 360) \div 12 = 1260 \div 12 = 105$ . By type: $a_{B} = 147.5$ and $a_{N} = 62.5$ , so $(885 + 375) \div 12 = 105$ . Both ways divide the total of all 12 prices by 12.
4. Tenzin simply averages the four friends' means. What does he get, and why is it wrong?
  rupees per kg
  Show answer
  Answer 104.79 rupees per kg
  $(86.67 + 107.5 + 105 + 120) \div 4 \approx 104.79$ . It is wrong because it gives Sana's 2 prices the same say as Rahul's 4.
3
NumericalMediumNCERT End-of-chapter Q3
A grocer has 40 kg of dal that cost him ₹120 per kg on average.
1. He buys 60 kg more at ₹100 per kg. What is his new average cost per kilogram?
  rupees per kg
  Show answer
  Answer 108 rupees per kg
  $(40 \times 120 + 60 \times 100) \div 100 = 10800 \div 100 = 108$ rupees per kg.
2. Another grocer has 20 kg at an average of ₹120 per kg. She buys some dal at ₹100 per kg and her average falls to ₹104 per kg. How many kilograms did she buy?
  kg
  Show answer
  Answer 80 kg
  $(2400 + 100 k) \div (20 + k) = 104$ , so $2400 + 100 k = 2080 + 104 k$ . Then $4 k = 320$ and $k = 80$ kg.
NumericalMediumNCERT End-of-chapter Q4
(After a problem in Pṛthūdakasvāmī's commentary on Brahmagupta, c. 864 CE.) A hasta is an old unit of length, about one forearm. A trench 24 hastas long is dug in four stretches of lengths 4, 6, 6 and 8 hastas, to depths of 9, 7, 4 and 3 hastas. Find the mean depth of the trench.
hastas
Show answer
Answer 5.25 hastas
$(9 \times 4 + 7 \times 6 + 4 \times 6 + 3 \times 8) \div 24 = 126 \div 24 = 5.25$ hastas.
5
NumericalMediumNCERT End-of-chapter Q5
A shopkeeper holds 10 kg of onions bought at ₹30 per kg. For each purchase below, first mark on a number line where you expect his new average cost to fall. Then calculate it.
1. He buys 10 kg at ₹50 per kg.
  rupees per kg
  Show answer
  Answer 40 rupees per kg
  Equal amounts, so halfway: $(300 + 500) \div 20 = 40$ rupees per kg.
2. He buys 30 kg at ₹50 per kg.
  rupees per kg
  Show answer
  Answer 45 rupees per kg
  Three times as much new stock pulls the average three-quarters of the way: $(300 + 1500) \div 40 = 45$ .
3. He buys 40 kg at ₹20 per kg.
  rupees per kg
  Show answer
  Answer 22 rupees per kg
  $(300 + 800) \div 50 = 22$ , close to 20 because most of the stock is now the cheap kind.
4. He buys 5 kg at ₹30 per kg.
  rupees per kg
  Show answer
  Answer 30 rupees per kg
  The new onions cost the same as the old ones, so the average does not move.
ProofMediumNCERT End-of-chapter Q6
Some values have weights $w_{1}, w_{2}, \dots, w_{n}$ . What happens to the weighted mean if every weight is multiplied by the same number $k$ , for example if all are tripled? Try an example, then prove your answer.
Show solution
AnswerNothing changes: only the ratio of the weights matters.
Example: 40 and 80 with weights 1 and 3 give $(40 + 240) \div 4 = 70$ . Tripled weights 3 and 9 give $(120 + 720) \div 12 = 70$ , the same.
In general every term on the top has a factor $k$ , and so does every term on the bottom:
$\frac{k (w_{1} x_{1} + \dots)}{k (w_{1} + \dots)}$
The $k$ cancels, so the weighted mean is unchanged.
7
Short answerMediumNCERT End-of-chapter Q7
The chart in Fig. 10.13 shows the members of a town library each year, split into students and adults.
1. About how many members were there in 2016? By which year had this roughly doubled?
  Show solution
  About 1200 in 2016. Double is about 2400. The total was 2200 in 2022 and 2800 in 2024, so it doubled between 2022 and 2024.
2. About how many members were students in 2026, and what share of all members was that?
  Show solution
  The student piece runs from about 1400 to 3500, so about 2100 students, which is $2100 \div 3500 = 60 %$ of the members.
3. How did the numbers of student and adult members change? How did their shares change?
  Show solution
  Both numbers grew: students from about 300 to 2100, adults from about 900 to 1400. Students grew much faster, so their share rose from 25% to 60%, while the adults' share fell from 75% to 40% even though there were more adults than ever.
8
Short answerMediumNCERT End-of-chapter Q8
The table gives the percentage of households with a tap-water connection in five districts (made-up data).
Table 10.7Tap-water connections by district
District With tap water
Amli 64%
Barod 81%
Chirai 75%
Dhanpur 92%
Eklera 58%
1. (a)
  Which of these can you conclude? Choose all that apply.
  Barod has more households with tap water than Amli.About 3 in every 4 households in Chirai have tap water.Dhanpur has the most households with tap water.If Dhanpur has more households than Barod, it also has more households with tap water.If Eklera has more households than Amli, it also has more households with tap water.
  Select all that apply
  Show answer
  Answer
  (B) About 3 in every 4 households in Chirai have tap water.
  (D) If Dhanpur has more households than Barod, it also has more households with tap water.
  75% is 3 in 4 (B). In D, Dhanpur has both more households and a higher percentage, so it must have more with tap water. A and C need the numbers of households. E can fail: if Eklera has 1000 households (580 with tap water) and Amli 950 (608), Amli has more.
2. Can you find the percentage of households with tap water in all five districts together? If not, what else do you need?
  Show solution
  Not from the percentages alone. You need the number of households in each district: the overall percentage is the weighted mean of the five percentages, weighted by those numbers.
9
Short answerMediumNCERT End-of-chapter Q9
A food app shows the star ratings of three eateries, P, Q and R. Fig. 10.14 shows them first as shares (a 100% chart) and then as numbers of ratings (a stacked chart).
1. Looking only at the 100% chart, which eatery would you choose? Why?
  Show solution
  P: 60% of its ratings are 5 stars, the largest share. Its mean rating, $(60 + 16 + 6 + 2 + 1) \div 20 = 4.25$ , is also the highest.
2. Now look at the numbers of ratings. Would you change your choice? Discuss.
  Show solution
  Many people would switch to Q. P's shares rest on only 20 ratings, so a few new reviews could change them a lot. Q's mean of 4.1 rests on 600 ratings and is far more reliable. R has a low mean, about 2.93, and 40% of its ratings are 1 star.
10
Short answerMediumNCERT End-of-chapter Q10
Three riders finished a three-stage cycle rally. Their times are in hours : minutes.
Table 10.8Stage times
Hill stage Plain stage City stage
Rider 1 2:10 3:05 1:20
Rider 2 2:25 2:35 1:30
Rider 3 2:05 3:20 1:35
1. Would you draw these times with bars that show the totals (stacked) or with equal-length bars (100% stacked)? Give a reason.
  Show solution
  A stacked bar chart. The total time decides the winner, and only a stacked chart shows the totals (6 h 35 min, 6 h 30 min and 7 h). A 100% chart would draw every rider's bar the same length.
2. (b)
  If a 100% stacked bar chart were drawn, which of these could you answer from it alone?
  Who finished first?Roughly what fraction of Rider 1's time was spent on the hill stage?Who took the longest on the city stage?
  Select all that apply
  Show answer
  Answer
  (B) Roughly what fraction of Rider 1's time was spent on the hill stage?
  Only B is a share within one bar: 130 of 395 minutes, about a third. The other two need actual times. Careful: Rider 2 has the largest city share (90 of 390 minutes, 23%), but Rider 3 took the longest (95 minutes).
11
Short answerMediumNCERT End-of-chapter Q11
A school asked students in three classes how they usually travel to school. The table gives percentages.
Table 10.9How students travel to school (%)
Walk Cycle School bus Car or two-wheeler
Class 6 40 10 35 15
Class 9 25 35 25 15
Class 12 15 30 20 35
1. Sketch a 100% stacked bar chart of this data.
  Show solution
  Draw three bars of equal length. Mark the boundaries between pieces at 40%, 50% and 85% for Class 6; at 25%, 60% and 85% for Class 9; and at 15%, 45% and 65% for Class 12.
2. Write two things you notice and two things you wonder about.
  Show solution
  Possible noticings: walking falls steadily as students get older (40% to 15%); cycling jumps between Class 6 and Class 9; Class 12 uses cars and two-wheelers most. Possible wonderings: do older students live farther away? Are younger students allowed to cycle? How many students are in each class? The chart cannot say.
12
ActivityEasyNCERT End-of-chapter Q12
Individual project. Do one of these.
1. Track your screen time for one week, split into study, entertainment and talking with people. Draw a stacked bar chart with one bar per day.
  Show solution
  Keep a small log each evening. Use the same order of pieces in every bar and label the scale in minutes. Then write two observations, for example which day had the largest total, and whether study or entertainment time varied more.
2. With your family, record your household's spending in 4 to 6 categories (food, travel, electricity and so on) for three months. Draw a 100% stacked bar chart with one bar per month.
  Show solution
  Agree on the categories first. For each month, divide every category by that month's total and multiply by 100. Finally, note any share that changed, and decide whether spending on that category changed or the total did.
13
ActivityMediumNCERT End-of-chapter Q13
Group project (3–4 students). Design a rating scheme for your school canteen, a park, a bus route or a clinic.
1. Choose 4 to 6 aspects to rate and agree on weights for them. Justify your weights.
  Show solution
  Sample for a canteen: taste 4, hygiene 5, price 3, speed of service 2. Hygiene gets the highest weight because unsafe food matters more than anything else.
2. Ask at least 10 people to score each aspect from 1 to 5. Work out every person's overall rating, then the mean of these.
  Show solution
  For one person rating taste 4, hygiene 5, price 3 and speed 2, the overall rating is $(16 + 25 + 9 + 4) \div 14 = 54 \div 14 \approx 3.86$ . The group's mean is the plain mean of the individual overall ratings.
3. Draw a 100% stacked bar chart showing, for each aspect, the shares of 5-star, 4-star … 1-star ratings. Write what your scheme captures well and what it misses.
  Show solution
  One bar per aspect, split by star rating. A scheme misses anything nobody thought to rate, and its weights reflect the group's opinion rather than a fact.
NumericalMediumNCERT End-of-chapter Q14
Whole-class project. Each student finds the mean daily water use per person at home and the number of people at home. How can the class find the mean water use per person across all their homes? Try it with this sample: homes of 4, 6, 3 and 5 people using 110, 90, 130 and 100 L per person per day.
L
Show answer
Answer 103.9 L
Weight each home by its number of people: $(440 + 540 + 390 + 500) \div 18 = 1870 \div 18 \approx 103.9$ L. The plain mean of the four homes' means, 107.5 L, is too high because the small home with the highest use counts as much as the big one.
ProofHardNCERT End-of-chapter Q15
(Challenge) What happens to a weighted mean if you add the same number $c$ to every weight? Try 40 and 80 with weights 1 and 3, and then with weights 3 and 5. Explain using algebra.
Show solution
AnswerThe mean moves towards the simple mean of the values (unless it already equals it).
With weights 1 and 3: $(40 + 240) \div 4 = 70$ . With weights 3 and 5: $(120 + 400) \div 8 = 65$ . The mean moved from 70 towards 60, the simple mean of 40 and 80.
In general, let the weights add up to $W$ and let the weighted mean be $m$ , so the weighted total is $W m$ . Adding $c$ to every weight adds $c (x_{1} + \dots + x_{n}) = c n \overset{ˉ}{x}$ to the top, where $\overset{ˉ}{x}$ is the simple mean, and $c n$ to the bottom:
$\frac{W m + c n \overset{ˉ}{x}}{W + c n}$
This is a weighted mean of $m$ and $\overset{ˉ}{x}$ with weights $W$ and $c n$ , so it lies between them. Adding to every weight pulls the result towards the simple mean.
16
Quick answerHardNCERT End-of-chapter Q16
(Challenge) Last year a town's vehicles were 50% two-wheelers, 30% cars and 20% autos. This year the number of every type went up. Which of these could have happened to the shares? Choose all that are possible.
All three shares went up.All three shares went down.All three shares stayed the same.The two-wheeler share fell, the car share rose and the auto share stayed the same.The two-wheeler and car shares rose and the auto share fell.
Select all that apply
Show answer
Answer
(C) All three shares stayed the same.
(D) The two-wheeler share fell, the car share rose and the auto share stayed the same.
(E) The two-wheeler and car shares rose and the auto share fell.
Shares always total 100%, so they cannot all rise or all fall (A and B are impossible). C happens if every type grows by the same percentage. From 500, 300 and 200: rising to 510, 350 and 215 gives 47.4%, 32.6% and 20% (D); rising to 600, 360 and 201 gives 51.7%, 31.0% and 17.3% (E).

District	With tap water
Amli	64%
Barod	81%
Chirai	75%
Dhanpur	92%
Eklera	58%

	Hill stage	Plain stage	City stage
Rider 1	2:10	3:05	1:20
Rider 2	2:25	2:35	1:30
Rider 3	2:05	3:20	1:35

	Walk	Cycle	School bus	Car or two-wheeler
Class 6	40	10	35	15
Class 9	25	35	25	15
Class 12	15	30	20	35

Recap

How Quantities Combine: Understanding Data: key points and summary

The mean of group means equals the true mean only when the groups are the same size (Section 10.1.1).
Mean × number of values = total, so two groups combine as $\frac{a n + b m}{n + m}$ ((10.1)).
A mixture's concentration is total substance ÷ total mixture. It lies between the concentrations mixed, nearer the larger amount (Section 10.1.2).
The weighted mean is $\frac{w_{1} x_{1} + \dots + w_{n} x_{n}}{w_{1} + \dots + w_{n}}$ (formula). Weights can be sizes, amounts or importance (Section 10.1.3).
Equal weights give the ordinary mean, and multiplying every weight by the same number changes nothing.
To add or remove values, rebuild the total, change it, and divide by the new count.
Clustered bar charts compare individual parts; the choice of cluster decides which comparisons are easy (Section 10.2.1).
Stacked bar charts show totals and their parts, but only the first piece of each bar starts at 0.
100% stacked bar charts, like pie charts, show shares, not amounts (Section 10.2.2).
From a stacked bar chart you can always draw the 100% chart; going back needs the totals (what a 100% chart can tell).
An overall average can hide large differences between groups (whose typical day?).

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Exercise Set 10.1, Q1	Practice 10.1 · Q1
NCERT Exercise Set 10.1, Q2	Practice 10.1 · Q2
NCERT Exercise Set 10.1, Q3	Practice 10.1 · Q3
NCERT Exercise Set 10.1, Q4	Practice 10.1 · Q4
NCERT Exercise Set 10.1, Q5	Practice 10.1 · Q5
NCERT Exercise Set 10.2, Q1	Practice 10.2 · Q1
NCERT Exercise Set 10.3, Q1	Practice 10.3 · Q1
NCERT Exercise Set 10.3, Q2	Practice 10.3 · Q2
NCERT Exercise Set 10.3, Q3	Practice 10.3 · Q3
NCERT Exercise Set 10.3, Q4	Practice 10.3 · Q4
NCERT Exercise Set 10.3, Q5	Practice 10.3 · Q5
NCERT Exercise Set 10.3, Q6	Practice 10.3 · Q6
NCERT Exercise Set 10.3, Q7	Practice 10.3 · Q7
NCERT Exercise Set 10.3, Q8	Practice 10.3 · Q8
NCERT Exercise Set 10.4, Q1	Practice 10.4 · Q1
NCERT Exercise Set 10.4, Q2	Practice 10.4 · Q2
NCERT Exercise Set 10.5, Q1	Practice 10.5 · Q1
NCERT End-of-chapter Q1	End-of-chapter exercises · Q1
NCERT End-of-chapter Q2	End-of-chapter exercises · Q2
NCERT End-of-chapter Q3	End-of-chapter exercises · Q3
NCERT End-of-chapter Q4	End-of-chapter exercises · Q4
NCERT End-of-chapter Q5	End-of-chapter exercises · Q5
NCERT End-of-chapter Q6	End-of-chapter exercises · Q6
NCERT End-of-chapter Q7	End-of-chapter exercises · Q7
NCERT End-of-chapter Q8	End-of-chapter exercises · Q8
NCERT End-of-chapter Q9	End-of-chapter exercises · Q9
NCERT End-of-chapter Q10	End-of-chapter exercises · Q10
NCERT End-of-chapter Q11	End-of-chapter exercises · Q11
NCERT End-of-chapter Q12	End-of-chapter exercises · Q12
NCERT End-of-chapter Q13	End-of-chapter exercises · Q13
NCERT End-of-chapter Q14	End-of-chapter exercises · Q14
NCERT End-of-chapter Q15	End-of-chapter exercises · Q15
NCERT End-of-chapter Q16	End-of-chapter exercises · Q16

Quick check: MCQs on How Quantities Combine: Understanding Data

10.1.1 Average of Averages3 questions
10.1.2 Mixtures3 questions
10.1.3 Custom Weights3 questions
Read the charts4 questions
10.2.1 Stacking Columns3 questions
Whose typical day?3 questions

Glossary · 9

Key terms (English–Hindi)

100% stacked bar chart: A stacked bar chart whose bars all have the same length, standing for 100%, so each piece shows a share.First seen in the chapter →

clustered bar chart: A bar chart that stands related bars side by side in groups (clusters), every bar starting at 0.First seen in the chapter →
concentrationसांद्रताsaandrata: The fraction or percentage of a mixture that is one particular substance, such as sugar in a drink.First seen in the chapter →

meanमाध्यmadhya: The sum of a set of values divided by the number of values; also called the average.First seen in the chapter →

pie chartपाई चार्टpai chart: A circle divided into slices whose angles are in proportion to the parts of a whole.First seen in the chapter →

share: A part written as a fraction or percentage of its whole; 180 L out of 500 L is a 36% share.First seen in the chapter →
stacked bar chart: A bar chart in which the parts of each whole are joined end to end, so the full bar shows the total.First seen in the chapter →

weightभारbhaar: In a weighted mean, the number that says how much a value counts: a group's size, an amount mixed, or an importance.First seen in the chapter →
weighted meanभारित माध्यbharit madhya: The sum of each value times its weight, divided by the sum of the weights.

10.1 Combining Things

10.1.1 Average of Averages

10.1.2 Mixtures

10.1.3 Custom Weights

10.2 Visualising and Interpreting Data

10.2.1 Stacking Columns

10.2.2 An Alternative to Pie Chart

From amounts to shares

What a 100% chart can and cannot tell

Whose typical day?

End-of-chapter exercises

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Quick check: MCQs on How Quantities Combine: Understanding Data

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