NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 10 How Quantities Combine: Understanding Data

Class 9 Ganita Manjari Chapter 10 Quick Links:

• Think and Reflect
• Exercise Set 10.1
• Exercise Set 10.2
• Exercise Set 10.3
• Exercise Set 10.4
• Exercise Set 10.5
• End of Chapter Exercises

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Class 9 Maths Ganita Manjari Chapter 10 Think and Reflect Solutions

Answer:
There are 8 seniors and only 3 juniors. Method 1 gives equal importance to the two group averages even though the groups have different numbers of trainees.
Total height of seniors = 1324 cm.
Total height of juniors = 448 cm.
Correct average = (1324 + 448) ÷ (8 + 3)
= 1772 ÷ 11 ≈ 161.09 cm.
Thus, the two averages must be weighted by 8 and 3. Taking their simple average gives the wrong answer here.

Answer:
A stacked bar chart shows actual quantities and their totals. A 100% stacked bar chart shows the fraction or percentage of the total belonging to each part. Equal-length 100% bars do not mean equal actual totals. Therefore, the two charts answer different kinds of questions.

Answer:
Yes, provided each total is non-zero and the values can be read.
For each bar, first add its parts to find its total. Then calculate:
Percentage of a part = (value of the part ÷ total of that bar) × 100.
Draw all the new bars with equal length, representing 100%, and divide them using these percentages.
Example: Parts 20, 30, 50 have total 100 and become 20%, 30%, 50%.

Answer:
Not uniquely from the percentages alone. We must also know the actual total represented by each bar.
For example, 40% and 60% could mean 40 and 60 out of 100 or 80 and 120 out of 200.
If the total is known, use:
Actual value of a part = (percentage ÷ 100) × total.

Answer:
From a stacked bar chart, we can compare actual totals, actual amounts in each category and how the parts contribute to a total.
From a 100% stacked bar chart, we can compare proportions within a group and the percentage share of a category across groups.
Without the group totals, a larger percentage in one group does not necessarily mean a larger actual quantity than in another group.

Answer:
Observations based on Fig. 10.6:
Sleep occupies a large part of the day in every age group: about 38% for children, 34% for youth, 32% for adults and 37% for the elderly.
Children spend about 22% of the day learning, whereas youth spend about 11%.
Adults spend about 14% in paid work and 15% in unpaid work and care.
The elderly have the largest share for leisure, social activities and travel, about 33%.
These are group averages, so they need not describe the routine of every individual.

Answer:
The general trend of children sleeping longer than younger and middle-aged adults agrees with this chart.
Children: 38% of 24 hours = 9.12 hours.
Youth: 34% of 24 hours = 8.16 hours.
Adults: 32% of 24 hours = 7.68 hours.
However, the elderly group rises to 37% or 8.88 hours. Thus, this chart does not show sleep decreasing steadily throughout all ages; it decreases up to adulthood and then increases for the elderly group.

Answer:
A possible explanation is that most children aged 6-14 spend much of their day in school and study. The 15-24 group includes students as well as people who have completed or left formal education and have started paid work or household responsibilities. This can lower the group’s average learning time.
The graph shows the difference; it does not by itself establish the reasons for it.

Answer:
Yes, approximately. For adults aged 25-59, paid work is about 14% and unpaid work and care about 15% of the day.
Paid work ≈ 14/100 × 24 = 3.36 hours.
Unpaid work and care ≈ 15/100 × 24 = 3.60 hours.
The difference is about 0.24 hour or 14.4 minutes. The averages are close, though individual people’s routines can differ greatly.

Answer:
Yes. Each full strip represents one complete day or 24 hours = 100%. Joining all the small pieces for the same activity makes one segment for that activity. Its length represents the activity’s share of the day.
Using the same activity order and colours for every strip gives a 100% stacked bar chart. This grouping preserves the total time for each activity but loses information about when during the day the activity happened.

Answer:
They may look similar because teachers can have similar work routines, but they need not be identical. A different teacher may spend different amounts of time sitting, standing, travelling, sleeping or doing other activities. Each person’s actual data determines the bar.

Answer:
They may look different because weekend routines can differ from weekday routines.
For each body-state:
Seven-day daily average = (5 × weekday daily average + Saturday hours + Sunday hours) ÷ 7.
If the weekend average matches the weekday average, the bar stays the same. Otherwise, its proportions change.

Answer:
Yes. Julie’s bar could also fit a student, accountant or tailor. Raghu’s bar could fit a shop assistant or someone whose work involves much standing and walking. Pavani’s bar could fit someone resting during recovery from an illness.
Different activities can produce similar amounts of time in the same body-states. Therefore, the bars alone cannot tell us a person’s occupation.

Class 9 Maths Ganita Manjari Chapter 10 Exercise Set 10.1 Solutions

Answer:
Total score of Section A = 72 × 30 = 2160.
Total score of Section B = 76 × 25 = 1900.
Total students = 30 + 25 = 55.
Combined average = (2160 + 1900) ÷ 55
= 4060 ÷ 55 = 812/11 ≈ 73.82.
So, the combined average score is approximately 73.82.

Answer:
The quantities are equal, so all three fractions have equal weights.
Fraction of nitrogen = (1/10 + 9/50 + 3/60) ÷ 3
= (10/100 + 18/100 + 5/100) ÷ 3
= (33/100) ÷ 3 = 11/100.
So, 11/100 of the mixture is nitrogen.

Answer:
Total quantity of alloy = 9 + 5 + 17 = 31 units.
Purity = (9 × 12 + 5 × 10 + 17 × 11) ÷ 31
= (108 + 50 + 187) ÷ 31
= 345/31 ≈ 11.13 varṇa.
So, the combined gold has a purity of approximately 11.13 varṇa.

Answer:
May has 31 days, June has 30 days, and July has 31 days.
Required expression = (3.5 × 31 + 10 × 30 + 8.7 × 31) ÷ (31 + 30 + 31).
Its value is (108.5 + 300 + 269.7) ÷ 92
= 678.2 ÷ 92 ≈ 7.37 mm per day.

(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.
Answer:
(i) Total quantity = 100 + 200 + 300 = 600 mL.
Concentration = (100 × 5 + 200 × 10 + 300 × 15) ÷ 600
= 7000/600% = 35/3% ≈ 11.67%.

(ii) Total quantity = 300 + 200 + 100 = 600 mL.
Concentration = (300 × 5 + 200 × 10 + 100 × 15) ÷ 600
= 5000/600% = 25/3% ≈ 8.33%.
The first mixture has a greater concentration because more of it comes from the 15% mixture.

Class 9 Maths Ganita Manjari Chapter 10 Exercise Set 10.2 Solutions

(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)
Answer:
First express all three scores on the same scale.
Internal tests = 35/50 × 100 = 70%.
Project = 44/60 × 100 = 220/3%.
Final exam = 80/100 × 100 = 80%.
Annual percentage = [70 × 3 + (220/3) × 4 + 80 × 5] ÷ 12
= [210 + 880/3 + 400] ÷ 12
= 1355/18% ≈ 75.28%.

Here, expressions (iii) and (iv) are correct.
Expression (i) combines marks with different maximum marks without converting them to a common scale. Expression (ii) incorrectly uses 100 as the maximum for every component and does not convert the resulting fraction into a percentage.

Class 9 Maths Ganita Manjari Chapter 10 Exercise Set 10.3 Solutions

Answer:
Profit from books = 30/100 × 8000 = ₹2400.
Profit from book covers = 50/100 × 1000 = ₹500.
Total profit = ₹2900; total sales = ₹9000.
Percentage of profit on sales = 2900/9000 × 100
= 290/9% ≈ 32.22%.
So, approximately 32.22% of the total sales is profit.

Answer:
Guess: The new average should be slightly above 44.5 km, perhaps about 45 km, because only one extra day is being added.

Distance in the first 20 days = 44.5 × 20 = 890 km.
Total distance = 890 + 55 = 945 km.
New average = 945 ÷ 21 = 45 km per day.

(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%
Answer:
Total solution = 600 + 300 = 900 mL.
Only the first solution contributes salt; only the second contributes sugar.
Salt concentration = (600 × 5 + 300 × 0) ÷ 900
= 10/3% ≈ 3.33%.
Sugar concentration = (600 × 0 + 300 × 8) ÷ 900
= 8/3% ≈ 2.67%.
So, Option (v).

Answer:
Required concentration = 3/4 × 8% = 6%.
Let x litres of regular water be added. It contains no spice.
The amount of spice stays the same, while the total mixture becomes 10 + x litres.
⇒ (10 × 8 + x × 0)/(10 + x) = 6
⇒ 80 = 60 + 6x
⇒ 6x = 20
⇒ x = 10/3 = 3⅓.
So, Add 3⅓ litres of regular water.

(i) Find out whose total is more without calculating.
(ii) What are their final marks?
Answer:
(i) Keerthi’s final marks are higher. Compared with Rashi, she loses 5 marks in strength but gains 5 in agility. Agility has the greater weight, so the gain counts more than the loss. Their flexibility marks are equal.

(ii) Rashi’s final marks = (60 × 4 + 65 × 5 + 70 × 6) ÷ 15
= (240 + 325 + 420) ÷ 15 = 985/15 = 65⅔ ≈ 65.67.
Keerthi’s final marks = (55 × 4 + 65 × 5 + 75 × 6) ÷ 15
= (220 + 325 + 450) ÷ 15 = 995/15 = 66⅓ ≈ 66.33.

What is the average rating if the metrics are combined with the weights food : ambience : service = 6 : 5 : 4?
Answer:
Food average = (5 × 5 + 3 × 4 + 2 × 3) ÷ 10 = 43/10 = 4.3.
Ambience average = (4 × 4 + 5 × 3 + 1 × 2) ÷ 10 = 33/10 = 3.3.
Service average = (1 × 5 + 2 × 4 + 2 × 3 + 4 × 2 + 1 × 1) ÷ 10 = 28/10 = 2.8.
Combined average = (4.3 × 6 + 3.3 × 5 + 2.8 × 4) ÷ 15
= (25.8 + 16.5 + 11.2) ÷ 15
= 53.5/15 = 107/30 ≈ 3.57.
So, approximately 3.57 out of 5.

(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?
Answer:
There are more female langurs. The combined average is nearer the female average, so the female group contributes more individuals.

(i) Let x be the number of males and y the number of females. Then x + y = 60.
Correct relation: (16.5x + 13.8y)/(x + y) = 14.925.
From the given options, only option (a) is correct.
Options (b), (c) and (d) show a minus sign before 14.925. Average weight cannot be negative here. If the minus sign in (b) is removed, (b) is also correct because x + y = 60. Options (c) and (d) still have incorrect denominators.

(ii) Using y = 60 − x:
16.5x + 13.8(60 − x) = 14.925 × 60
16.5x + 828 − 13.8x = 895.5
2.7x = 67.5
x = 25.
Thus, there are 25 males and 60 − 25 = 35 females.

(iii) Original total female weight = 35 × 13.8 = 483 kg.
New total = 483 + 15.2 = 498.2 kg.
New number = 36.
New female average = 498.2 ÷ 36 ≈ 13.839 kg, or 13.84 kg.

(iv) Original total male weight = 25 × 16.5 = 412.5 kg.
Weight removed = 16.9 + 16.1 = 33 kg.
Remaining total weight = 379.5 kg; remaining number = 23.
New male average = 379.5 ÷ 23 = 16.5 kg.

(v) Continuing after part (iv), the number remains 23 and the total weight falls by 1 kg.
New average = (379.5 − 1) ÷ 23
= 378.5/23 ≈ 16.457 kg, or 16.46 kg.

(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?
Answer:
The given volumes add and concentration is weighted by volume. The starting percentages are approximate, so the answers are approximate too.

(i) Salinity = (1 × 34 + 2 × 0.001) ÷ 3
= 34.002/3% = 11.334%.

(ii) Yes. The required 0.01% lies between 0.001% and 34%.
Let x litres of purified drinking water be added.
(34 + 0.001x)/(1 + x) = 0.01
34 + 0.001x = 0.01 + 0.01x
33.99 = 0.009x
x = 33.99/0.009 ≈ 3776.67 litres.
So, about 3776.67 litres of purified drinking water.

(iii) No. Both waters have salinity greater than 0.001%. Mixing them cannot produce a salinity below the smaller starting concentration, 0.01%.
Algebra confirms this:
(34 + 0.01x)/(1 + x) = 0.001
0.009x = −33.999
x ≈ −3777.67 litres.
A negative amount cannot be added. Therefore, no non-negative quantity of groundwater can give the required salinity.

Class 9 Maths Ganita Manjari Chapter 10 Exercise Set 10.4 Solutions

(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?
Answer:
(i) It represents the total number of animal species across all the classes shown in the chart for 2019: 14,234 species.
(ii) Approximately 1,000 reptile species (about 1,000-1,200 is a reasonable reading of this small chart).
(iii) Mammals and birds show relatively small increases. Their segment heights change only a little compared with those of groups such as fish, reptiles and insects.

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

Answer:
(i) Stack the segments in the order Test, ODI, T20.
Home total = 62 + 100 + 32 = 194 wickets.
Home boundaries: 0, 62, 162, 194.
Overseas total = 172 + 49 + 71 = 292 wickets.
Overseas boundaries: 0, 172, 221, 292.
Thus, more wickets were taken overseas.

(ii) Stack Home first, then Overseas.
Test total = 62 + 172 = 234; boundaries: 0, 62, 234.
ODI total = 100 + 49 = 149; boundaries: 0, 100, 149.
T20 total = 32 + 71 = 103; boundaries: 0, 32, 103.
Thus, the greatest number of wickets was taken in Test matches.

Class 9 Maths Ganita Manjari Chapter 10 Exercise Set 10.5 Solutions

Table 10.3: Average number of hours spent per day (Monday–Friday) lying down, sitting, and standing or moving around, etc., of 5 people

(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.

Answer:
(i) My guess is: lying down 8 hours, sitting 10 hours, standing/moving around 6 hours.
Total: 8 + 10 + 6 = 24 hours.
This allows time for sleep, seated lessons and homework and movement, play and daily activities.

(ii) According to me, Julie could be an office worker who spends much of the working day seated. About 7 of my classmates agree with me but rest are guessing a tailor or a student or something else.

(iii) For each state, percentage of day = hours/24 × 100.
Person | Lying down | Sitting | Standing/moving
Pavani | 83⅓% | 12½% | 4⅙%
Raghu | 29⅙% | 16⅔% | 54⅙%
Zakir | 33⅓% | 25% | 41⅔%
Sahana | 33⅓% | 41⅔% | 25%
Julie | 33⅓% | 41⅔% | 25%

For Sahana, the first segment ends at 33⅓%, the second at 75% and the third at 100%.

Class 9 Maths Ganita Manjari Chapter 10 End-of-Chapter Exercises Solutions

(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?
Answer:
(i) Total runs after 2 overs = 6 + 12 = 18.
Run rate = 18 ÷ 2 = 9 runs per over.

(ii) Runs after 19 overs = 19 × 6 = 114.
Runs after 20 overs = 114 + 12 = 126.
Run rate = 126 ÷ 20 = 6.3 runs per over.

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?
Answer:
Each friend can report the sum of the prices collected and the number of prices. Adding all the sums and dividing by the total number of prices. Equivalently, combine their averages using these numbers as weights.

(i) Yusuf collected 6 prices, Srikanth 4, Kashvi 5, Prasanna 4 and Gracy 4.
Total number = 6 + 4 + 5 + 4 + 4 = 23.
Combined average = (6 × a_y + 4 × a_s + 5 × a_k + 4 × a_p + 4 × a_g)/23.
Now, their price totals are ₹917, ₹493, ₹766, ₹897 and ₹990.
Combined average = (917 + 493 + 766 + 897 + 990) ÷ 23
= 4063/23 ≈ ₹176.65.

(ii) There are 13 nylon prices and 10 feather prices.
Required expression = (13 × a_n + 10 × a_f)/23.
Nylon price total = ₹1429; feather price total = ₹2634.
Thus, a_n = 1429/13 and a_f = 2634/10.
Combined average = (1429 + 2634) ÷ 23 = 4063/23 ≈ ₹176.65.
Yes, both methods give the same result because they combine the same 23 observations, grouped in different ways.

(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?
Answer:
(i) Original cost = 25 × 150 = ₹3750.
New purchase cost = 10 × 30 = ₹300.
Total cost = ₹4050; total shares = 35.
Average price = 4050/35 = 810/7 ≈ ₹115.71 per share.

(ii) Let x be the number of shares bought.
⇒ (5 × 150 + 30x)/(5 + x) = 70
⇒ 750 + 30x = 350 + 70x
⇒ 400 = 40x
⇒ x = 10.
So, Vaishnavi bought 10 shares.

Answer:
The depths must be weighted by the lengths over which they occur.
Mean depth = (4 × 9 + 5 × 7 + 6 × 7 + 7 × 3 + 8 × 2) ÷ 30
= (36 + 35 + 42 + 21 + 16) ÷ 30
= 150 ÷ 30 = 5 hastas.
So, the mean depth is 5 hastas.

(i) Purchase 1g gold at ₹30k
(ii) Purchase 2g gold at ₹30k
(iii) Purchase 1g gold at ₹10k
(iv) Purchase 10g gold at ₹10k
(v) Purchase 0.5g gold at ₹30k
(vi) Purchase 0.5g gold at ₹15k
Answer:
Average price per gram = total cost ÷ total grams.

(i) Equal quantities at 15 and 30 give the midpoint.
Average = (1 × 15 + 1 × 30)/(1 + 1) = 22.5.
So, the new average = ₹22,500 per gram.

(ii) More gold is bought at 30, so the result is nearer 30.
Average = (1 × 15 + 2 × 30)/(1 + 2) = 75/3 = 25.
So, the new average = ₹25,000 per gram.

(iii) Equal quantities at 15 and 10 give the midpoint.
Average = (15 + 10)/2 = 12.5.
So, the new average = ₹12,500 per gram.

(iv) Most of the gold is bought at 10, so the average is just above 10.
Average = (1 × 15 + 10 × 10)/(1 + 10) = 115/11 ≈ 10.45.
So, the new average (approximately) = ₹10,454.55 per gram.

(v) Only half as much is bought at 30 as was originally held at 15.
Average = (1 × 15 + 0.5 × 30)/(1 + 0.5) = 30/1.5 = 20.
So, the new average = ₹20,000 per gram.

(vi) Both purchases have the same rate, so the average does not change.
Average = (1 × 15 + 0.5 × 15)/(1 + 0.5) = 22.5/1.5 = 15.
So, the new average = ₹15,000 per gram.

Answer:
The weighted average remains unchanged.
Example:
For values 10 and 20 with weights 1 and 3,
Original average = (1 × 10 + 3 × 20)/(1 + 3) = 70/4 = 17.5.
After doubling the weights,
New average = (2 × 10 + 6 × 20)/(2 + 6) = 140/8 = 17.5.

Algebraic justification:
Let M = (w₁x₁ + w₂x₂ + ⋯ + wₙxₙ)/(w₁ + w₂ + ⋯ + wₙ).
After doubling every weight,
M′ = (2w₁x₁ + 2w₂x₂ + ⋯ + 2wₙxₙ)/(2w₁ + 2w₂ + ⋯ + 2wₙ)
= 2(w₁x₁ + w₂x₂ + ⋯ + wₙxₙ)/[2(w₁ + w₂ + ⋯ + wₙ)]
= M.
The factor 2 cancels. The relative weights remain the same.

(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?

Answer:
(i) The 2003 column is approximately 10,000 objects. Twice this is about 20,000, which the graph reaches around 2021.

(ii) A reading for 2025 is about 33,000 objects in total: approximately 17,000 payload objects and 16,000 other objects.
Payload share ≈ 17,000/33,000 × 100 ≈ 51.5% or about 52%.
Other-object share ≈ 16,000/33,000 × 100 ≈ 48.5%, or about 48%.

(iii) Both cumulative counts increase over time. Payload objects rise particularly sharply in the most recent years. Other objects also increase overall, with a noticeable rise around 2007.
The payload share is relatively small through much of the earlier period and becomes much larger in recent years, reaching about half of the total. The other-object share correspondingly becomes smaller in recent years, even though its actual count remains large and grows overall.

(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?
Answer:
(i) Correct inferences: (b) and (d).

• (a) Cannot be inferred. Haryana has a higher percentage (90%) than Uttarakhand (78%), but their total numbers of schools are not given.
• (b) Correct. Arunachal Pradesh is shown at about 68%, close to 2/3 × 100 = 66⅔%.
• (c) Cannot be inferred. A high percentage does not establish the highest number of schools with playgrounds.
• (d) Correct. Maharashtra has a higher percentage, 93%, than Telangana, 74%. It is also given to have more schools. Therefore, it must have more schools with playgrounds.

If their totals are M and T, then M > T, so 0.93M > 0.93T > 0.74T.

(ii) No. We need the total number of schools in each state and union territory, along with the corresponding percentages (or the actual numbers with playgrounds).
National percentage = (total schools with playgrounds in India ÷ total schools in India) × 100.
If N₁, N₂, … are the total school counts and p₁, p₂, … the numerical percentages,
National percentage = (N₁p₁ + N₂p₂ + ⋯)/(N₁ + N₂ + ⋯).
The state percentages must be weighted by their numbers of schools.

(i) Which of the plays would you choose to watch based on the following chart?.

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

Answer:
(i) I would choose Play A. Its combined share of 4-star and 5-star ratings is high and its 5-star share is slightly greater than Play B’s. Play C has a larger 5-star share but also a much larger 1-star share, so opinions about it are more divided.

(ii) I might change my choice to Play B. The second chart shows that Play B has far more ratings, roughly 900 or more, compared with about 200 for A and about 400 for C. Its high proportion of favourable ratings is therefore supported by more responses.

(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?
Answer:
(i) A stacked bar chart using actual time is more suitable for comparing race performance. It shows both each segment’s time and each athlete’s total time.
Athlete 1: 1 h 08 min + 5 h + 3 h 15 min = 9 h 23 min = 563 min.
Athlete 2: 1 h 05 min + 5 h 10 min + 3 h 35 min = 9 h 50 min = 590 min.
Athlete 3: 1 h 22 min + 5 h 55 min + 3 h 50 min = 11 h 07 min = 667 min.
The shortest total bar belongs to Athlete 1, the fastest finisher.

(ii) Only (b) can be answered from a 100% stacked bar chart alone.
Athlete 1’s cycling fraction = 300/563 ≈ 0.533 or about 53.3%, slightly more than half.
(a) cannot be answered because equal-length 100% bars hide the actual total times.
(c) cannot be answered because the greatest percentage of time spent running need not be the greatest actual running time.
The original table shows that Athlete 3 took longest for running, but this information is not available from the percentage chart alone.

Answer:
What I notice:

• Each age-group bar totals 100%, so the chart compares the types of disability within each age group.
• The shares for seeing and movement difficulties are larger in the oldest group than in the youngest group.
• Movement difficulties occupy roughly one-quarter of the oldest group’s bar and form one of its largest parts.
• The share for speech difficulties is smaller in the oldest group than in the youngest group.

What I wonder:

• How many people are represented by each bar? How might access to support differ across age groups? How might the distribution have changed since the 2011 Census?

Inferences:

• The distribution of disability types varies with age group.
• The graph does not show the total number of persons with disabilities in each age group, so we cannot decide which group contains the largest number from these percentages alone.
• It also does not show what percentage of all people in each age group have a disability; its percentages are within the population represented by each bar.
• The chart alone does not establish the causes of the differences.

(i) Recall the previous day and fill in the approximate time spent lying down, sitting, and standing/moving around, etc. Ask at least 2 of your family members about their day and fill it in. Alternatively, you can choose to track just your own body-state for any one of the weekdays, Saturday, and Sunday. Visualise it using a stacked bar 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% stacked bar chart.
(a) First, identify the expense categories in your family budget.
(b) Discuss with your family members to decide how you will collect and organise the data.
(c) Collect expense data for at least 3 months.
(d) Represent the data in a 100% stacked bar chart, with each bar showing how the total monthly expenditure is divided across the different categories.
(e) Finally, write your observations and inferences based on the chart.
Answer:
(i) Body-state project

Method of estimating:
Making a time diary from waking to bedtime and include sleep. Classifying each time interval by body-state, rather than counting overlapping activities separately. For example, seated travel counts as sitting. Add the intervals for each state and checking that each day totals 24 hours. Asking the two family members to check their own estimates.

To draw the chart, using a horizontal scale of 0 to 24 hours and stack lying down, sitting and standing/moving in that order.
My (Student) boundaries: 0, 8, 18, 24.
Family member A boundaries: 0, 8, 16, 24.
Family member B boundaries: 0, 7, 12, 24.

Sample observations:

• I (student) spend the most time sitting: 10 hours.
• Family member B spends the most time standing or moving: 12 hours.
• All three bars have the same total, 24 hours, but their parts differ.

(ii) Monthly expenditure project
(a) Categories: housing, food, education, transport, healthcare and other expenses.

(b) Agree with family members to record expenses in a notebook or spreadsheet. Use bills and payment records, place each expense in one category only and add the amounts at the end of each month.

(c) Data for three months, in rupees:

(d) Converting each amount into a percentage of its own month’s total.
Example: Month 1 housing = 10000/30000 × 100 ≈ 33.33%.

Drawing three equal bars, each representing 100%, divided in the same category order. Use unrounded values to set the segment lengths.

(e) Observations:

• Total monthly spending rises from ₹30,000 to ₹35,000, as shown by the data table.
• Housing spending stays at ₹10,000, but its share falls from about 33.33% to 28.57% because the total rises.
• Healthcare increases both in rupees and as a share of the budget.
• Food rises from ₹7,500 to ₹8,400 while its share falls from 25% to 24%.
• The percentage chart alone does not show the rise in total spending; the totals must be supplied separately.

(i) Discuss and arrive at 4–6 aspects to rate. For each aspect chosen, justify why it matters for an overall experience.
(ii) Discuss and decide the relative weights and justify.
(iii) Collect or imagine ratings from at least 10 people on a scale of 1–5 (5 being very good).
(iv) Compute each individual rating. Compute the overall average across individuals.
(v) Make a 100% stacked bar chart using your data.
(vi) Write a short note on the observations and inferences, what your rating system captures well and what it misses.
Answer:
Chosen scenario: Travel experience in a bus.
(i) and (ii) Aspects and weights:

The weights total 10. Safety receives the greatest importance, followed by punctuality, comfort and cleanliness.

(iii) Ratings from 10 passengers:

(iv) Individual rating R = (4S + 3P + 2C + L)/10.
Passenger 1: (20 + 12 + 8 + 3)/10 = 4.3.
Passenger 2: (16 + 12 + 6 + 4)/10 = 3.8.
Passenger 3: (20 + 15 + 8 + 4)/10 = 4.7.
Passenger 4: (12 + 12 + 6 + 3)/10 = 3.3.
Passenger 5: (16 + 9 + 8 + 3)/10 = 3.6.
Passenger 6: (20 + 12 + 10 + 4)/10 = 4.6.
Passenger 7: (16 + 15 + 8 + 5)/10 = 4.4.
Passenger 8: (12 + 9 + 6 + 4)/10 = 3.1.
Passenger 9: (16 + 12 + 8 + 4)/10 = 4.0.
Passenger 10: (20 + 12 + 6 + 5)/10 = 4.3.
Sum of individual ratings = 40.1.
Overall average = 40.1 ÷ 10 = 4.01 out of 5.
Each passenger is given equal weight in final average.

(v) Making one 100% stacked bar for each aspect, showing the proportions of 1-, 2-, 3-, 4- and 5-star ratings. Since there are 10 passengers, each passenger represents 10%.

Each row totals 100%. The chart displays the distribution of the raw ratings for each aspect; the weights are used separately in calculating the individual overall ratings.

(vi) Observations and inferences:

• Safety gets the highest share of 5-star ratings, 40%.
• Comfort has the highest share of 3-star ratings, 40%, suggesting room for improvement in this sample.
• No passenger gives a rating below 3 to any aspect.
• The overall weighted rating is 4.01/5, indicating a generally favourable experience in the imagined data.

I have captured four important parts of the journey and gives safety greater importance. I missed aspects such as fare, accessibility and staff behaviour. Different weights could produce different overall ratings.

Answer:
To find the average age of all family members represented, use each family’s size as the weight.

• Step 1: Record each family’s average age and number of members.
• Step 2: Multiply these to obtain the sum of ages in that family.
• Step 3: Add these sums for all families.
• Step 4: Divide by the total number of family members.

If family averages are a₁, a₂, …, aₙ and family sizes are n₁, n₂, …, nₙ,
Combined average age = (n₁a₁ + n₂a₂ + ⋯ + nₙaₙ)/(n₁ + n₂ + ⋯ + nₙ).

Observation:

• Family A: 4 members, average age 25 years; sum of ages = 100 years.
• Family B: 6 members, average age 30 years; sum of ages = 180 years.
• Family C: 5 members, average age 28 years; sum of ages = 140 years.

Combined average = (100 + 180 + 140)/(4 + 6 + 5)
= 420/15 = 28 years.

Answer:
Unlike multiplying all weights by the same number, adding 1 does not generally preserve their ratios. The weighted mean moves towards the ordinary arithmetic mean. It may increase, decrease or remain unchanged.

Example of an increase:
Values 10 and 20, weights 3 and 1:
Original mean = (30 + 20)/4 = 12.5.
After adding 1 to each weight: new mean = (40 + 40)/6 = 13⅓.

Example of a decrease:
Values 10 and 20, weights 1 and 3:
Original mean = (10 + 60)/4 = 17.5.
New mean = (20 + 80)/6 = 16⅔.

Example of no change:
Values 10 and 20, equal weights 1 and 1 give mean 15. New equal weights 2 and 2 also give mean 15.

Justification:
Let there be n values x₁, x₂, …, xₙ with total weight W = w₁ + w₂ + ⋯ + wₙ > 0.
Let their weighted mean be M, so w₁x₁ + w₂x₂ + ⋯ + wₙxₙ = WM.
Let their ordinary arithmetic mean be A, so x₁ + x₂ + ⋯ + xₙ = nA.
After adding 1 to each weight,
M′ = [(w₁ + 1)x₁ + (w₂ + 1)x₂ + ⋯ + (wₙ + 1)xₙ]/(W + n)
= (WM + nA)/(W + n).
Therefore, M′ − M = n(A − M)/(W + n).

If A > M, the weighted mean increases.
If A < M, it decreases.
If A = M, it remains unchanged.
For a positive constant c added to every weight, the same argument gives M′ = (WM + cnA)/(W + cn).

(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.
Answer:
(iii), (iv) and (v) are possible.
The three shares must always add to 100%. Thus, all three cannot decrease together and all three cannot increase together. Options (i) and (ii) are impossible.
However, all three actual populations can decrease while their percentage shares change differently.
To demonstrate the possible cases, suppose last year there were 200 animals:
Cows = 120, sheep = 50, chickens = 30.
These are 60%, 25% and 15%, respectively.

(iii) New numbers: 60 cows, 25 sheep, 15 chickens; total 100.
New shares: 60%, 25%, 15%.
Every actual number has decreased and every share is unchanged.

(iv) New numbers: 55 cows, 30 sheep, 15 chickens; total 100.
New shares: 55%, 30%, 15%.
Cows’ share decreases, sheep’s share increases and chickens’ share stays the same. All three actual numbers are below their original numbers.

(v) New numbers: 65 cows, 26 sheep, 9 chickens; total 100.
New shares: 65%, 26%, 9%.
Cows’ and sheep’s shares increase, while chickens’ share decreases. Again, all three actual numbers have decreased.