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NCERT Solutions · Class 9 Maths · Ganita Manjari Part II · Chapter 10

Chapter 10: How Quantities Combine: Understanding Data (Weighted Averages and Data)

Step-by-step solutions to Exercise Sets 10.1 to 10.5 and the End-of-Chapter Exercises of Ganita Manjari Part II Chapter 10 (NCERT Class 9 Maths, 2026-27): weighted mean, average of averages, mixtures and concentrations, stacked bar charts and 100% stacked bar charts. All 33 questions are answered, with the key answer highlighted.

Weighted mean of $x_{1}, x_{2}, \dots, x_{n}$ with weights $w_{1}, w_{2}, \dots, w_{n}$ : $\overset{x}{ˉ} = \frac{w _{1} x _{1} + w _{2} x _{2} + \dots + w _{n} x _{n}}{w _{1} + w _{2} + \dots + w _{n}}$ . To combine averages of groups, use the group sizes as weights; to combine concentrations, use the quantities as weights.

Exercise Set 10.1

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?

Solution

Total marks of A $= 72 \times 30 = 2160$ ; of B $= 76 \times 25 = 1900$ .

$Combined average = \frac{2160 + 1900}{30 + 25} = \frac{4060}{55} \approx 73.82$

It is closer to 72 than to 76 because Section A has more students.

The combined average is about 73.82.

A farmer mixes three equal quantities of fertilisers. The first one contains

\frac{1}{10}

nitrogen, the second contains

\frac{9}{50}

nitrogen, and the third contains

\frac{3}{60}

nitrogen. What is the fraction of nitrogen in the mixture?

Solution

With equal quantities, all weights are equal, so the weighted mean is the ordinary mean:

$\frac{1}{3} (\frac{1}{10} + \frac{9}{50} + \frac{3}{60}) = \frac{1}{3} (0.10 + 0.18 + 0.05) = \frac{0.33}{3} = 0.11 = \frac{11}{100}$

The mixture is $\frac{11}{100}$ (11%) nitrogen.

(Śrīdharācārya, Pāṭīgaṇita, c. 750 CE) In ancient India, Varṇa was the measure of gold purity. A purity of 16 varṇa meant pure gold; in general, a purity of k varṇa meant that the gold-alloy was k/16 gold and the rest impurities. Suppose a goldsmith melts together three pieces of gold: 9 units at 12 varṇa, 5 units at 10 varṇa, and 17 units at 11 varṇa. Find the purity in varṇa of the combined gold.

Solution

Weight each purity by the amount of gold:

$\frac{9 \times 12 + 5 \times 10 + 17 \times 11}{9 + 5 + 17} = \frac{108 + 50 + 187}{31} = \frac{345}{31} \approx 11.13 varṇa$

About 11.13 varṇa ( $\frac{345}{31}$ ).

The average rainfall per day in the months of May, June, and July in a certain location are 3.5 mm, 10 mm and 8.7 mm respectively. Write an expression that gives their combined average.

Solution

The months have 31, 30 and 31 days, so these are the weights:

$Combined average = \frac{31 \times 3.5 + 30 \times 10 + 31 \times 8.7}{31 + 30 + 31} = \frac{108.5 + 300 + 269.7}{92} = \frac{678.2}{92} \approx 7.37 mm per day$

(The simple mean $\frac{3.5 + 10 + 8.7}{3} = 7.4$ is slightly different, because June has one day fewer.)

$\frac{31 \times 3.5 + 30 \times 10 + 31 \times 8.7}{92} \approx 7.37$ mm per day

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.

Solution

$(i) \frac{100 \times 5 + 200 \times 10 + 300 \times 15}{600} = \frac{7000}{600} \approx 11.67%$

$(ii) \frac{300 \times 5 + 200 \times 10 + 100 \times 15}{600} = \frac{5000}{600} \approx 8.33%$

The same three concentrations give different results because the larger quantity pulls the average towards its own concentration.

(i) about 11.67% (ii) about 8.33%

Exercise Set 10.2

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)? (i)

\frac{35 \times 3 + 44 \times 4 + 80 \times 5}{3 + 4 + 5}

(ii)

\frac{\frac{35}{100} \times 3 + \frac{44}{100} \times 4 + \frac{80}{100} \times 5}{3 + 4 + 5}

(iii)

\frac{\frac{35}{50} \times 3 + \frac{44}{60} \times 4 + \frac{80}{100} \times 5}{3 + 4 + 5} \times 100

(iv)

\frac{( \frac{35}{50} \times 100 ) \times 3 + ( \frac{44}{60} \times 100 ) \times 4 + ( \frac{80}{100} \times 100 ) \times 5}{3 + 4 + 5}

Solution

The marks are out of different totals, so first convert each to a fraction (or percentage) of its maximum: $\frac{35}{50} = 70%$ , $\frac{44}{60} \approx 73.33%$ , $\frac{80}{100} = 80%$ . Then take the weighted mean with weights 3, 4, 5.

(i) uses the raw marks 35, 44, 80 as if they were all out of 100: wrong.
(ii) also treats every mark as out of 100: wrong.
(iii) weighted mean of the fractions, then × 100: correct.
(iv) weighted mean of the percentages: correct (it is the same as (iii)).

$Annual score = \frac{70 \times 3 + 73.33 \times 4 + 80 \times 5}{12} = \frac{903.33}{12} \approx 75.28%$

Expressions (iii) and (iv); her annual score is about 75.28%.

Exercise Set 10.3

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?

Solution

$Profit = 0.30 \times 8000 + 0.50 \times 1000 = 2400 + 500 = ₹2900$

$Profit % = \frac{2900}{9000} \times 100 \approx 32.22%$

(Not 40%, the simple average of 30% and 50%, because book sales are much larger.)

About 32.22% of the total sales.

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.

Solution

Guess: one longer day among 21 days will raise the average only a little, to a bit above 44.5 km.

$\frac{20 \times 44.5 + 55}{21} = \frac{890 + 55}{21} = \frac{945}{21} = 45 km$

The new average is 45 km per day.

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

Solution

Salt $= 5%$ of 600 mL $= 30$ mL; sugar $= 8%$ of 300 mL $= 24$ mL; total volume $= 900$ mL.

$Salt = \frac{30}{900} \times 100 \approx 3.33%, Sugar = \frac{24}{900} \times 100 \approx 2.67%$

Each substance is now spread through a larger volume, so both concentrations go down.

Option (v): salt 3.33%, sugar 2.67%.

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

\frac{3}{4}

th of the original concentration?

Solution

The target concentration is $\frac{3}{4} \times 8% = 6%$ . The amount of spice stays the same: $8%$ of 10 L $= 0.8$ L. Let x litres of water be added.

$\frac{0.8}{10 + x} = 0.06 \Rightarrow 10 + x = \frac{0.8}{0.06} = \frac{40}{3} \Rightarrow x = \frac{10}{3} \approx 3.33 L$

About $3 \frac{1}{3}$ litres of water.

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?

Solution

(i) Compared with Rashi, Keerthi has 5 marks less in strength (weight 4), the same in flexibility, and 5 marks more in agility (weight 6). Agility carries more weight, so Keerthi's final mark is higher.

(ii)

$Rashi = \frac{60 \times 4 + 65 \times 5 + 70 \times 6}{15} = \frac{985}{15} \approx 65.67, Keerthi = \frac{55 \times 4 + 65 \times 5 + 75 \times 6}{15} = \frac{995}{15} \approx 66.33$

Keerthi's is higher: Rashi ≈ 65.67, Keerthi ≈ 66.33.

A restaurant collected ratings from 10 customers on a scale of 1 to 5. The number of customers giving each rating (5, 4, 3, 2, 1) was: 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?

Solution

First find the average rating of each metric (10 customers each):

$Food = \frac{5 \times 5 + 4 \times 3 + 3 \times 2}{10} = \frac{43}{10} = 4.3$

$Ambience = \frac{4 \times 4 + 3 \times 5 + 2 \times 1}{10} = \frac{33}{10} = 3.3$

$Service = \frac{5 \times 1 + 4 \times 2 + 3 \times 2 + 2 \times 4 + 1 \times 1}{10} = \frac{28}{10} = 2.8$

$Overall = \frac{6 \times 4.3 + 5 \times 3.3 + 4 \times 2.8}{6 + 5 + 4} = \frac{25.8 + 16.5 + 11.2}{15} = \frac{53.5}{15} \approx 3.57$

The combined average rating is about 3.57.

The following table shows the weight data of langurs in an animal facility: average weight of male langurs 16.5 kg, of female langurs 13.8 kg, of all 60 langurs 14.925 kg. Without doing any computations, can you tell whether there are more male langurs or more female langurs? Or are they equal in number? (i) Which of the following expression(s) describes the given scenario? (a)

\frac{16.5 x + 13.8 y}{x + y} = 14.925

(b)

\frac{16.5 x + 13.8 y}{60} = 14.925

(c)

\frac{16.5 x + 13.8 y}{2} = 14.925

(d)

\frac{16.5 x + 13.8 y}{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?

Solution

Without computing: the overall average 14.925 kg is closer to the female average 13.8 kg (difference 1.125) than to the male average 16.5 kg (difference 1.575). A weighted average lies closer to the group with more members, so there are more female langurs.

(i) Let there be x males and y females, with x + y = 60. Total weight $= 16.5 x + 13.8 y$ , and dividing by the number of langurs gives the overall average. So (a) is correct, and (b) is also correct because x + y = 60. (c) and (d) are wrong: they do not divide by the number of langurs.

(ii) Put y = 60 − x in (b):

$16.5 x + 13.8 (60 - x) = 14.925 \times 60 = 895.5 \Rightarrow 2.7 x + 828 = 895.5 \Rightarrow x = 25$

So 25 males and 35 females.

(iii) Females' total $= 35 \times 13.8 = 483$ kg. With the new langur: $\frac{483 + 15.2}{36} = \frac{498.2}{36} \approx 13.84$ kg.

(iv) Males' total $= 25 \times 16.5 = 412.5$ kg. Removing 16.9 + 16.1 = 33 kg: $\frac{412.5 - 33}{23} = \frac{379.5}{23} = 16.5$ kg. (The two released langurs averaged exactly 16.5 kg, so the average does not change.)

(v) The total drops by 1 kg: $\frac{379.5 - 1}{23} = \frac{378.5}{23} \approx 16.46$ kg.

More females. (i) (a) and (b) (ii) 25 males (iii) ≈ 13.84 kg (iv) 16.5 kg (v) ≈ 16.46 kg

Dorjee has collected 1 litre of water from the Dead Sea! 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?

Solution

(i)

$\frac{1 \times 34 + 2 \times 0.001}{1 + 2} = \frac{34.002}{3} \approx 11.33%$

(ii) Yes. The salinity of a mixture always lies between the salinities of the two parts. Groundwater's 0.01% lies between 0.001% and 34%, so it can be reached by adding enough purified water. Let x litres be added:

$\frac{34 + 0.001 x}{1 + x} = 0.01 \Rightarrow 34 + 0.001 x = 0.01 + 0.01 x \Rightarrow 0.009 x = 33.99 \Rightarrow x \approx 3777 L$

That is about 3,777 litres of purified water for 1 litre of Dead Sea water!

(iii) No. Both the Dead Sea water (34%) and groundwater (0.01%) are saltier than purified water (0.001%). A mixture's salinity can never be below the lower of the two (0.01%), so no amount of groundwater can bring it down to 0.001%.

(i) ≈ 11.33% (ii) Yes, with about 3,777 L of purified water (iii) No; the mixture can never be less salty than groundwater.

Exercise Set 10.4

The following stacked column chart shows the number of animal species in the IUCN Red List, by class, over time (totals: 2007: 7,851; 2010: 9,618; 2013: 11,212; 2016: 12,630; 2019: 14,234). (i) What does the number 14,234 on the top of the column 2019 represent? (ii) Approximately how many reptile species were in the list in 2016? (iii) Which class(es) of species have seen a relatively small increase in count between 2007 and 2019?

Solution

(i) It is the height of the whole 2019 column: the total number of animal species of all classes (mammals, reptiles, birds, insects, amphibians, molluscs, fish and others) on the IUCN Red List in 2019.

(ii) In the 2016 column the reptiles segment (just above mammals) runs from about 1,050 to about 2,150 on the scale, so there were roughly 1,100 reptile species.

(iii) Reading each segment's height: mammals stay at about 1,000 to 1,100 and birds grow only from about 1,200 to 1,500. These two classes show only a small increase. Insects, molluscs and fish grow much more, which is why the total almost doubles.

(i) The total number of listed animal species in 2019 (ii) about 1,100 (iii) mammals and birds (small increases).

The wickets taken by a bowler in International Cricket matches till 2025 are: Home: Test 62, ODI 100, T20 32; Overseas: Test 172, ODI 49, T20 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.

Solution

(i) Home total $= 62 + 100 + 32 = 194$ ; overseas total $= 172 + 49 + 71 = 292$ .

Wickets taken at home vs overseas (stacked by format)

(ii) Test $= 62 + 172 = 234$ ; ODI $= 100 + 49 = 149$ ; T20 $= 32 + 71 = 103$ .

Wickets taken in each format (stacked by home/overseas)

(i) Home 194 (62 + 100 + 32), overseas 292 (172 + 49 + 71). (ii) Test 234, ODI 149, T20 103, each split into home and overseas.

Exercise Set 10.5

The smart watch data of 5 people was tracked from Monday to Friday (average hours per day lying down, sitting, standing or moving around): Pavani (Patient) 20, 3, 1; Raghu (Nurse) 7, 4, 13; Zakir (Teacher) 8, 6, 10; Sahana (Student) ?; Julie (____) 8, 10, 6. (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. (iii) Complete the 100% stacked bar chart based on the tabular data.

Solution

(i) A reasonable guess for a school student: about 9 hours lying down (sleep), 9 hours sitting (classes, homework, meals, travel) and 6 hours standing or moving (play, chores, walking). The three must add up to 24 hours.

(ii) Julie sits for 10 hours a day, more than anyone else, and moves for 6. She probably has a desk job, for example an office worker, bank employee, computer programmer or a shopkeeper who sits at the counter. (Other answers are possible.)

(iii) Convert each row to percentages of 24 hours (for example, Pavani lies down for $\frac{20}{24} \approx 83%$ of the day):

100% stacked bar chart of the smart-watch data (Sahana = our guess)

(i) e.g. Sahana: 9 h, 9 h, 6 h (ii) e.g. a desk job (office worker) (iii) Each bar shows the three times as percentages of 24 hours, as drawn above.

End-of-Chapter Exercises

In cricket, the run rate is the average number of runs scored per over. In a T20 match, a team scored 6 runs in the first over making the run rate 6. (i) In the second over they scored 12 runs. What is the run rate now? (ii) After Over 19, their run rate was 6. What is the run rate after 20 overs given the team made 12 runs in the last over?

Solution

(i) $\frac{6 + 12}{2} = 9$ runs per over.

(ii) After 19 overs the total is $19 \times 6 = 114$ runs. After 20 overs:

$\frac{114 + 12}{20} = \frac{126}{20} = 6.3$

(One good over raises the average only a little once many overs have been bowled.)

(i) 9 (ii) 6.3

Five friends surveyed the city and collected the price of shuttlecocks (in ₹; N = nylon, F = feather). 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

a_{y}, a_{s}, a_{k}, a_{p}, a_{g}

respectively. Write an expression that gives the combined average. (ii) Suppose

a_{n}, a_{f}

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?

Solution

The friends collected different numbers of prices (6, 4, 5, 4 and 4, in all 23), so they cannot simply average their five averages. Each friend's average must be weighted by the number of prices they collected.

(i)

$Average = \frac{6 a _{y} + 4 a _{s} + 5 a _{k} + 4 a _{p} + 4 a _{g}}{23}$

(ii) There are 13 nylon prices and 10 feather prices, so

$Average = \frac{13 a _{n} + 10 a _{f}}{23}$

Yes, both give the same answer: each numerator is just the sum of all 23 prices, grouped in a different way. Here the sum is ₹4063, so the average price is $\frac{4063}{23} \approx ₹176.65$ .

(i) $\frac{6 a _{y} + 4 a _{s} + 5 a _{k} + 4 a _{p} + 4 a _{g}}{23}$ (ii) $\frac{13 a _{n} + 10 a _{f}}{23}$ ; both equal the mean of all 23 prices, about ₹176.65.

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?

Solution

(i)

$\frac{25 \times 150 + 10 \times 30}{35} = \frac{3750 + 300}{35} = \frac{4050}{35} \approx ₹115.71$

(ii) Let her buy x shares:

$\frac{5 \times 150 + 30 x}{5 + x} = 70 \Rightarrow 750 + 30 x = 350 + 70 x \Rightarrow 40 x = 400 \Rightarrow x = 10$

(i) about ₹115.71 per share (ii) 10 shares.

(Pṛthūdakasvāmī, commentary on Brahmagupta's Brāhmasphuṭasiddhānta, c. 864 CE) 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.

Solution

The mean depth is the depths weighted by the lengths of the sections:

$\frac{4 \times 9 + 5 \times 7 + 6 \times 7 + 7 \times 3 + 8 \times 2}{30} = \frac{36 + 35 + 42 + 21 + 16}{30} = \frac{150}{30} = 5 hastas$

The mean depth is 5 hastas.

Suvarna had purchased 1 g gold at ₹15k. 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. (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

Solution

New average $= \frac{1 \times 15 + m \times p}{1 + m}$ (in ₹ thousand per gram) for m grams bought at ₹p k:

(i) $\frac{15 + 30}{2} = 22.5$
(ii) $\frac{15 + 60}{3} = 25$
(iii) $\frac{15 + 10}{2} = 12.5$
(iv) $\frac{15 + 100}{11} \approx 10.45$
(v) $\frac{15 + 15}{1.5} = 20$
(vi) $\frac{15 + 7.5}{1.5} = 15$ (no change: the price is the same as her average)

Average price (₹ thousand per gram) after each transaction

The more gold she buys at a price, the closer her average moves to that price.

(i) ₹22.5k (ii) ₹25k (iii) ₹12.5k (iv) about ₹10.45k (v) ₹20k (vi) ₹15k

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.

Solution

Experiment: values 10 and 20 with weights 1 and 3 give $\frac{10 + 60}{4} = 17.5$ ; with weights 2 and 6 they give $\frac{20 + 120}{8} = 17.5$ . No change.

Algebra: with all weights doubled,

$\frac{2 w _{1} x _{1} + 2 w _{2} x _{2} + \dots + 2 w _{n} x _{n}}{2 w _{1} + 2 w _{2} + \dots + 2 w _{n}} = \frac{2 ( w _{1} x _{1} + \dots + w _{n} x _{n} )}{2 ( w _{1} + \dots + w _{n} )} = \frac{w _{1} x _{1} + \dots + w _{n} x _{n}}{w _{1} + \dots + w _{n}}$

Only the ratio of the weights matters, and doubling all of them keeps the ratio the same.

The weighted average does not change; the factor 2 cancels from numerator and denominator.

Answer the following questions based on the graph (cumulative number of objects currently orbiting Earth, by year of launch/separation, payload objects and other objects). (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?

Solution

(Readings from the graph are approximate.)

(i) The 2003 bar reaches about 10,000 objects. Double this is about 20,000, which the bars reach in about 2021.

(ii) The 2025 bar reaches about 33,000 in all; the payload part (bottom) is about 17,000 and the other objects about 16,000. So payload objects are about $\frac{17000}{33000} \approx 52%$ and other objects about 48%.

(iii) Numbers: both kinds keep increasing (the graph is cumulative). Other objects (debris, rocket stages) grew steadily from the 1960s, with sudden jumps (for example around 2007 and 2009, when satellites were broken up). Payload objects grew slowly for decades and then very rapidly after about 2019, when thousands of small satellites began to be launched.

Shares: for most of the time other objects made up the large majority (about three-quarters or more). In recent years the payload share has risen sharply, to about half, and the share of other objects has fallen correspondingly.

(i) about 10,000; doubled around 2021 (ii) payload ≈ 17,000 (≈ 52%), other ≈ 16,000 (≈ 48%) (iii) both numbers rise; the payload share was small for decades but has risen to about half recently.

Observe the infographic "What Percentage of Schools have a Playground?" (state-wise percentages on a map of India). (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?

Solution

The map gives percentages, not numbers of schools (for example Haryana 90%, Uttarakhand 78%, Arunachal Pradesh 68%, Punjab 98%, Maharashtra 93%, Telangana 74%).

(i)

(a) Cannot be inferred. Haryana has a higher percentage, but we do not know how many schools each state has.
(b) Correct. 68% is close to $\frac{2}{3} \approx 67%$ .
(c) Cannot be inferred. Punjab has the highest percentage, not necessarily the highest number.
(d) Correct. Number with a playground = percentage × number of schools. Maharashtra has both a higher percentage (93% vs 74%) and more schools, so its number is larger.

(ii) No. The national percentage is a weighted average of the state percentages, with the number of schools in each state as weights. We need the number of schools in each state (and UT).

(i) (b) and (d) are correct (ii) No; we also need the number of schools in each state.

Decision Dilemma: (i) Which of the plays would you choose to watch based on the "Share of ratings of three plays" (100% stacked bar) chart? (ii) The corresponding stacked bar chart (number of ratings) is shown. Would you change your decision after looking at this chart? Why/Why not? Discuss.

Solution

(Readings are approximate.)

(i) From the shares: Play A has about 28% five-star and 43% four-star ratings (about 71% "4 stars or more"); Play B about 24% and 48% (about 72%); Play C has the most five-star ratings (about 40%) but also a large share of one-star ratings (about one-third). So A or B seems the safer choice, with very similar shares; C is a gamble ("people love it or hate it").

(ii) The stacked chart shows how many people rated each play: Play A about 210, Play B about 925, Play C about 380. Play B's high share of good ratings comes from far more viewers, so it is more reliable; Play A's equally good share is based on very few ratings. So Play B is the best choice. The 100% chart hides the number of raters, and this information can change the decision.

(i) Play A or B (highest share of 4- and 5-star ratings) (ii) Choose Play B: it has a similar share of good ratings from many more viewers (about 925 vs 210), so its rating is more trustworthy.

A triathlon consists of swimming 3.8 km, cycling 180 km, and running 42.2 km. Finish times (hh:mm): Athlete 1: 01:08, 05:00, 03:15; Athlete 2: 01:05, 05:10, 03:35; Athlete 3: 01:22, 05:55, 03:50. (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?

Solution

Total times: Athlete 1: 9 h 23 min; Athlete 2: 9 h 50 min; Athlete 3: 11 h 07 min.

(i) A stacked bar chart. The race is won on total time, and a stacked bar shows both the total (length of the whole bar) and each segment. A 100% stacked bar makes all bars the same length and hides the totals.

(ii)

(a) No: all bars are 100%, so the totals cannot be compared.
(b) Yes: the cycling part of Athlete 1's bar shows its share, $\frac{300}{563} \approx 53%$ , about half.
(c) No: the chart shows shares of each athlete's own time, not the actual running times.

(i) A stacked bar chart, because totals decide the race. (ii) Only (b) can be answered (about half, ≈ 53%).

Look at the graph "Distribution of disabled persons by age group and type of disability in India (Census 2011)" (a 100% stacked bar for each age group). What do you notice? What do you wonder? Write your inferences.

Solution

Readings are approximate; each bar shows percentages within one age group, not numbers of people.

Seeing disability is a bigger share in older people: about 18% in the 0–19 group but about 25% in the 60+ group.
Movement disability is the largest single type for adults: its share grows from about 13% (0–19) to about 25% (60+), as joints and bones weaken with age.
Mental retardation is a bigger share among the young (about 8% in 0–19) than among the elderly (about 2%).
Speech disability is a larger share among the young (about 9%) and very small for the elderly.
Multiple disability has a larger share at the two ends of life (0–19 and 60+).

Things to wonder about: how many people are in each age group (the chart does not show counts)? How do the patterns differ for men and women, or rural and urban areas? What is included in "any other"?

Shares of seeing and movement disabilities increase with age; mental retardation and speech disabilities form larger shares among the young. The chart shows shares within each age group, not numbers of people.

Individual project: (i) Recall the previous day and fill in the approximate time spent lying down, sitting, and standing/moving around, etc., for yourself and at least 2 family members, and visualise it using a stacked bar chart. (ii) Visualise your family's monthly expenditure for at least 3 months using a 100% stacked bar chart, and write your observations.

Solution

This is a project, so every answer will be different. A guide:

(i) Make a table with one row per person and three columns that add up to 24 hours (as in Exercise Set 10.5). Draw one bar per person of length 24 units, split into the three parts.
(ii) Choose 5–7 categories (for example food, housing/rent, education, travel, health, others). For each month, divide each category's amount by the month's total and multiply by 100. Draw one bar per month of the same length (100%), split by these percentages. Then compare: which share is largest? Which share changed most from month to month, and why (festivals, school fees, travel)?

Project work: make sure each bar of (i) totals 24 hours and each bar of (ii) totals 100%, then write what you notice.

Small-group project: Choose one scenario (shopping at a cloth store, bus travel, a tourist spot, or a clinic/hospital) and design a custom rating scheme with weights: choose 4–6 aspects, decide weights, collect at least 10 ratings on a scale of 1–5, compute individual and overall ratings, draw a 100% stacked bar chart and write your observations.

Solution

Example for bus travel: aspects punctuality, cleanliness, comfort, safety, staff behaviour, with weights 3 : 2 : 2 : 4 : 1 (safety matters most). If a passenger rates them 4, 3, 3, 5, 4, their overall rating is

$\frac{3 \times 4 + 2 \times 3 + 2 \times 3 + 4 \times 5 + 1 \times 4}{3 + 2 + 2 + 4 + 1} = \frac{48}{12} = 4.0$

Average such ratings over all 10 people for the overall score. For the 100% stacked bar, show for each aspect the share of people who gave 5, 4, 3, 2 and 1 stars.

Project work: weighted rating = Σ(weight × rating) ÷ Σ(weights), then average over all people.

Whole class project: Each student shares the average age of their family and the number of family members. Discuss among the class and come up with a way to find out the average age of all the families of the class.

Solution

Since families have different sizes, we must not simply average the family averages. Use the family sizes as weights. If family i has $n_{i}$ members with average age $a_{i}$ , then

$Average age = \frac{n _{1} a _{1} + n _{2} a _{2} + \dots}{n _{1} + n _{2} + \dots}$

Each $n_{i} a_{i}$ is the total age of family i, so the numerator is the total age of all the people and the denominator is the total number of people.

Average age = Σ(family size × family average) ÷ Σ(family sizes).

Given some data with corresponding weights, what would happen to the weighted average if all the weights are increased by a constant value, say 1? If required, experiment with some data. What do you observe? Justify your answer using algebra.

Solution

Experiment: values 10 and 20 with weights 1 and 3 give $\frac{10 + 60}{4} = 17.5$ . Increase both weights by 1 (to 2 and 4): $\frac{20 + 80}{6} \approx 16.67$ . The average has changed: it moved towards the ordinary mean, 15.

Algebra: let $W = \sum w_{i}$ and $\overset{x}{ˉ} = \frac{\sum w _{i} x _{i}}{W}$ . Adding 1 to each of the n weights:

$\frac{\sum ( w _{i} + 1 ) x _{i}}{W + n} = \frac{W x ˉ + n \cdot m}{W + n}, where m = \frac{\sum x _{i}}{n} is the ordinary mean.$

So the new value is a weighted average of the old weighted mean $\overset{x}{ˉ}$ (weight W) and the ordinary mean m (weight n). It lies between them, and it equals the old value only when $\overset{x}{ˉ} = m$ (for example, when all the weights were equal to begin with).

Generally it changes: adding the same number to all weights pulls the weighted average towards the ordinary mean (unlike doubling, which changes nothing).

A farm has some cows, sheep, and chickens. Last year the cows made up 60%, the sheep 25%, and the chickens 15%. There was a decrease in the number of all three animals' population over the year. Choose the possibilities for the change in their respective shares of the population. (i) % of cows decreased, % of sheep decreased, % of chickens decreased (ii) % of cows increased, % of sheep increased, % of chickens increased (iii) % of cows remained the same, % of sheep remained the same, % of chickens remained the same (iv) % of cows decreased, % of sheep increased, % of chickens remained the same (v) % of cows increased, % of sheep increased, % of chickens decreased.

Solution

The three shares must always add up to 100%. So they cannot all go down, and they cannot all go up. Anything else is possible, depending on how much each population decreased.

(i) Impossible (the total would be less than 100%).
(ii) Impossible (the total would be more than 100%). Take last year's farm as 120 cows, 50 sheep and 30 chickens (200 animals: 60%, 25%, 15%). In each example below, all three numbers go down.

(iii) Possible: 96 cows, 40 sheep, 24 chickens (160 animals) give 60%, 25%, 15% again.
(iv) Possible: 104 cows, 49 sheep, 27 chickens (180 animals) give about 57.8%, 27.2% and exactly 15%.
(v) Possible: 114 cows, 48 sheep, 18 chickens (180 animals) give about 63.3%, 26.7% and 10%.

(iii), (iv) and (v) are possible; (i) and (ii) are impossible because the shares must still total 100%.

← Chapter 9: Propositions and their Converses Chapter 11: The World of Algorithms →

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