NCERT Solutions Class 9 Maths Chapter 10 How Quantities Combine Understanding Data

F R E E S T U D Y M AT E R I A L F O R E V E R Y S T U D E N T

C L A S S 9 · M AT H S

NCERT Solutions
+ ×
Chapter 10: How
Quantities Combine:
Understanding Data

NCERT Textbook — Ganita Manjari

BOOK PAGES SECTIONS QUESTIONS MEDIUM

Part II, 8 – 36 16 60 English

Every question Step by step Exam ready
Activities, in-text questions The working shown, not Written to match how
and all exercises. just the final answer. marks are awarded.

Solutions, notes, sample papers & more at 66 pages

Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

CLASS 9 · MATHS · GANITA MANJARI

NCERT Solutions — Chapter 10: How Quantities
Combine: Understanding Data
Chapter 10 asks what happens to an average when groups of data, mixtures or marks are combined.
Averaging the averages works only when the groups are the same size; otherwise each part must count
according to its size or importance, which gives the weighted mean (Brahmagupta, c. 628 CE). The second
half turns to pictures of combined data: stacked bar charts, which show totals and their parts, and 100%
stacked bar charts, which show shares. You will learn to compute weighted means and to judge what each
chart can and cannot tell you.

TEXTBOOK BOOK PAGES
Ganita Manjari (Class 9) Part II, 8 – 36

SECTIONS QUESTIONS
16 60

MEDIUM
English

Think and Reflect — Page 8
Section 10.1.1 Average of Averages (Example 1)
THINK AND REFLECT

Q1 In a badminton academy there are 11 trainees — 8 seniors and 3 juniors. Their heights
(in cm) are:

HEIGHTS (IN CM) AVERAGE HEIGHT (IN CM)

SENIORS 165, 169, 164, 167, 170, 159, 164, 166 165.5

JUNIORS 146, 149, 153 149.33

Method 1 found the average height of the whole group as (165.5 + 149.33) ÷ 2 = 157.415
cm. Method 2 added all eleven heights and divided by 11 to get 1772 ÷ 11 = 161.09 cm,
which is correct.
Why did Method 1 not work?

Method 1 gives the two group averages equal importance, as if there were as many juniors as
seniors. But there are 8 seniors and only 3 juniors. Each senior’s height should count once, so
the seniors’ average must count 8 times and the juniors’ average only 3 times.

Page 1 of 66

Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

Correct average = (165.5 × 8 + 149.33 × 3) ÷ (8 + 3)
= (1324 + 448) ÷ 11 = 1772 ÷ 11 ≈ 161.09 cm

Method 1 pulls the answer down towards the juniors’ average, so it is too small (157.4 cm).

Why: an average is “total ÷ number of values”. Averaging two averages divides
by 2, the number of groups, not by 11, the number of trainees. That is only right
when both groups have the same size.

Check it yourself: the correct average 161.09 cm lies between 149.33 and 165.5
but much closer to 165.5, because the larger group (seniors) pulls it that way.

Tip: before averaging averages, always ask: “Are the groups the same size?” If
not, weight each average by its group size.

Exercise Set 10.1 — Page 13
Sections 10.1.1 Average of Averages and 10.1.2 Mixtures

Q1 The average score of students on a test in Section A is 72 and that of students in
Section B is 76. What is the combined average of both the sections given that Section
A has 30 students and Section B has 25 students?

Total score of Section A = 72 × 30 = 2160

Total score of Section B = 76 × 25 = 1900

Combined average = (2160 + 1900) ÷ (30 + 25)
= 4060 ÷ 55 ≈ 73.82

The combined average is about 73.82. It is closer to 72 than to 76 because Section A has more
students.

Page 2 of 66

Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

Why: the simple average (72 + 76) ÷ 2 = 74 would treat the two sections as
equal in size. Section A has 30 students and Section B has 25, so 72 must count
more.

Check it yourself: 73.82 lies between 72 and 76, and slightly below the
midpoint 74, as expected.

Q2 A farmer mixes three equal quantities of fertilisers. The first one contains 1⁄10 nitrogen,
the second contains 9⁄50 nitrogen, and the third contains 3⁄60 nitrogen. What is the
fraction of nitrogen in the mixture?

The quantities are equal, so the fraction in the mixture is the simple average of the three
fractions.

1⁄ 30⁄ 9⁄ 54⁄ 3⁄ 15⁄
10 = 300, 50 = 300, 60 = 300

Sum = 99⁄300 = 33⁄100

Fraction of nitrogen = 33⁄100 ÷ 3 = 11⁄100 (that is, 11%)

Why: if each quantity is q, nitrogen = q/10 + 9q/50 + 3q/60 and the mixture
weighs 3q. The q cancels, leaving the ordinary average of the three fractions.

Check it yourself: in decimals: (0.10 + 0.18 + 0.05) ÷ 3 = 0.33 ÷ 3 = 0.11.

Page 3 of 66

as e
ag l
Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

(Śrīdharācārya, Pāṭīgaṇita, c. 750 CE) In ancient India, Varṇa was the measure . comof gold
em meant that
Q3
m a s
purity. A purity of 16 varṇa meant pure gold; in general, a purity of k varṇa
.co was k/16 gold and the rest impurities. (Now the term a gl used is karat; 16
em = 24 karat.)
the gold-alloy
a s
agl Suppose a goldsmith melts together three pieces of gold: 9 units at 12 varṇa, 5 units at
Varṇa

10 varṇa, and 17 units at 11 varṇa. Find the purity in varṇa of the combined gold.
co m
m . ag
la se
agmean of the purities, with the amounts as weights.
Purity of the mixture is the weighted

com
m.
Purity = (9 × 12 + 5 × 10 + 17 × 11) ÷ (9 + 5 + 17)

m as e
.co
= (108 + 50 + 187) ÷ 31
agl
a em÷ 31 = 114⁄31 ≈ 11.13 varṇa
=s345
agl
m a s
.co agl

a s em
l
Why: a piece of w units aatgk varṇa contains w × k/16 units of pure gold. Total
gold = (9×12 + 5×10 + 17×11)/16 = 345/16 units in 31 units of alloy, so the purity
is (345/31)/16, that is, 345/31 varṇa. com .
m
c om gtol 11, because the ase
. a
se m
Check it yourself: 11.13 lies between 10 and 12 and close
a
agl largest piece (17 units) is at 11 varṇa.

se m
com× 24/16 ≈ 16.7 karat. g l a
.
Did you know? in karats this is 11.13
a
a sem
agl
m
Q4 The average rainfall per day in the months of May, June, and m .coin a certain location
se that gives their
July
om la
c ag
are 3.5 mm, 10 mm and 8.7 mm respectively. Write an expression

sem. combined average.
la
ag c
m .
m 30 days, so each month’s average is weighted by its gla s e
c o
May and July have 31 days and June has
number of days. m. e a
g l as
a

co m
m .
m as e
.co

ag l Page 4 of 66

Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

Combined average = (3.5 × 31 + 10 × 30 + 8.7 × 31) ÷ (31 + 30 + 31)
= (108.5 + 300 + 269.7) ÷ 92 = 678.2 ÷ 92
≈ 7.37 mm per day

Why: 3.5 × 31 is the total May rainfall, 10 × 30 the total June rainfall and 8.7 × 31
the total July rainfall. Total rainfall ÷ total days gives the average per day for the
whole period.

Tip: the simple average (3.5 + 10 + 8.7) ÷ 3 ≈ 7.40 mm is close but not exact,
because the months are not all the same length.

Q5 Calculate the concentration of spice mix in these two scenarios.
(i) A 100 mL kashayam/kadha with 5% spice mix, a 200 mL one with 10% spice mix,
and a 300 mL one with 15% spice mix are combined.
(ii) A 300 mL kashayam/kadha with 5% spice mix, a 200 mL one with 10% spice mix,
and a 100 mL one with 15% spice mix are mixed.

(i) Spice = 100 × 0.05 + 200 × 0.10 + 300 × 0.15 = 5 + 20 + 45 = 70 mL

Concentration = 70 ÷ 600 ≈ 11.67%

(ii) Spice = 300 × 0.05 + 200 × 0.10 + 100 × 0.15 = 15 + 20 + 15 = 50 mL

Concentration = 50 ÷ 600 ≈ 8.33%

The same three strengths give different results because the amounts are swapped: in (i) most
of the drink is 15%, in (ii) most of it is 5%.

Page 5 of 66

Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

Why: the weighted mean is pulled towards the value with the biggest weight. In
both cases the total is 600 mL, but the weights 100, 200, 300 are attached to
different concentrations.

Check it yourself: the two answers are equally far from 10% (11.67 − 10 = 10 −
8.33 = 1.67), because scenario (ii) is the mirror image of scenario (i).

Exercise Set 10.2 — Pages 14–15
Section 10.1.3 Custom Weights

Q1 Savitri’s marks in Kashmiri are as follows: 35 out of 50 in internal tests, 44 out of 60 in
the project, and 80 out of 100 in the final exam. The annual percentage score is
calculated by combining the internals, project, and final exam in the ratio 3 : 4 : 5.
Which of the following expression(s) gives her annual score (as a percentage) in
Kashmiri?

(i) (35 × 3 + 44 × 4 + 80 × 5) ÷ (3 + 4 + 5)

(ii) [35⁄100 × 3 + 44⁄100 × 4 + 80⁄100 × 5] ÷ (3 + 4 + 5)

(iii) {[35⁄50 × 3 + 44⁄60 × 4 + 80⁄100 × 5] ÷ (3 + 4 + 5)} × 100

(iv) [(35⁄50 × 100) × 3 + (44⁄60 × 100) × 4 + (80⁄100 × 100) × 5] ÷ (3 + 4 + 5)

(iii) and (iv) are correct.
First change each mark to a percentage of its own maximum: 35/50 = 70%, 44/60 ≈ 73.33%,
80/100 = 80%. Then take the weighted mean with weights 3, 4, 5.

Page 6 of 66

Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

Annual score = (70 × 3 + 73.33 × 4 + 80 × 5) ÷ 12

= (210 + 293.33 + 400) ÷ 12 ≈ 75.28%

(i) and (ii) are wrong: they treat 35 and 44 as if they were out of 100, but they are out of 50 and
60. (ii) also gives a fraction, not a percentage.

Why: (iii) multiplies by 100 at the end and (iv) multiplies each fraction by 100
first. Multiplying by 100 before or after taking the weighted mean gives the
same result, so (iii) = (iv) ≈ 75.28.

Check it yourself: (i) gives 681 ÷ 12 = 56.75 and (ii) gives 0.5675. Both are far
below every one of her actual percentages (70%, 73.3%, 80%), so they cannot
be her annual score: a weighted mean must lie between the smallest and largest
values.

Exercise Set 10.3 — Pages 15–18
Section 10.1 Combining Things: weighted mean in different situations

Q1 A stationery shop owner made ₹8000 selling books, of which 30% is the profit
amount, and ₹1000 selling book covers, of which 50% is the profit amount. What is the
percentage of profit on the total sales?

Profit on books = 30% of ₹8000 = ₹2400
Profit on covers = 50% of ₹1000 = ₹500

Total profit = ₹2900 on total sales of ₹9000

Profit percentage = 2900 ÷ 9000 × 100 ≈ 32.22%

It is close to 30% because most of the sales (₹8000 of ₹9000) came from books.

Page 7 of 66

Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

Why: this is the weighted mean of 30% and 50% with weights 8000 and 1000:
(8000 × 30 + 1000 × 50) ÷ 9000 = 32.22%.

Tip: the simple average (30 + 50) ÷ 2 = 40% is wrong, because the two kinds
of sales are very different in size.

Q2 A white stork’s migration is tracked. The average daily distance travelled, calculated
over 20 days, is 44.5 km. On the 21st day, it flew 55 km. What is the average daily
distance travelled over these 21 days? Make a guess before you calculate.

Guess: a little more than 44.5 km, because one day out of 21 is above the average.

Distance in 20 days = 44.5 × 20 = 890 km

Distance in 21 days = 890 + 55 = 945 km
New average = 945 ÷ 21 = 45 km per day

Why: adding one value is like combining two collections: 20 values with
average 44.5 and 1 value of 55. The weighted mean is (44.5 × 20 + 55 × 1) ÷ 21.

Tip: the new day is 55 − 44.5 = 10.5 km above the old average. Shared over 21
days, it raises the average by 10.5 ÷ 21 = 0.5 km, so 44.5 + 0.5 = 45 km.

Page 8 of 66

as e
ag l
Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

. comWhat
A 600 mL solution with 5% salt is mixed with a 300 mL solution with 8% sugar.
em
Q3
m
are the concentrations of salt and sugar in the mixture?
a s
.co5%, Sugar: 8% (ii) Salt: 13%, Sugar: 3% agl
emSalt: 6%, Sugar: 6% (iv) Salt: 5.55%, Sugar: 8.88%
(i) Salt:
a s
agl (v) Salt: 3.33%, Sugar: 2.67% (vi) Salt: 4.1%, Sugar: 7.08%
(iii)

co m
. ag

se m
The correct option is (v).
la
ag
m
Salt = 5% of 600 = 30 mL; sugar = 8% of 300 = 24 mL
co
em.
as
Total mixture = 600 + 300 = 900 mL

.co÷m900 × 100 ≈ 3.33% agl
em= 24 ÷ 900 × 100 ≈ 2.67%
Salt = 30
a s
agl
Sugar

mvolume, so both concentrations fall. a s
.co agl
Each substance is now spread through a larger

se m
la
ag

c om has
Why: the 300 mL sugar solution has 0% salt, and the 600 mL salt solution
.
s e
0% sugar. So salt = (600 × 5 + 300 × 0) ÷ 900 and sugar = (600m × 0 + 300 × 8)
. om
÷c900. gla a
e m
g las
a Tip: options (i), (iv) and (vi) cannot be right: mixing with something that has
se m
com a
none of a substance can never keep its concentration the same or raise it.
. a g l
e m
ag las

co m
m .
m la se
.co ag
a s em
agl c
m .
m a s e
. co agl
a sem
agl

co m
m .
m as e
.co

ag l Page 9 of 66

Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

Q4 At a panipuri (golgappa) stall, the concentration of spice in the pani (spiced water)
was 8%. Many customers complained that it was too spicy. What quantity of regular
water should be mixed into the 10 litres of pani so that the spice level is reduced to
(3⁄4)th of the original concentration?

Spice in the pani = 8% of 10 L = 0.8 L

Required concentration = 3⁄4 × 8% = 6%

Let w litres of water be added (water has 0% spice):
0.8 ÷ (10 + w) = 0.06

10 + w = 0.8 ÷ 0.06 = 13.33…

w = 3.33… = 10⁄3 litres ≈ 3.33 L

Why: adding plain water does not change the amount of spice (0.8 L); it only
increases the total volume. Concentration falls to 3/4 when the volume rises to
4/3 of 10 L, that is, 40/3 L.

Check it yourself: 0.8 ÷ 13.33 = 0.06 = 6%, which is 3/4 of 8%.

Q5 A physical fitness evaluation is being undertaken. The final marks are calculated by
combining the marks for strength, flexibility, and agility in the ratio 4 : 5 : 6. Rashi has
scored 60, 65, and 70. Keerthi has scored 55, 65, and 75 respectively.
(i) Find out whose total is more without calculating.
(ii) What are their final marks?

(i) Keerthi’s. Both have 65 in flexibility. Keerthi has 5 less in strength (weight 4) but 5 more in
agility (weight 6). Agility has the larger weight, so her gain outweighs her loss.

Page 10 of 66

Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

(ii) Rashi = (60 × 4 + 65 × 5 + 70 × 6) ÷ 15 = (240 + 325 + 420) ÷ 15

= 985 ÷ 15 ≈ 65.67

Keerthi = (55 × 4 + 65 × 5 + 75 × 6) ÷ 15 = (220 + 325 + 450) ÷ 15
= 995 ÷ 15 ≈ 66.33

Why: compared with Rashi, Keerthi’s weighted total changes by −5 × 4 + 0 × 5
+ 5 × 6 = −20 + 30 = +10. Dividing by 15, her final mark is 10/15 ≈ 0.67 higher,
which matches 66.33 − 65.67.

Tip: “without calculating” means reasoning about the weights. You only need to
see that the subject in which Keerthi gains carries more weight than the one in
which she loses the same amount.

Page 11 of 66

Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

Q6 A restaurant collected ratings from 10 customers on a scale of 1 to 5. The resulting
data is shown in the table below (number of customers giving each rating).

5★ 4★ 3★ 2★ 1★

FOOD 5 3 2 0 0

AMBIENCE 0 4 5 1 0

SERVICE 1 2 2 4 1

What is the average rating if the metrics are combined with the weights food :
ambience : service = 6 : 5 : 4?

Food = (5×5 + 4×3 + 3×2) ÷ 10 = 43 ÷ 10 = 4.3
Ambience = (4×4 + 3×5 + 2×1) ÷ 10 = 33 ÷ 10 = 3.3

Service = (5×1 + 4×2 + 3×2 + 2×4 + 1×1) ÷ 10 = 28 ÷ 10 = 2.8
Combined = (4.3 × 6 + 3.3 × 5 + 2.8 × 4) ÷ (6 + 5 + 4)
= (25.8 + 16.5 + 11.2) ÷ 15 = 53.5 ÷ 15 ≈ 3.57

The restaurant’s average rating is about 3.57 out of 5.

Why: there are two weighted means here. First, each metric’s average uses the
number of customers as weights for the star values. Then the three metric
averages are combined using the weights 6 : 5 : 4.

Check it yourself: each row adds up to 10 customers, so dividing by 10 is right.
3.57 lies between the lowest (2.8) and highest (4.3) metric averages, nearer 4.3
because food has the largest weight.

Page 12 of 66

Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

Q7 The following table shows the weight data of langurs in an animal facility. Without
doing any computations, can you tell whether there are more male langurs or more
female langurs? Or are they equal in number?

MALE LANGURS FEMALE LANGURS ALL LANGURS

AVERAGE WEIGHT 16.5 kg 13.8 kg 14.925 kg

NUMBER OF LANGURS 60

(i) Which of the following expression(s) describes the given scenario?

(a) (16.5x + 13.8y) ÷ (x + y) = 14.925
(b) (16.5x + 13.8y) ÷ 60 = – 14.925
(c) (16.5x + 13.8y) ÷ 2 = – 14.925

(d) (16.5x + 13.8y) ÷ (16.5 + 13.8) = – 14.925

(ii) Find out how many male langurs are present.
(iii) A female langur weighing 15.2 kg is admitted to the facility. What is the average
weight of the female langurs after this?
(iv) Two male langurs weighing 16.9 kg and 16.1 kg are released from the facility. What
is the average weight of the male langurs after this?
(v) Now, suppose one of the male langurs lost 1 kg of weight. What is the average
weight of all the male langurs after this?

More females. The overall average 14.925 kg is closer to the female average (13.8) than to the
male average (16.5), so the females carry more weight in the mean.
(i) (a) is correct (x males, y females). Since x + y = 60, (b) is also correct if read as “= 14.925”.
(c) and (d) are wrong.

(ii) 16.5x + 13.8(60 − x) = 14.925 × 60 = 895.5
828 + 2.7x = 895.5 ⇒ x = 25 males (35 females)

(iii) (35 × 13.8 + 15.2) ÷ 36 = 498.2 ÷ 36 ≈ 13.84 kg
(iv) (25 × 16.5 − 16.9 − 16.1) ÷ 23 = 379.5 ÷ 23 = 16.5 kg
(v) (379.5 − 1) ÷ 23 = 378.5 ÷ 23 ≈ 16.46 kg

So there are 25 male langurs.

Page 13 of 66

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Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

co m
m.

o m l ase
Why:m(c) ag
.c divides by 2 and (d) divides by the sum of the two averages;
s e neither is
a number of langurs. In (iv) the two langurs released have average (16.9 +
agl
the
16.1) ÷ 2 = 16.5 kg, equal to the male average, so removing them leaves the
average unchanged.
co m
m . ag
About the book: options (b),ag(c)
se
laand (d) are printed as “= – 14.925”. A weighted
mean of positive weights cannot be negative, so the minus sign looks like a
m
.co
printing slip. With “=”, (b) is true because x + y = 60.
se m
om l a
g = 2.7 into 1.125
m .c it yourself: the overall average splits the gap 16.5 −a13.8
Check
e
ag las(above the female average) and 1.575 (below the male average). The numbers
of langurs are in the reverse ratio: females : males = 1.575 : 1.125 = 7 : 5, that is,
35 : 25.
m a s
em
.co agl
as to the original 25 males instead, (412.5 − 1)
Tip: in (v), if the 1 kg loss gislapplied
÷ 25 = 16.46 kg as well, aso either reading gives about 16.46 kg.

co m
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as e
. com agl
a se m
agl
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m . a
e
ag las

co m
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m la se
.co ag
a s em
agl c
m .
m a s e
. co agl
a sem
agl

co m
m .
m as e
.co

ag l Page 14 of 66

Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

Q8 Dorjee has collected 1 litre of water from the Dead Sea! Using the information in the
table, answer the following questions. A calculator can be used if necessary.

WATER SOURCE SALINITY

DEAD SEA ≈ 34%

OTHER SEAS AND OCEANS ≈ 3.5%

GROUND WATER ≈ 0.01%

PURIFIED DRINKING WATER ≈ 0.001%

(i) What is the salinity of the mixture if he mixes 1 litre of water from the Dead Sea with
2 litres of purified drinking water?
(ii) Is it possible to mix water from the Dead Sea and purified drinking water to get a
mixture with the salinity of groundwater? Why/Why not? What quantity of purified
drinking water should Dorjee mix with 1 litre of water from the Dead Sea to get a
mixture having the salinity of groundwater?
(iii) Is it possible to mix water from the Dead Sea and groundwater to get a mixture with
the salinity of purified drinking water? Why/Why not? What quantity of groundwater
should he mix with 1 litre of water from the Dead Sea to get a mixture having the
salinity of purified drinking water?

(i) (1 × 34 + 2 × 0.001) ÷ 3 = 34.002 ÷ 3 ≈ 11.33%

(ii) Yes, because 0.01% lies between 0.001% and 34%. Let w litres be added:

(34 + 0.001w) ÷ (1 + w) = 0.01
34 + 0.001w = 0.01 + 0.01w
33.99 = 0.009w ⇒ w ≈ 3777 litres

(iii) No. A mixture’s salinity always lies between the salinities of the parts, here between 0.01%
and 34%. It can never fall to 0.001%, which is below both. So no quantity of groundwater
works.

Page 15 of 66

Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

Why: a weighted mean of two values, with positive weights, is never smaller
than the smaller value or larger than the larger value. Purified water (0.001%) is
saltier than nothing but less salty than groundwater (0.01%), so it cannot be
made from groundwater and sea water.

Check it yourself: in (ii), 3776.67 L of water plus 1 L of sea water is 3777.67 L
containing 34 + 3.78 ≈ 37.78 units of salt; 37.78 ÷ 3777.67 ≈ 0.01%.

Did you know? that is nearly 3.8 thousand litres — about 190 buckets of 20 L —
to bring just 1 litre of Dead Sea water down to the saltiness of groundwater.

In-text Questions — Pages 18–20

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Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

Section 10.2.1 Stacking Columns (cluster-column charts, Fig. 10.1)
IN-TEXT

Q1 The table shows the average weekly expenditure, in rupees (₹), of three families
across categories.

HOUSING FOOD EDUCATION TRANSPORTATION HEALTHCARE RECREATION

FAMILY 1940 1700 1280 1200 1770 535
A

FAMILY 1750 1546 1500 1280 1210 0
B

FAMILY 950 1700 1540 1400 1300 150
C

Table 10.2, page 18

Choice 1: Comparing expenditure of families across categories
Family A Family B Family C
2000

1500

1000

500

0
ng od n n e n
ousi Fo c atio r t atio thcar e atio
H Ed
u o al cr
ransp He Re
T

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Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

Choice 2: Comparing category-wise expenditure of each family
Housing Food Education Transportation Healthcare
Recreation
2000

1500

1000

500

0
Family A Family B Family C

Fig. 10.1, page 19 — Two cluster-column graphs showing different ways of visualising the
data in Table 10.2 (expenditure in ₹)

Answer each question by looking at Choice 1 first, and then Choice 2.
1. Which family spent the least on housing?

Family C (₹950). In Choice 1 the three housing bars stand side by side and Family C’s bar is
clearly the shortest (Family A ₹1940, Family B ₹1750).

Why: Choice 1 groups the bars by category, so all three families’ housing bars
are next to each other and easy to compare. In Choice 2 you have to find the
red bar in three different clusters.

Q2 2. Which family spent the most on healthcare? (Use Table 10.2 and Fig. 10.1 (shown in
Q1 of this section).)

Family A (₹1770, against ₹1210 for Family B and ₹1300 for Family C). Choice 1 shows this at a
glance: Family A’s healthcare bar is the tallest in the healthcare cluster.

Page 18 of 66

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Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

comand Fig.
3. Approximately how much did Family A spend on education? (Use Table .10.2
sem
Q3

com
10.1 (shown in Q1 of this section).)
a
m . agl
la se
ag
About ₹1300 (the table gives ₹1280). On either chart the education bar for Family A ends a little
m
above the 1000 line, roughly halfway to 1500.
. co ag
se m
la
ag
m
Tip: when reading a value off a bar chart, look at where the top of the bar falls
between two grid lines and estimate the fraction of the gap. Here it is.caolittle
m from 1000 to 1500. as em
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Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

Q1–5 Try to answer each of Questions 1–5 by looking at Choice 1 first, and then Choice 2.
Which was easier?

Choice 1 was easier for Questions 1 and 2, because they compare families within one
category, and Choice 1 puts those bars side by side.
Choice 2 was easier for Questions 4 and 5, because they compare categories within one
family, and Choice 2 keeps each family’s bars together.
Question 3 asks for one value, so both charts work equally well.

Why: a chart makes the comparison you need easy only if the bars you want to
compare stand next to each other. So the choice of cluster depends on the
question you want to answer.

Q6 6. Which family spent the least amount overall? (Use Table 10.2 and Fig. 10.1 (shown in
Q1 of this section).)

Family A: 1940 + 1700 + 1280 + 1200 + 1770 + 535 = ₹8425

Family B: 1750 + 1546 + 1500 + 1280 + 1210 + 0 = ₹7286
Family C: 950 + 1700 + 1540 + 1400 + 1300 + 150 = ₹7040

Family C spent the least overall (₹7040).

Why: neither cluster chart shows totals. You must add six bar heights for each
family, which is slow and error-prone if you read the values off the chart.

Page 20 of 66

Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

Q7 7. Approximately how much did Family C spend overall? (Use Table 10.2 and Fig. 10.1
(shown in Q1 of this section).)

About ₹7000 (exactly ₹950 + ₹1700 + ₹1540 + ₹1400 + ₹1300 + ₹150 = ₹7040).

Q8 8. Approximately what percentage of Family C’s expenses were on housing? (Use
Table 10.2 and Fig. 10.1 (shown in Q1 of this section).)

Housing ÷ total = 950 ÷ 7040 × 100 ≈ 13.5%

About 13.5% (roughly one-seventh) of Family C’s expenses were on housing.

Tip: a quick estimate: 950 is a little less than 1000, and 1000 ÷ 7000 ≈ 14%, so
the answer is a little under 14%.

In-text Questions — Page 20

Page 21 of 66

Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

Section 10.2.1 Stacking Columns (stacked bar chart, Fig. 10.2)
IN-TEXT

Q6–8 Was it as straightforward to answer Questions 6–8 (least total spending, Family C’s
total, and Family C’s percentage on housing) as the earlier questions? Is there a
visualisation that makes it simpler to answer such questions? Look at the following
visualisation of the same data. Try to understand how the chart is organised and
answer the three questions above.

Expenditure of three families in a month
Housing Food Education Transportation Healthcare
Recreation

Family A

Family B

Family C
0 1000 2000 3000 4000 5000 6000 7000 8000 9000

Fig. 10.2, page 20 — A stacked bar chart of the data in Table 10.2 (₹)

No, the cluster charts made us add six values per family. The stacked bar chart (Fig. 10.2)
makes it simpler: each family’s bars are joined end to end, so the length of the whole bar is the
family’s total.
6. Family C’s bar is the shortest, so Family C spent the least.
7. Family C’s bar ends at about ₹7000.
8. The housing (red) piece is about 950 of the 7000-long bar, roughly one-seventh, about 13–
14%.

Why: the stacked bar is built by placing the six category bars one after another
in the same order for every family. Its total length is the sum of the six values,
so totals can be compared directly by comparing bar lengths.

Page 22 of 66

Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

About the book: the text calls Table 10.2 “average weekly expenditure”, while
the chart title says “in a month”. The numbers are the same; it does not change
any answer.

Q1–5 Now, use the stacked bar chart (Fig. 10.2) to answer Questions 1–5 shown earlier. Is it
possible? Is it easy? If not, what made it harder than before?

It is possible but harder.
Only the first piece (housing) starts at 0. Every other piece starts where the previous one ends,
so to read, say, Family A’s education you must subtract the start from the end (about 4920 −
3640 = 1280). Also, the same category starts at different points in different bars, so comparing,
for example, healthcare across families means comparing pieces that are not lined up.
Q1 (housing) is still easy, since housing starts at 0 in every bar. The others take more effort.
Stacked bars make totals easy but individual parts harder.

Why: our eyes compare lengths best when the bars share a common starting
line. A stacked bar gives that only to the first piece and to the totals.

Tip: put the category you most want to compare first in the stack, so that it
starts from zero in every bar.

Exercise Set 10.4 — Pages 21–22

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Section 10.2.1 Stacking Columns
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corals and arachnids (spiders, scorpions). Source: IUCN Red List
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Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

2019, that is, the full height of the 2019 column.
(ii) About 1,000 reptile species. The dark red reptile piece in 2016 is about one-third of the gap
between 0 and 3,000.
(iii) Mammals, birds and amphibians. Their pieces are only a little taller in 2019 than in 2007 (a
rise of roughly a tenth to a quarter), while reptiles, insects, molluscs, fish and others roughly
doubled or more.

Why: for (ii), the reptile piece starts at the top of the mammal piece (about
1,200) and ends at about 2,200, so its length is about 2,200 − 1,200 = 1,000.
Read the length of the piece, not where it ends.

Check it yourself: rough readings from the chart (2007 → 2019): mammals ≈
1,100 → 1,200; birds ≈ 1,200 → 1,500; amphibians ≈ 1,800 → 2,200; reptiles ≈
400 → 1,400; insects ≈ 600 → 1,650.

Tip: “relatively small” means small compared with its own starting size. Reptiles
grew by about 1,000 species, to more than three times their 2007 count;
mammals grew by only about 100–150 species, roughly a tenth.

Page 25 of 66

Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

Q2 The wickets taken by a bowler in International Cricket matches till 2025 are shown in
the table.

TEST ODI T20

HOME 62 100 32

OVERSEAS 172 49 71

Complete the given stacked bar charts (the bar lengths can be approximate),
(i) comparing the total wickets taken at home vs. overseas:
(ii) the total wickets taken in each format:

Wickets taken at home vs. overseas
Test ODI T20

Home

Overseas
0 50 100 150 200 250 300

(i) To complete: one stacked bar each for Home and Overseas

Wickets taken in different formats
Home Overseas

Test

ODI

T20
0 50 100 150 200 250 300

(ii) To complete: one stacked bar for each format

(i) Home: 62 + 100 + 32 = 194; Overseas: 172 + 49 + 71 = 292
(ii) Test: 62 + 172 = 234; ODI: 100 + 49 = 149; T20: 32 + 71 = 103

Page 26 of 66

Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

Wickets taken at home vs. overseas
Test ODI T20

Home 62 100 32

Overseas 172 49 71
0 50 100 150 200 250 300

(i) Completed: Home total 194, Overseas total 292

Wickets taken in different formats
Home Overseas

Test 62 172

ODI 100 49

T20 32 71
0 50 100 150 200 250 300

(ii) Completed: Test 234, ODI 149, T20 103

The bowler took more wickets overseas (292) than at home (194), and the most in Test
matches (234).

Why: in (i) each bar is a venue, split into the three formats; in (ii) each bar is a
format, split into home and overseas. The same six numbers make both charts,
but each chart makes a different total easy to see.

Tip: draw each piece starting where the previous one ends: for Home, Test runs
0–62, ODI 62–162 and T20 162–194.

In-text Questions — Page 22

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Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

Section 10.2.2 An Alternative to Pie Chart
IN-TEXT

Q1 Try answering the following question based on the expenditure stacked bar chart (Fig.
10.2) you saw earlier: Which family has the smallest share of their total expenditure
towards healthcare? Was it easy? If not, what visualisation can help answer such
questions easily?

Family A: 1770 ÷ 8425 ≈ 21.0%
Family B: 1210 ÷ 7286 ≈ 16.6%
Family C: 1300 ÷ 7040 ≈ 18.5%

Family B has the smallest share for healthcare. It was not easy from Fig. 10.2: a “share”
compares a piece with its own total, and the bars have different totals. Pie charts, or better a
100% stacked bar chart, show each family’s spending as parts of the same whole, so shares
can be compared directly.

Relative expenditure of families across categories
Housing Food Education Transportation Healthcare
Recreation

Family A 23% 20% 15% 14% 21% 6%

Family B 24% 21% 21% 18% 17%

Family C 13% 24% 22% 20% 18%
0% 25% 50% 75% 100%

Fig. 10.4, page 23 — A 100% stacked bar chart of the data in Table 10.2 (segment labels added
here, rounded)

Why: in a 100% stacked bar every family’s bar has the same length (100%), and
each piece’s length is its percentage. The pink healthcare piece is visibly
shortest for Family B.

Page 28 of 66

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. com
Tip: Family C spends the least rupees on housing, but Family B spends the
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Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

Section 10.2.2 Powering Up! (Example 7)
IN-TEXT

Q1 The average category-wise daily electricity consumption of a house, in kWh, in 2005
and 2025 is shown in the table, with the corresponding stacked bar chart and 100%
stacked bar chart.

LIGHTING COOLING KITCHEN APPLIANCES OTHER APPLIANCES

2005 360 300 120 220

2025 220 800 460 520

Lighting Cooling Kitchen appliances Other appliances

2005

2025
0 500 1000 1500 2000

Stacked bar chart, page 23 (daily consumption in kWh, as given in the book)

Lighting Cooling Kitchen appliances Other appliances

2005 36% 30% 12% 22%

2025 11% 40% 23% 26%
0% 25% 50% 75% 100%

The corresponding 100% stacked bar chart, page 24

What do you observe? What do you find interesting in the data/graphs? Discuss with
others.

Some observations:
• Total use doubled, from 1000 units to 2000 units.
• Lighting fell from 360 to 220 units, and its share from 36% to 11%. More efficient bulbs (such
as LEDs) could explain this.
• Cooling rose from 300 to 800 units and is now the biggest share (40%), perhaps from more
air conditioners and hotter summers.
• Kitchen appliances rose almost four times (120 → 460) and their share nearly doubled (12%
→ 23%).

Page 30 of 66

Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

Interesting: cooling’s share rose only from 30% to 40%, but its amount rose 2.7 times. Shares
hide how much the total grew; amounts hide how the mix changed. We need both charts.

Why: a 100% stacked bar shows each part as a percentage of its own total.
Since the 2025 total is double, a share that stays the same would still mean
double the units.

About the book: the book calls these “daily” figures in kWh, but 1000 kWh in a
day is far more than any house uses (a typical Indian home uses a few kWh a
day). The numbers would fit better as yearly figures. The observations about
change and shares are the same either way.

Your own answer: discuss what has changed in your own home in the last 20
years: refrigerator, mixer, washing machine, fans, coolers or AC, phone
chargers, TV. Which of these would fall under “other appliances”?

Think and Reflect — Page 26

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Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

Section 10.2.2 What Can We Infer? (Example 8)
THINK AND REFLECT

Q1 In Example 8, a 100% stacked bar compares the proportion of flowers blooming in
Fatima’s and Naveen’s gardens across seasons. The book then shows two possible
stacked bar charts (Case 1 and Case 2) that give exactly the same 100% stacked bar.

Proportion of flowers that bloomed across seasons
Summer Monsoon Winter

Fatima 50% 30% 20%

Naveen 40% 25% 35%
0% 25% 50% 75% 100%

Example 8, page 24 — the 100% stacked bar chart

NUMBER OF BLOOMS SUMMER MONSOON WINTER TOTAL

CASE 1: FATIMA 105 63 42 210

CASE 1: NAVEEN 48 30 42 120

CASE 2: FATIMA 90 54 36 180

CASE 2: NAVEEN 96 60 84 240

What does this say about the scope of the stacked bars and 100% stacked bars?

A stacked bar shows actual amounts. From it we can compare totals, and parts within and
across bars (for example, Fatima’s 105 summer blooms against Naveen’s 48).
A 100% stacked bar shows only proportions within each bar. It tells us that 50% of Fatima’s
flowers bloomed in summer, but not how many. The same 100% chart fits Case 1 (Fatima has
more summer blooms) and Case 2 (Naveen has more). So it cannot be used to compare actual
numbers across bars.

Why: converting to percentages divides each bar by its own total. That
information about the total is lost, so many different data sets give the same
100% chart.

Page 32 of 66

Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

About the book: on page 25 the text says Naveen’s bar is “35% in monsoon,
and 25% in winter”, but the chart and both Case charts show 25% in monsoon
and 35% in winter (30 and 42 out of 120). The chart values are used here.

Q2 Given a stacked bar chart can we make a corresponding 100% stacked bar chart?

Yes. A stacked bar chart gives each part’s actual value, so we can find each bar’s total and
work out every part as a percentage of it. For example, Fatima in Case 1: 105 ÷ 210 = 50%, 63 ÷
210 = 30%, 42 ÷ 210 = 20%.

Check it yourself: Example 7 did exactly this: 360 ÷ 1000 = 36% for lighting in
2005.

Q3 Given a 100% stacked bar chart can we make a corresponding stacked bar chart?

No, not unless we are also told the total for each bar. The 100% chart gives only the
percentages. Case 1 (totals 210 and 120) and Case 2 (totals 180 and 240) both give the same
percentages, so we cannot tell which stacked chart it came from.

Tip: if the totals are given, then yes: part = percentage × total. For example,
40% of 240 = 96.

Page 33 of 66

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Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

Section 10.2.2 A Typical Day (Example 9)
IN-TEXT

Q1 The following chart shows the time spent on different activities in a day by the average
Indian. What do you notice? What do you wonder about?

The Average Indian’s day
Sleep Personal Care Paid Work
Unpaid Work & Care Learning
Leisure, Social & Travel
100% 24h

27%

75% 18h
5%
11%
yad fo %

50% 10% 12h

13%

25% 6h
34%

0% 0h
Fig. 10.5, page 26 — Time spent on different activities in a day by an average Indian (values
rounded off). Source: National Time Use Survey 2024

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Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

I notice:

Sleep 34% ≈ 8.2 h; Personal care 13% ≈ 3.1 h
Paid work 10% ≈ 2.4 h; Unpaid work & care 11% ≈ 2.6 h
Learning 5% ≈ 1.2 h; Leisure, social & travel 27% ≈ 6.5 h

Sleep is the biggest part (about one-third of the day), then leisure. Paid and unpaid work take
about the same time. Learning is small.
I wonder: why is learning only about 1 hour when students spend 6–7 hours in school? Why is
paid work only about 2.4 hours when working adults work about 8 hours? The reason is that
this is an average over all Indians — children, workers, homemakers and the elderly — so it
may not describe any real person well.

Why: each 1% of a day is 24 × 0.01 = 0.24 hours, so 34% ≈ 8.2 hours. The strip
adds to 100% = 24 hours.

Tip: other “wonder” questions: Does this differ between men and women,
villages and cities, weekdays and weekends? How much of “leisure, social &
travel” is travel?

Think and Reflect — Page 27

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Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

Section 10.2.2 A Typical Day (Fig. 10.6)
THINK AND REFLECT

Q1 When the same data is grouped by age, we get the following stacked column chart.

Time use by age group
Sleep Personal Care Paid Work Unpaid Work & Care
Learning Leisure, Social & Travel
100% 24h

26% 26% 26%
33%

75% 18h
11% 15%
22% 9%
9%
7%
yad fo %

50% 7% 14% 12h
13% 14%
13% 13%

25% 6h
38% 34% 37%
32%

0% 0h
Children (6–14) Youth (15–24) Adults (25–59) Elderly (60+)

Fig. 10.6, page 27 — Age-wise grouping of the same data (values rounded off; the thin pink
strip for children is about 1%). Source: National Time Use Survey 2024

What do you find interesting in the chart above? What can you infer? Discuss.

Interesting points and inferences:
• Children spend 22% (about 5.3 h) on learning and almost nothing on paid work; they also
sleep the most (38%, about 9 h).
• Youth spend only 11% on learning but 7% on paid work and 9% on unpaid work: many have
started working or helping at home.
• Adults spend about 29% (about 7 h) on paid plus unpaid work and almost no time learning.
• The elderly spend the most on leisure, social and travel (33%) and less on work.
• Personal care (13–14%) is almost the same for every age group.
So the “average Indian” of Fig. 10.5 hides big differences between age groups.

Page 37 of 66

Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

About the book: the left axis is labelled 0%, 25%, 55%, 75%, 100% (and 0h,
6h, 12h, 18h, 24h on the right). The middle label should read 50%, matching 12
h.

Your own answer: pick the age group you belong to and compare it with your
own day.

Q2 Do you remember the sleep time over age trend that you studied last year? Does that
trend align with this chart?

The usual trend is that sleep falls as people grow from childhood to adulthood: young children
need the most sleep, teenagers a little less, and adults less still.
The chart mostly agrees: children 38% (about 9.1 h), youth 34% (about 8.2 h), adults 32%
(about 7.7 h). The elderly (37%, about 8.9 h) do not follow the falling trend. This may be
because many have retired and spend more time resting, and not because older people need
more sleep.

Why: the chart shows time actually spent sleeping, which depends on habits
and free time, not only on the body’s need for sleep.

Your own answer: if your notes from last year give different hours, compare
those numbers with 9.1 h, 8.2 h, 7.7 h and 8.9 h.

Q3 Can you explain why the learning time of people in the age group 15 – 24 has reduced
significantly compared to that of the age group 6 – 14?

Learning drops from 22% (about 5.3 h) to 11% (about 2.6 h) because:

Page 38 of 66

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Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

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Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

Section 10.2.2 A Typical Day
THINK AND REFLECT

Q1 Do you remember the ‘What Can A Strip Say?’ activity from last year? In each strip, if
we club the tiny strips belonging to each activity together, will we get a 100% stacked
bar chart like the one in Fig. 10.6?

Yes. A strip for one day has a fixed length (24 hours), with small pieces for each activity in the
order they happened. If we join all the pieces of the same activity (for example, all the sleep
pieces) into one block, and keep the blocks in the same order in every strip, each strip
becomes one stacked bar of length 24 h = 100%.
Several such strips side by side form a 100% stacked bar chart like Fig. 10.6. We lose only the
time order of the activities, not how much time each takes.

Why: the length of each block is the total time for that activity, and the full strip
is always 24 hours, so the block’s length is also its share of the day.

Exercise Set 10.5 — Page 29

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Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

Section 10.2.2 An Alternative to Pie Chart

Q1 The smart watch data of 5 people was tracked from Monday to Friday. The recorded
data tells us how many hours each person spent lying down, sitting, and standing or
moving around, etc., as shown in the following table.

LYING SITTING STANDING OR MOVING
DOWN AROUND ETC.

PAVANI (PATIENT) 20 3 1

RAGHU (NURSE) 7 4 13

ZAKIR (TEACHER) 8 6 10

SAHANA
(STUDENT)

JULIE 8 10 6
(___________)

Table 10.3, page 29 — Average number of hours spent per day (Monday – Friday)

(i) How much time does Sahana spend per day in each of these body-states? Make a
reasonable guess and fill her row in the table.
(ii) Guess what activity or work Julie could be engaged in based on the data. Find out
what your classmates’ guesses are.
(iii) Complete the chart below based on the tabular data.

Lying down Sitting Standing/Moving around, etc.
Pavani
(Patient)
Raghu
(Nurse)
Zakir
(Teacher)
Sahana
(Student)
Julie
(________)
0% 25% 50% 75% 100%

Chart for part (iii), page 29 — Sahana’s bar is left for you to complete

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Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

(i) Sample guess: lying down 9 h (sleep), sitting 10 h (about 6 h in class, plus homework, meals
and travel), standing/moving 5 h (games, walking, chores). Total = 24 h.
(ii) Julie sits for 10 hours and moves for 6, so she probably has a desk job, such as an office
worker, bank clerk or computer operator. She could also be someone who sits while working,
such as a tailor or a driver.
(iii) Each value ÷ 24 × 100 gives the percentage (Sahana: 9 h = 37.5%, 10 h ≈ 41.7%, 5 h ≈
20.8%).

Lying down Sitting Standing/Moving around, etc.
Pavani 83% 13%
(Patient)
Raghu 29% 17% 54%
(Nurse)
Zakir 33% 25% 42%
(Teacher)
Sahana 38% 42% 21%
(Student)
Julie 33% 42% 25%
(________)
0% 25% 50% 75% 100%

Completed chart (Sahana’s bar uses the sample guess 9 h : 10 h : 5 h)

Why: a 100% stacked bar needs each person’s hours as a share of 24 h. Since
every row adds up to 24 h, the chart could also be read in hours: 25% = 6 h,
50% = 12 h, 75% = 18 h.

Your own answer: use your own school day for Sahana’s row: hours of sleep,
hours sitting (class, tuition, homework, screen time) and hours moving. Check
that the three add to 24.

Think and Reflect — Page 30

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Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

Section 10.2.2 (after Exercise Set 10.5)
THINK AND REFLECT

Q1 Will the bars look similar if a different teacher’s data is considered instead of Zakir’s?

Roughly similar, but not the same. Most teachers sleep about 7–8 hours, stand or walk while
teaching, and sit while checking work, so the bar will have a similar shape. But the details
depend on the person: a teacher who teaches mostly from a chair, travels a long way by bus, or
plays sport after school will have a different mix.

Why: one person’s data is a single example. To describe “teachers”, we would
need the average of many teachers, and even then individuals would differ.

Q2 Will the bars look different if the data for all 7 days of the week is considered?

Probably yes. Weekends are different from weekdays: people may sleep longer, students and
teachers do not sit in class, a nurse may have a day off, and some people move around more
doing chores or outings. The 7-day average would mix these days in, so the shares of lying
down, sitting and moving would change, for example a little more lying down for most people.

Q3 Some bars in the chart may also match people doing other kinds of work/activities.
Can you think of any? Discuss.

Yes. For example:
• Raghu (nurse), mostly standing/moving: a shopkeeper who stands at the counter, a farmer, a
delivery person, a traffic police officer, a waiter, a construction worker.
• Zakir (teacher), a mix: a salesperson, a doctor in a clinic, a chef.
• Julie, mostly sitting: an office worker, a driver, a tailor, a cashier, a software engineer.
• Pavani (patient), mostly lying down: a person recovering from an illness or injury, or a very
elderly person who needs rest.

Page 43 of 66

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Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

co m
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Tip: amchart
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co m
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End-of-Chapter Exercises — Pages 30–36
co m
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team made 12 runs in the last over?.com
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a s
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g l
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c
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a sem
agl

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m .
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.co

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Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

Tip: shortcut: the last over is 12 − 6 = 6 runs above the old run rate; shared over
20 overs this adds 6 ÷ 20 = 0.3, giving 6.3.

Q2 Five friends who play badminton surveyed the city and collected the price of
shuttlecocks across brands and types (Nylon and Feather). The following is the list of
the prices, in rupees (₹), that each one compiled.
Yusuf: 40(N), 105(N), 383(F), 108(N), 165(F), 116(F)
Srikanth: 194(N), 85(N), 93(N), 121(N)
Kashvi: 49(N), 297(F), 105(N), 275(F), 40(N)
Prasanna: 124(N), 333(F), 182(N), 258(F)
Gracy: 220(F), 458(F), 129(F), 183(N)
Can you describe a way they can work together to find the average price of a
shuttlecock across types?
(i) Each one calculated the average of the prices they had gathered. Suppose these are
ay, as, ak, ap, ag respectively. Write an expression that gives the combined average.
(ii) Suppose an, af are the average prices of the nylon shuttlecocks and the feather
shuttlecocks, respectively. Write an expression that gives the average price of a
shuttlecock. Will this be equal to the answer we got using the method in the previous
part?

Way: each friend tells the others the total of their prices and how many prices they have (or
their average and count). Add all the totals and divide by the total count, 6 + 4 + 5 + 4 + 4 = 23.

(i) Combined average = (6ay + 4as + 5ak + 4ap + 4ag) ÷ 23

(ii) There are 13 nylon and 10 feather prices:
Average = (13an + 10af) ÷ 23

Yes, the two are equal. Both numerators are the sum of all 23 prices (₹4063), so both give
4063 ÷ 23 ≈ ₹176.65.

Why: 6ay is Yusuf’s total, 13an is the total of all nylon prices, and so on.
Grouping the same 23 prices by friend or by type does not change their sum or
their count.

Page 45 of 66

Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

Check it yourself: friends’ averages: ay = 917 ÷ 6 ≈ 152.83, as = 493 ÷ 4 =
123.25, ak = 766 ÷ 5 = 153.2, ap = 897 ÷ 4 = 224.25, ag = 990 ÷ 4 = 247.5.
Types: an = 1429 ÷ 13 ≈ 109.92, af = 2634 ÷ 10 = 263.4.

Tip: a common mistake is (an + af) ÷ 2 ≈ ₹186.66 or (ay + as + ak + ap + ag) ÷ 5.
These ignore that the groups have different sizes. Also, if two friends noted the
same shuttlecock at the same shop (both Yusuf and Kashvi list 105(N)), the
friends might decide to count it only once.

Q3 Shreyas holds 25 shares of a company at an average price of ₹150 and Vaishnavi holds
5 shares of the same company at an average price of ₹150.
(i) Shreyas buys 10 shares of this company at a price of ₹30 each. What is the average
price per share for him after the purchase?
(ii) Vaishnavi buys some shares of this company at a price of ₹30 each and the
average price per share for her after the purchase is ₹70. How many shares did she
buy?

(i) Average = (25 × 150 + 10 × 30) ÷ (25 + 10)
= (3750 + 300) ÷ 35 = 4050 ÷ 35 ≈ ₹115.71

(ii) Let her buy n shares:
(5 × 150 + 30n) ÷ (5 + n) = 70
750 + 30n = 350 + 70n

400 = 40n ⇒ n = 10 shares

Check it yourself: (ii) (750 + 300) ÷ 15 = 1050 ÷ 15 = 70 ✓.

Page 46 of 66

Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

Why: the new average is a weighted mean of ₹150 and ₹30, weighted by the
number of shares at each price. Vaishnavi needs ₹70, which splits the gap 150 −
30 = 120 into 40 below and 80 above. So she needs twice as many shares at
₹30 as at ₹150: 2 × 5 = 10.

Q4 (Pṛthūdakasvāmī, commentary on Brahmagupta’s Brāhmasphuṭasiddhānta, c. 864 CE)
A “hasta” (meaning “forearm”) refers to a unit of length measuring around 18 inches.
A pool 30 hastas long is dug to different depths along its length. It is divided into 5
sections having lengths 4, 5, 6, 7, and 8 hastas, and is dug to depths of 9, 7, 7, 3, and 2
hastas, respectively. Find the mean depth of the pool.

Use Brahmagupta’s weighted mean with the section lengths as weights.

Mean depth = (4 × 9 + 5 × 7 + 6 × 7 + 7 × 3 + 8 × 2) ÷ (4 + 5 + 6 + 7 + 8)
= (36 + 35 + 42 + 21 + 16) ÷ 30 = 150 ÷ 30
= 5 hastas

Why: a longer section affects the average depth more. The simple average (9 +
7 + 7 + 3 + 2) ÷ 5 = 5.6 hastas is wrong because it treats the 4-hasta section
like the 8-hasta section.

Did you know? the sum 150 is the area of the side view (in square hastas).
Mean depth × length = area, so a flat-bottomed pool 30 hastas long and 5
hastas deep would hold the same amount of water per unit width.

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Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

Q5 Suvarna had purchased 1 g gold at ₹15k (short for ₹15,000). This is shown by point O
denoting the average price of gold that she possesses. Six different scenarios are
given below for her next transaction. For each scenario estimate and mark the average
price of gold she will have after the transaction.

O

0 5 10 15 20 25 30

Page 31 — average price of gold (in ₹ thousand per gram); O marks ₹15k

(i) Purchase 1 g gold at ₹30k (ii) Purchase 2 g gold at ₹30k
(iii) Purchase 1 g gold at ₹10k (iv) Purchase 10 g gold at ₹10k
(v) Purchase 0.5 g gold at ₹30k (vi) Purchase 0.5 g gold at ₹15k

(i) (15 + 30) ÷ 2 = 22.5k
(ii) (15 + 2 × 30) ÷ 3 = 25k
(iii) (15 + 10) ÷ 2 = 12.5k
(iv) (15 + 10 × 10) ÷ 11 ≈ 10.45k
(v) (15 + 0.5 × 30) ÷ 1.5 = 20k
(vi) (15 + 0.5 × 15) ÷ 1.5 = 15k

O

0 5 10 15 20 25 30
(iv) 10.45 (v) 20
(iii) 12.5 (i) 22.5
(vi) 15 (ii) 25
Average price after each transaction (₹ thousand per gram)

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Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

co m
m.

o m l ase
.c new average lies between ₹15k and the new price,aand
Why:meach g closer to
la se
ag
whichever has more grams. Equal grams (i, iii) give the midpoint; more new
grams (ii, iv) pull the point towards the new price; fewer (v) pull it less. Buying at
the same price (vi) does not move it.
co m
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la gap in the ratio of the grams. In (ii) the gap 15
agthe
Tip: to estimate quickly, split
→ 30 is split 2 : 1, so the point is 2/3 of the way: 15 + 10 = 25.
co m
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m l as
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a sem
agl
Q6 Given some data with corresponding weights, how would the weighted average
change if all the weights are doubled? If required, experiment with some data. What do
you observe? Justify your answer using algebra.
m a s
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ag
It does not change.

m
Experiment: values 10, 20 with weights 1, 3 give (10 + 60) ÷ 4 = 17.5; with weights 2, 6 they
.co
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give (20 + 120) ÷ 8 = 17.5.
m as
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la se
ag c
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m a s e
. co agl
a sem
agl

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m .
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.co

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Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

Q7 Answer the following questions based on the graph.

Cumulative number of objects currently orbiting Earth,
by year of launch/separation (as of Apr. 24, 2026)
Payload objects* Other objects**
35,000

30,000

25,000

20,000

15,000

10,000

5,000

0
1960 1970 1980 1990 2000 2010 2020

EoC Q7, page 32 (redrawn; bar heights are approximate, read from the printed chart). *
Former transport good of a rocket (e.g., spaceship, satellite). ** Space mission components,
debris, rocket stages. Source: J. McDowell, General Catalogue of Artificial Space Objects,
Release 1.8.0 (2026)

(i) In 2003, approximately how many total objects were found orbiting Earth in space?
In what year did this number double?
(ii) Find the approximate number and share of payload objects and other objects in the
year 2025.
(iii) What can you say about the number of payload objects over time, and the number
of other objects over time? What can you say about the share of payload objects over
time, and the share of other objects over time?

(i) About 10,000 objects in 2003. The total reached about 20,000 (double) in about 2021; it
passes 20,000 between 2021 and 2022.
(ii) In 2025 the total is about 33,000: payload ≈ 17,000 (about 51%) and other objects ≈ 16,000
(about 49%), roughly half each.
(iii) Both numbers only increase, since the chart is cumulative. Payload objects grew slowly
until about 2019 (about 4,000) and then shot up to about 18,000 by 2026. Other objects grew
steadily, with jumps around 2007 and 2009, and kept rising after that.

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Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

For decades payload objects were only about one-fifth to one-quarter of the total; since 2019
their share has risen sharply to about half, so the share of other objects has fallen from about
three-quarters to about half.

Why: in 2003 payload ≈ 2,500 of 10,000 = 25%; in 2019 ≈ 3,900 of 17,900 ≈
22%; in 2025 ≈ 17,000 of 33,000 ≈ 51%. The share of payload rose even though
the number of other objects also kept rising, because payload grew much
faster.

Did you know? the jump in 2007 is generally linked to an anti-satellite test that
broke an old weather satellite into thousands of pieces, and the jump in 2009 to
an accidental collision of two satellites. The rapid rise after 2019 comes from
launches of very large groups of small internet satellites.

Note: strictly, each bar counts the objects launched up to that year that are still
in orbit in April 2026. Objects that have since fallen back to Earth are not
counted, so the bar for 2003 is a little less than the number actually in orbit in
2003.

Tip: all readings are approximate; answers within about 1,000 of these are fine.

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Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

Q8 Observe the following infographic.

What Percentage of Schools
have a Playground?
76%
62% Ladakh 45% 100%
J&K
87%
98% Himachal Pradesh
Punjab
78%
Uttarakhand Arunachal Pradesh
90% Sikkim
96% 86% 68%
Haryana
Delhi Nagaland
80% 76% 78% 73%
Rajasthan Uttar Pradesh Assam
50%
Bihar 44% 87% Manipur
Meghalaya

81% 71% 66% 78% 80%
88% Madhya Pradesh Jharkhand West Bengal Tripura
Mizoram
Gujarat
79%
99.8% Chhattisgarh 64%
DNH & DD Odisha
93%
Maharashtra
74%
Telangana

89% 79%
Goa 81% Andhra Pradesh
Karnataka

92% 96%
A & N Islands
97% 87% Puducherry
Lakshadweep 88% Tamil Nadu
Kerala

Data: UDISE+ 2021–22, as printed in the book

EoC Q8, page 32 — percentage of schools in each State/UT that have a playground (redrawn
on the official map of India; small States/UTs are marked with a dot; DNH & DD = Dadra &
Nagar Haveli and Daman & Diu)

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Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

− The same values as a list

STATE/UT % STATE/UT % STATE/UT %

Jammu & Kashmir 62% Ladakh 76% Himachal 87%
Pradesh

Punjab 98% Uttarakhand 78% Haryana 90%

Delhi 96% Rajasthan 80% Uttar Pradesh 76%

Bihar 50% Sikkim 86% Arunachal 68%
Pradesh

Assam 78% Nagaland 73% Meghalaya 44%

Manipur 87% Mizoram 80% Tripura 78%

West Bengal 66% Jharkhand 71% Odisha 64%

Chhattisgarh 79% Madhya Pradesh 81% Gujarat 88%

Dadra & Nagar Haveli 99.8% Maharashtra 93% Telangana 74%
and Daman & Diu

Andhra Pradesh 79% Goa 89% Karnataka 81%

Kerala 88% Tamil Nadu 87% Puducherry 92%

Lakshadweep 97% Andaman & 96%
Nicobar Islands

(i) Identify the correct inference(s) from the statements given.
(a) More schools have a playground in Haryana compared to Uttarakhand.
(b) Approximately every 2 out of 3 schools in Arunachal Pradesh have a playground.
(c) Punjab has the highest number of schools with a playground.
(d) Suppose it is given that Maharashtra has more schools than Telangana. Then the
number of schools having a playground is more in Maharashtra.
(ii) Using the information given, can we find the nation-wide percentage of schools
with a playground? If not, what additional information is needed?

(i) Correct: (b) and (d).
(a) Not correct: Haryana has a higher percentage (90% vs 78%), but we do not know how
many schools each state has, so we cannot compare numbers.
(b) Correct: 68% ≈ 66.7% = 2/3.

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(c) Not correct: the map gives percentages, not numbers. Punjab’s 98% is the highestm
o
among
m .c
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the states, but Dadra & Nagar Haveli and Daman & Diu shows 99.8%.

c o m
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g as schools, so
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m . number is more than 74% of a smaller number. a
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93% of a larger
asWe need the number of schools in each State/UT. The national percentage is the
l
(ii)gNo.
a
weighted mean of the State/UT percentages, weighted by these numbers.

co m
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la se
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Why: this is like the 100% stacked bars: a percentage tells us a share of that
o m as
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c
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a sem a simple average of the percentages on the map would be wrong, because
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Tip:

m a s
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co m
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as e
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a se m
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e
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co m
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m la se
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a s em
agl c
m .
m a s e
. co agl
a sem
agl

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Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

Q9 Decision Dilemma:
(i) Which of the plays would you choose to watch based on the following chart?

Share of ratings of three plays
★★★★★ ★★★★ ★★★ ★★ ★

Play A

Play B

Play C
0% 25% 50% 75% 100%

EoC Q9 (i), page 33 (redrawn from the printed chart; segment lengths approximate)

(ii) The corresponding stacked bar chart is shown below. Would you change your
decision after looking at this chart? Why/Why not? Discuss.

Ratings of three plays
★★★★★ ★★★★ ★★★ ★★ ★

Play A

Play B

Play C
0 250 500 750 1000

EoC Q9 (ii), page 33 (redrawn; number of ratings, segment lengths approximate)

(i) Play A or Play B. Each has about 71% of ratings at 4★ or 5★ and only about 14% at 1★ or
2★. Play A has a slightly bigger 5★ share, so Play A looks a touch better. Play C has the
largest 5★ share (about 40%) but also about one-third 1★ ratings, so it is risky: people either
love it or hate it.
(ii) Yes, I would choose Play B. The stacked chart shows that Play B has about 925 ratings, Play
A only about 210 and Play C about 380. Play B’s good shares come from many more people, so
they are more reliable; it is also clearly the most popular.

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Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

Why: a 100% stacked bar hides how many people each bar represents. A play
rated by only a few people can look good or bad by chance.

Check it yourself: approximate average ratings from the chart: A ≈ 3.74, B ≈
3.70, C ≈ 3.16. A and B are almost tied, so the number of ratings decides it.

Your own answer: someone who loves bold, unusual plays might still pick Play
C for its many 5★ ratings. Any choice is fine if you give a reason from the
charts.

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Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

Q10 A triathlon is an endurance race consisting of swimming 3.8 km, cycling 180 km, and
running 42.2 km. Athletes compete for the fastest overall time, completing each
segment in that order. The following table shows the finish time of 3 athletes in each
segment.

SWIMMING CYCLING RUNNING

ATHLETE 1 01:08 05:00 03:15

ATHLETE 2 01:05 05:10 03:35

ATHLETE 3 01:22 05:55 03:50

Table 10.4, page 34 — Finish time of three athletes in each segment of a triathlon race
(values in hh : mm)

(i) What is a suitable representation of this data — a stacked bar chart or a 100%
stacked bar chart? Why do you think so?
(ii) Suppose a 100% stacked bar chart is drawn. Which of the following question(s)
can be answered by looking at just that chart?
(a) Who finished the race first?
(b) Approximately what fraction of their race time did Athlete 1 spend cycling?
(c) Which athlete took the longest for running?

(i) A stacked bar chart. The race is decided by total time, and a stacked bar shows each
athlete’s total as the bar length, while still showing the three segments.

Totals: Athlete 1 = 1:08 + 5:00 + 3:15 = 9 h 23 min
Athlete 2 = 1:05 + 5:10 + 3:35 = 9 h 50 min
Athlete 3 = 1:22 + 5:55 + 3:50 = 11 h 07 min

(ii) Only (b): 300 ÷ 563 ≈ 0.53, a little more than half. (a) and (c) need actual times, which a
100% chart does not show.

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Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

Why: in a 100% chart all three bars have the same length, so the winner cannot
be seen. For (c), running shares are Athlete 1: 195/563 ≈ 35%, Athlete 2:
215/590 ≈ 36%, Athlete 3: 230/667 ≈ 34%. The chart would wrongly suggest
Athlete 2, though Athlete 3 ran longest (3 h 50 min).

Check it yourself: Athlete 1 finished first (9 h 23 min) — easy to see on the
stacked bar chart.

Q11 Look at the following graph. What do you notice? What do you wonder? Write your
inferences.

Distribution of disabled persons by age group and
type of disability in India (Census 2011)
seeing hearing speech movement mental retardation
mental illness any other multiple disability

0–19

20–39

40–59

60 & above
0 20 40 60 80 100
% of disabled persons in the age group

EoC Q11, page 34 (redrawn; segment lengths are approximate, read from the printed chart).
Rows are age groups.

I notice: seeing and movement disabilities form a bigger share as age increases (seeing about
18% → 25%, movement about 13% → 25%). Multiple disability is largest for 60 & above (about
12%). The shares of mental retardation (about 8% → 2%) and speech (about 9% → 4%) fall
with age. Hearing stays about the same (18–20%).
Inferences: problems of eyesight, joints and walking become more common in old age, while
conditions present from birth or childhood make up a larger part among the young.

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Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data AglaSem · NCERT Solutions

I wonder: how many disabled persons are in each age group? These are only shares, m
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falling share does not mean fewer people. What is included in “any other”, which is.clarge for
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c om row adds to 100% of the disabled
Why: this is a 100% stacked bar chart: each
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em the mix of disabilities, not the numbers.
persons in that age group. It compares
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Tip: today the term intellectual disability is used for what the 2011 Census
listed as “mental retardation”. com .
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Individual project: Do at least one of the following.
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m approximate time spent lying down, sitting, a
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(i) Recall the previous day and fill in the
m .Ask at least 2 of your family members about their
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as you can choose to track just your own body-state for
and standing/moving around, etc.
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day and fill it in. Alternatively,
a Saturday, and Sunday. Visualise it using a stacked bar
any one of the weekdays,

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chart. Discuss with your class how you arrived at the estimated time spent in each
state. .
(ii) Visualise your family’s monthly expenditure using a 100%sstacked
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co(a) First, identify the expense categories in your family abudget.
a s em (b) Discuss with your family members to decide how you will collect and organise the
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(c) Collect expense data for at least 3 months.
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(d) Represent the data in a 100%ostacked
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total monthly expenditure ism . a
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(e) Finally, write your observations and inferences based on the chart.

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Sample for (i): I went through the day in time blocks (sleep, school, travel,
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meals) lasto 24 h. ag
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LYING DOWN SITTING STANDING/MOVING
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9h 10 h 5h
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MOTHER 6h 11 h
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