How Quantities Combine:
Understanding Data
Chapter 10 • Complete exercise solutions • Your step-by-step virtual teacher
Based on the supplied Ganita Manjari, Grade 9, Part II chapter, printed pages 8–36. Includes Exercise Sets 10.1–10.5, all 16 end-of-chapter questions, model projects and supplementary reflection answers. Questions are restated briefly; the numbering follows your chapter.
Before you begin: the teacher’s toolkit
An average is a total shared equally. Whenever you combine groups, first reconstruct their totals. Ask: “How many observations does each average represent?”
| Situation | Correct weight |
|---|---|
| Combining class averages | Number of students in each class |
| Mixing solutions | Quantity of each solution |
| Combining assessments | Assigned weight, after putting marks on the same scale |
| Average daily rainfall | Number of days |
| Reading a stacked chart | Segment amount = end position − start position |
Mixture convention: As in the textbook, quantities are assumed to add and concentrations use a common basis. The exercises are mathematical models. Salinity alone does not establish whether water is safe to drink.
Exercise Set 10.1
Combining averages and mixtures • Printed page 13
Average marks of two sections
Step 1 — Recover the marks totals. A mean of 72 means the total marks would be unchanged if every student in A scored 72.
Step 2 — Add totals and divide by the total number of students.
Nitrogen in three equal quantities
Step 1 — Recognise equal weights. Let each quantity be 1 unit. Total mixture = 3 units.
Gold purity in varṇa
Step 1 — Convert purity to actual gold. A purity of k varṇa means k/16 of the alloy is gold.
Average daily rainfall
Step 1 — Choose days as weights. May has 31 days, June 30 and July 31. We need a mean per day, not per month.
Spice concentrations in two mixtures
For each mixture, find the spice amount in every portion, then divide the total spice by the total volume.
(i) More of the stronger mixture
(ii) More of the weaker mixture
Exercise Set 10.2
Custom weights • Printed pages 14–15
Which expressions give Savitri’s percentage?
Step 1 — Put the scores on the same scale. Raw marks cannot be compared directly when their maximum marks differ.
Step 2 — Apply the assigned weights.
| Option | Verdict and reason |
|---|---|
| (i) | Incorrect: weights raw marks 35, 44 and 80 despite different maximum marks. |
| (ii) | Incorrect: treats 35 and 44 as marks out of 100 and also omits conversion to percentage. |
| (iii) | Correct: weights the fractions scored, then multiplies by 100. |
| (iv) | Correct: first converts each score to a percentage, then weights them. |
Exercise Set 10.3
Applications of weighted mean • Printed pages 15–18
Profit as a percentage of total sales
Step 1 — Find the actual profit amounts. Here the percentages are explicitly fractions of sales.
A stork’s new daily average
First, estimate. The new mean must rise a little above 44.5 km because 55 km is above the old mean. One new day should not have the same influence as the previous 20 days.
Salt and sugar must be tracked separately
Assume the first solution contains no sugar and the second contains no salt. Mixing increases the total volume but does not create additional salt or sugar.
Diluting panipuri water
Step 1 — Find the target concentration.
Step 2 — Let x litres of plain water be added. Spice stays 0.8, while the mixture becomes 10 + x litres.
Rashi versus Keerthi
(i) Compare without calculating both final scores. Keerthi loses 5 points in strength but gains 5 in agility. The gain carries weight 6 and the loss weight 4; flexibility is unchanged. Therefore Keerthi scores more.
(ii) Calculate their final marks.
Restaurant ratings: two layers of averaging
Step 1 — Convert each frequency row into a mean rating. Multiply each star rating by the number of customers who gave it.
Step 2 — Apply the importance weights to these means.
Langurs: recovering group sizes and updating means
Initial comparison. The overall mean is closer to the female mean, so there are more females. Equal group sizes would put the mean exactly halfway between the two means.
(i) Choose the equation. Let x be the number of males and y the number of females.
Only option (a) is correct as printed. In the supplied PDF, (b), (c) and (d) have −14.925 on the right. An average of positive weights cannot be negative. If the minus sign in (b) is corrected, its denominator 60 is valid too.
(ii) Solve for the number of males.
(iii) Add a female weighing 15.2 kg.
(iv) Release males of 16.9 kg and 16.1 kg.
(v) Then one remaining male loses 1 kg. The number of males stays 23.
Dead Sea water and dilution
(i) Mix with 2 litres of purified water. Use the volume-weighted mean of the percentages.
(ii) Can we reach groundwater salinity? Yes: 0.01% lies strictly between 0.001% and 34%. Let x litres of purified water be added.
(iii) Can groundwater bring it down to 0.001%? No. Groundwater itself has salinity 0.01%, so combining it with saltier water cannot produce 0.001%. Algebra confirms this:
Exercise Set 10.4
Stacked charts • Printed pages 21–22
Reading the animal-species chart
(i) Meaning of 14,234. It is the total represented by the 2019 column: the sum of all eight displayed animal categories. It is not the count of just the top category and should not be interpreted as the size of the entire IUCN database.
(ii) Reptiles in 2016. Identify the red reptile band, above the dark-blue mammal band. Its lower boundary is roughly 1200 and its upper boundary roughly 2300.
A reading of about 1000–1200 species is reasonable from this small graph; 1100 is an estimate, not an exact table value.
(iii) Relatively small increases. The mammal and bird bands change comparatively little in thickness from 2007 to 2019. Other bands, such as fish and molluscs, expand much more.
Complete both cricket charts
(i) Home versus overseas. Keep the order Test → ODI → T20 in each bar. Start each new segment where the previous one ends.
| Bar | Test interval | ODI interval | T20 interval | Total |
|---|---|---|---|---|
| Home | 0–62 | 62–162 | 162–194 | 194 |
| Overseas | 0–172 | 172–221 | 221–292 | 292 |
Wickets by location
(ii) Compare formats. Now each bar represents a format. Use Home → Overseas within every bar.
| Format | Home interval | Overseas interval | Total |
|---|---|---|---|
| Test | 0–62 | 62–234 | 234 |
| ODI | 0–100 | 100–149 | 149 |
| T20 | 0–32 | 32–103 | 103 |
Wickets by format
Exercise Set 10.5
Body-state data • Printed page 29
Complete the table and chart
(i) A reasonable guess for Sahana. Choose 8 hours lying down, 9 hours sitting and 7 hours standing/moving. This might represent sleep/rest, classes/homework and other activities. It is a sample, not measured smartwatch data.
(ii) Julie’s possible occupation. An office worker or accountant is plausible because she spends 10 hours sitting. This is a guess: the chart cannot establish her occupation.
(iii) Convert hours to percentages. Every complete day is 24 hours, so each hour is 100/24 = 4⅙% of the bar.
| Person | Lying down | Sitting | Standing / moving |
|---|---|---|---|
| Pavani | 20 h ≈ 83.33% | 3 h ≈ 12.50% | 1 h ≈ 4.17% |
| Raghu | 7 h ≈ 29.17% | 4 h ≈ 16.67% | 13 h ≈ 54.17% |
| Zakir | 8 h ≈ 33.33% | 6 h ≈ 25.00% | 10 h ≈ 41.67% |
| Sahana · sample | 8 h ≈ 33.33% | 9 h ≈ 37.50% | 7 h ≈ 29.17% |
| Julie | 8 h ≈ 33.33% | 10 h ≈ 41.67% | 6 h ≈ 25.00% |
Completed 100% stacked chart
How to draw by hand: Take a 12 cm bar for every person. Each hour equals 0.5 cm. For Pavani, draw consecutive lengths 10 cm, 1.5 cm and 0.5 cm. For Sahana’s sample row, use 4 cm, 4.5 cm and 3.5 cm.
End-of-Chapter Exercises
All questions 1–16, including both starred questions • Printed pages 30–36
Cricket run rate
(i) After two overs. A run rate is total runs divided by overs completed.
(ii) After twenty overs. The rate of 6 represents all 19 earlier overs.
Combining shuttlecock price surveys
Work together: Each friend reports both the sum of their prices and the number of prices. Add these sums and counts. Treat each listed observation as one entry, as the exercise intends.
| Friend | Count | Sum (₹) | Mean (₹) |
|---|---|---|---|
| Yusuf | 6 | 917 | 917/6 ≈ 152.83 |
| Srikanth | 4 | 493 | 123.25 |
| Kashvi | 5 | 766 | 153.20 |
| Prasanna | 4 | 897 | 224.25 |
| Gracy | 4 | 990 | 247.50 |
(i) Weight each friend’s mean by their count.
(ii) Regroup by material. There are 13 nylon entries totalling ₹1429 and 10 feather entries totalling ₹2634.
Average purchase price of shares
(i) Shreyas buys 10 shares.
(ii) Vaishnavi’s new mean is ₹70. Let n be the number of shares bought.
Check: (₹750 + ₹300) ÷ 15 = ₹70.
Mean depth of an uneven pool
Each depth applies to a different length of the pool. Therefore use lengths as weights. For a uniform-width pool this also gives the depth of a flat-bottomed pool with the same volume.
Six independent gold purchases
Use total money spent ÷ total grams. A higher purchase price pulls the mean right; a lower price pulls it left. The larger purchase has a stronger pull.
| Part | New purchase | Calculation: ₹k per g | Position on 0–30 scale |
|---|---|---|---|
| (i) | 1 g at ₹30k/g | (15 + 1 × 30) ÷ 2 = 22.5 | Halfway between 20 and 25 |
| (ii) | 2 g at ₹30k/g | (15 + 2 × 30) ÷ 3 = 25 | At 25 |
| (iii) | 1 g at ₹10k/g | (15 + 10) ÷ 2 = 12.5 | Halfway between 10 and 15 |
| (iv) | 10 g at ₹10k/g | (15 + 100) ÷ 11 = 115/11 ≈ 10.45 | Just to the right of 10 |
| (v) | 0.5 g at ₹30k/g | (15 + 0.5 × 30) ÷ 1.5 = 20 | At 20 |
| (vi) | 0.5 g at ₹15k/g | (15 + 0.5 × 15) ÷ 1.5 = 15 | At the original point O |
Detailed example, part (iv): The first gram costs ₹15k and the next 10 grams cost ₹100k altogether. Total cost is ₹115k for 11 grams. Because most of the gold was bought at ₹10k/g, the mean is close to 10.
22.5030
25030
12.5030
≈10.45030
20030
15030
Doubling every weight
Prediction: Nothing changes because relative importance stays the same. Every contribution and the total weight are both doubled.
Example: Values 10 and 20 with weights 1 and 3 have mean 70/4 = 17.5. With weights 2 and 6 the mean is 140/8 = 17.5.
Objects orbiting Earth: count versus share
Graph-reading note: Values below are visual estimates. This graph counts objects still in orbit at the snapshot date by their launch/separation years; it is not a complete historical census of everything that was orbiting in each past year.
(i) Total around 2003 and its doubling. The column is near 10,000. Double that is about 20,000, reached around 2021 (a nearby year may be reasonable when reading the small print).
(ii) Approximate counts and shares for 2025. The 2025 column, just before the final 2026 column, is about 33,000 in total, with roughly 17,000 payload objects and 16,000 other objects.
(iii) Trends. Both cumulative counts increase across the chart. Payload growth is especially steep in the last few years. Other-object counts also grow, but their proportion falls in the recent years as payloads grow faster. Earlier in the chart, other objects dominate. Do not claim that the payload share rises steadily in every single year.
Schools with playgrounds
(i)(a) More schools in Haryana than Uttarakhand have playgrounds. Not inferable. A greater percentage does not prove a greater count when the total numbers of schools are unknown. For example, 90% of 1000 = 900, while 78% of 2000 = 1560.
(i)(b) Approximately two out of three Arunachal Pradesh schools have playgrounds. Correct: 2/3 × 100% ≈ 66.67%, close to the shown 68%.
(i)(c) Punjab has the highest number of such schools. Not inferable: the map gives percentages, not state school totals.
(i)(d) If Maharashtra has more schools than Telangana, it has more schools with playgrounds. Correct: it has both a larger school total and a larger proportion, 93% versus 74%. If M > T > 0, then 0.93M > 0.93T > 0.74T.
(ii) National percentage. The state percentages alone are insufficient. We need the total school count Nᵢ in each state/UT (and coverage of all states/UTs).
Here each pᵢ is the numeric percentage, such as 90, so do not multiply this final expression by 100 again. Exact counts of schools with playgrounds would avoid rounding in map percentages.
Decision dilemma: choosing a play
(i) One reasoned choice: Play A. A and B both have roughly 71% four- or five-star reviews. A has a slightly larger five-star share than B and relatively few very low ratings. Play C has the largest five-star share, but also a much larger one-star share, so its reviews are more divided.
Approximate shares read from the printed chart are shown below. These are estimates, not exact source data.
| Play | 5★ | 4★ | 3★ | 2★ | 1★ |
|---|---|---|---|---|---|
| A | 29% | 42% | 14% | 3% | 12% |
| B | 24% | 47% | 14% | 4% | 11% |
| C | 39% | 14% | 4% | 10% | 33% |
(ii) A defensible revision: choose Play B. The second chart shows about 210 reviews for A, about 920 for B and about 380 for C. B’s similarly favourable pattern is based on many more reviews, which may make it more persuasive if the reviews were collected comparably.
You could also keep A if your preference is its slightly larger five-star share. There is no unique compulsory choice: explain your criterion. A larger sample does not guarantee unbiased or authentic reviews.
Triathlon finish times
(i) Choose an ordinary stacked bar chart. It shows both the total race time and the time spent in each segment. The shortest complete bar identifies the quickest athlete. Convert every time to minutes first.
| Athlete | Swim (min) | Cycle (min) | Run (min) | Total |
|---|---|---|---|---|
| 1 | 68 | 300 | 195 | 563 min = 9 h 23 min |
| 2 | 65 | 310 | 215 | 590 min = 9 h 50 min |
| 3 | 82 | 355 | 230 | 667 min = 11 h 7 min |
Finish times in minutes
(ii) What can a 100% chart alone answer?
| Question | Can it be answered? |
|---|---|
| (a) Who finished first? | No. Normalising bars hides the unequal total times. |
| (b) Fraction of Athlete 1’s time spent cycling? | Yes. The cycling segment gives its share directly. |
| (c) Who took longest to run? | No. A share of a smaller total may represent less time. |
Interpreting disability distributions
Read the denominator correctly. Each bar represents 100% of persons with disabilities within that age group. It does not show what percentage of all people of that age have a disability.
| Observation from the chart | What it means |
|---|---|
| Movement share rises from roughly 13% at ages 0–19 to roughly 25% at 60+. | Movement disability occupies a larger fraction of the older group’s displayed distribution. |
| Seeing share is about 18% at ages 0–19 and about 25% at 60+. | The oldest group has a larger seeing-disability share than the youngest; the trend is not strictly increasing at every step. |
| Speech share falls from about 9% in the youngest group to about 4% in the oldest. | Speech represents a smaller proportion in the oldest age group. |
| The category labelled “mental retardation” in the source is smaller in older groups. | Use the respectful current term “intellectual disability” when discussing this category. |
| Hearing occupies roughly one-fifth of several bars. | Its shares appear broadly comparable across the age groups. |
What we cannot conclude: The chart does not identify which age group has the most persons with disabilities, the exact count in any category, or why differences occur. We would need age-group counts to convert percentages to numbers.
Questions worth investigating: How many people does each bar represent? How were disability types classified? How does access to services differ across age groups? Would a newer census or survey show a similar distribution?
Individual project: two complete models
This is an individual project, so there is no single numerical answer. Both options are modelled below with invented sample data. Replace them with your own observations. You need to complete at least one option.
(i) Body-state diary: sample solution. Record the start and end of each activity, assign it to one body-state, and add durations. Count each minute once; sleeping is included in lying down.
| Person | Lying down | Sitting | Standing / moving | Total |
|---|---|---|---|---|
| Student | 8 h | 9 h | 7 h | 24 h |
| Parent A | 8 h | 8 h | 8 h | 24 h |
| Parent B | 7 h | 5 h | 12 h | 24 h |
Sample one-day body-state diary (hours)
Construction: Use 1 cm = 2 hours, so every complete bar is 12 cm long. The student’s sections are 4 cm, 4.5 cm and 3.5 cm. All totals are equal, so this graph can also be read as a 100% chart.
Sample observations: Parent B spends the most time standing/moving. The student spends more time sitting than either parent. These describe only the recorded day; they do not establish permanent habits. To improve estimates, use a diary during the day rather than relying only on memory.
(ii) Three-month family budget: sample solution. Identify consistent categories, collect bills or a daily expense log, agree how to classify mixed purchases, and calculate monthly totals. These sample figures are invented.
| Month | Housing | Food | Education | Transport | Other | Total (₹) |
|---|---|---|---|---|---|---|
| Month 1 | 10000 | 8000 | 4000 | 3000 | 5000 | 30000 |
| Month 2 | 10000 | 9000 | 4000 | 4000 | 5000 | 32000 |
| Month 3 | 10000 | 10000 | 4000 | 4000 | 7000 | 35000 |
Sample monthly expense shares
| Month | Housing % | Food % | Education % | Transport % | Other % |
|---|---|---|---|---|---|
| Month 1 | 33.33 | 26.67 | 13.33 | 10.00 | 16.67 |
| Month 2 | 31.25 | 28.12 | 12.50 | 12.50 | 15.62 |
| Month 3 | 28.57 | 28.57 | 11.43 | 11.43 | 20.00 |
Sample observations: Housing expense stays ₹10,000 but its share falls from 33.33% to 28.57% because total spending rises. Food spending rises from ₹8000 to ₹10,000 and its share rises from 26.67% to 28.57%. The table shows total spending rises from ₹30,000 to ₹35,000; equal-length percentage bars alone cannot reveal that rise.
Small-group project: a custom rating system
Model project: a bus journey. The following ratings are imagined for practice, as the question permits. A real group should agree on clear descriptions for scores 1–5 before surveying people.
(i)–(ii) Choose aspects and justify weights.
| Aspect | Weight | Reason |
|---|---|---|
| Safety | 4 | Safe driving and safe boarding matter most. |
| Punctuality | 3 | Reliability helps passengers reach commitments on time. |
| Cleanliness | 2 | Clean surroundings contribute to comfort. |
| Staff behaviour | 1 | Courtesy supports a pleasant journey. |
Total weight = 10. Thus safety contributes 40%, punctuality 30%, cleanliness 20% and staff behaviour 10%. These choices are a judgement; a different group may justify a different scheme.
(iii)–(iv) Ten sample ratings and individual weighted means.
| Person | Safety | Punctuality | Cleanliness | Behaviour | Final /5 |
|---|---|---|---|---|---|
| 1 | 5 | 4 | 4 | 3 | 4.3 |
| 2 | 4 | 4 | 3 | 4 | 3.8 |
| 3 | 5 | 5 | 4 | 4 | 4.7 |
| 4 | 3 | 4 | 3 | 3 | 3.3 |
| 5 | 4 | 3 | 4 | 4 | 3.7 |
| 6 | 5 | 4 | 5 | 4 | 4.6 |
| 7 | 4 | 5 | 4 | 3 | 4.2 |
| 8 | 3 | 3 | 3 | 4 | 3.1 |
| 9 | 4 | 4 | 5 | 5 | 4.3 |
| 10 | 5 | 4 | 4 | 4 | 4.4 |
(v) Make a 100% stacked chart. One useful chart shows the distribution of 5-, 4-, 3-, 2- and 1-star scores within each aspect. Each bar represents ten people, so one person is 10%. This chart displays the raw rating distribution; the custom weights are used separately in the final-score calculation.
| Aspect | 5★ count | 4★ count | 3★ count | 2★ count | 1★ count |
|---|---|---|---|---|---|
| Safety | 4 | 4 | 2 | 0 | 0 |
| Punctuality | 2 | 6 | 2 | 0 | 0 |
| Cleanliness | 2 | 5 | 3 | 0 | 0 |
| Staff behaviour | 1 | 6 | 3 | 0 | 0 |
Sample rating distribution by aspect
(vi) Sample interpretation. All sample ratings are 3 or higher. Safety has the most 5-star ratings (four of ten). Each person contributes equally to the final overall mean, but the aspects have different weights. The scheme captures the chosen aspects and emphasises safety. It misses fare, accessibility, crowding and route convenience. Ten imagined responses cannot establish how passengers generally feel; a real survey also needs a fair sample.
Whole-class project: average family age
Step 1 — Collect two quantities per family. Record its mean age aᵢ and size nᵢ. The family’s total age is nᵢaᵢ.
Step 2 — Add age totals and member counts separately.
| Sample family | Members | Mean age | Sum of ages |
|---|---|---|---|
| A | 4 | 30 years | 120 years |
| B | 5 | 24 years | 120 years |
| C | 3 | 40 years | 120 years |
Adding 1 to every weight
It does not generally remain unchanged. Adding 1 changes the relative weights, unlike multiplying all weights by 2.
Example showing a decrease: Values 10 and 20, weights 1 and 3.
Example showing an increase: Reverse the original weights to 3 and 1. The old mean is 12.5. New weights 4 and 2 give 80/6 ≈ 13.33.
Algebraic explanation. Let there be n values, let W be the sum of the original weights, let M be their weighted mean, and let A be their ordinary arithmetic mean. Then the weighted numerator is WM and the unweighted sum of values is nA.
Therefore: if A > M the mean increases; if A < M it decreases; if A = M it stays unchanged. Since M′ is a weighted mean of M and A, it moves towards A without passing it.
Generalisation: Adding the same positive constant c to every weight gives:
Fewer animals, but what happens to their shares?
Key idea: Counts and shares are different. The three shares must still total 100%, so they cannot all rise or all fall.
Take an initial population of 200 animals: 120 cows, 50 sheep and 30 chickens. Test the options with actual smaller counts.
| Option | Possible? | Reason / concrete example |
|---|---|---|
| (i) All shares decrease | No | Their sum would become less than 100%. |
| (ii) All shares increase | No | Their sum would exceed 100%. |
| (iii) All shares unchanged | Yes | 60 cows, 25 sheep, 15 chickens: total 100; shares remain 60%, 25%, 15%. |
| (iv) Cow share falls, sheep share rises, chicken share unchanged | Yes | 55 cows, 30 sheep, 15 chickens: shares 55%, 30%, 15%. All counts are below 120, 50, 30. |
| (v) Cow and sheep shares rise, chicken share falls | Yes | 65 cows, 30 sheep, 5 chickens: shares 65%, 30%, 5%. Again all counts have fallen. |
Think & Reflect + in-text practice
Extra teacher-guided answers to the chapter’s discussion prompts. These supplement every formal exercise solved above.
Why averages need weights
Badminton: Eight seniors contribute 1324 cm of height; three juniors contribute 448 cm. The exact combined mean is 1772/11 ≈ 161.09 cm. Averaging the two group means equally wrongly gives the three juniors the same influence as the eight seniors.
The displayed 149.33 is rounded: 149.33 × 3 = 447.99, not exactly 448. Use the original sum or the exact mean for intermediate work.
Cycling: Every week has five weekdays, so the three weekly means have equal weights. Their average is (12.8 + 15.8 + 18) ÷ 3 = 46.6/3 ≈ 15.53 minutes. The printed numerator “46.63” is a typo; it should be 46.6.
When is an average of averages justified? Equal group sizes make it valid automatically. Unequal sizes require weighting in general, although special cases can coincide—for example, if every group mean is the same.
Lemonade: Equal quantities at 10% and 20% give 15%. If 500 mL at 10% is mixed with 200 mL at 20%, the result is below 15%, because the weaker mixture contributes more.
Eight family-expenditure questions
| Question | Answer and reason |
|---|---|
| 1. Least housing expense? | Family C: ₹950. |
| 2. Most healthcare expense? | Family A: ₹1770. |
| 3. A’s education expense? | ₹1280, approximately ₹1300 from the graph. |
| 4. B’s largest category? | Housing: ₹1750. |
| 5. Does A spend more on food or healthcare? | Healthcare ₹1770 exceeds food ₹1700 by ₹70. |
| 6. Least total expense? | Family C: ₹7040; B totals ₹7286 and A totals ₹8425. |
| 7. C’s approximate total? | ₹7040, about ₹7000 from the graph. |
| 8. Housing share for C? | 950 ÷ 7040 × 100% ≈ 13.49%, about 13.5%. |
Totals and components from Table 10.2 (₹)
Which display is easier? Category clusters help compare families for one expense; family clusters help compare expenses within one family. Stacked bars make total comparisons easy but make middle segments harder to compare because they start at different positions.
Smallest healthcare share: A ≈ 21.01%, B ≈ 16.61%, C ≈ 18.47%. Thus Family B has the smallest healthcare share.
Electricity and flower-chart observations
Electricity: Total use rises from 1000 to 2000 in the supplied units. Lighting falls from 360 to 220 and from 36% to 11%. Cooling rises from 300 to 800 and from 30% to 40%. Kitchen appliances rise from 120 to 460 (12% to 23%), and other appliances rise from 220 to 520 (22% to 26%). Both absolute use and percentage share can be investigated when totals are known.
Flowers: The graph shows Fatima: 50% summer, 30% monsoon, 20% winter; Naveen: 40% summer, 25% monsoon, 35% winter. A paragraph on printed page 25 accidentally swaps Naveen’s last two percentages; the chart and its numerical examples agree on 25% and 35%.
| Statement | Conclusion |
|---|---|
| (i) Fatima has more summer than monsoon blooms. | True: both are fractions of the same annual total, and 50% > 30%. |
| (ii) Fatima has more summer blooms than Naveen. | Cannot infer without comparing their annual totals. |
| (iii) Fatima has fewer winter blooms than Naveen. | Cannot infer from percentages alone. |
| (iv) Both gardens have equal annual totals. | Cannot infer: 100% bars always have equal length. |
Counterexample for summer: With totals 210 and 120, summer counts are 105 and 48. With totals 180 and 240, summer counts are 90 and 96. The same percentages permit opposite comparisons.
Four reflections on chart types
| Prompt | Teacher-guided answer |
|---|---|
| 1. Scope of each chart? | Ordinary stacks retain quantities and totals. Percentage stacks emphasise each category’s fraction of its own total. |
| 2. Can an ordinary stack become a 100% stack? | Yes, provided each total is positive: divide every segment by its bar total and multiply by 100. |
| 3. Can a 100% stack become an ordinary stack? | Not uniquely without the totals. With each total supplied, multiply it by each segment’s percentage ÷ 100. |
| 4. What comparisons are allowed? | Ordinary stacks compare total amounts and category amounts. Percentage stacks compare shares. Within one bar, a larger share also means a larger amount. Across bars, amounts require their totals. |
Time use across age groups
Average day: The supplied percentages are sleep 34%, personal care 13%, paid work 10%, unpaid work/care 11%, learning 5%, and leisure/social/travel 27%. They sum to 100%. Convert any share to hours by multiplying it by 24/100.
Age-chart reflection 1: Children have a much larger learning share than older groups. Adults have larger paid and unpaid work shares than children. Sleep accounts for about one-third or more of the day in each age group. These are group averages; they do not describe every person.
Reflection 2: Sleep shares are about 38%, 34%, 32% and 37% from children through the elderly. They decrease through adulthood, then rise in the oldest group. This agrees with a decline from childhood to adulthood, but not with a strictly falling trend across all ages.
Reflection 3: Learning falls from about 22% for ages 6–14 to about 9% for ages 15–24. A plausible explanation is that the older group includes people who have finished schooling or started work. The graph itself does not prove the cause.
Reflection 4: Adults’ paid and unpaid work shares are about 14% and 15%, respectively. They are close: 3.36 hours and 3.60 hours, differing by about 14 minutes. These are averages across adults, not a claim that each adult divides their own time equally.
Strips activity: Yes. Joining all pieces for the same activity preserves their total length and creates a stacked bar. A whole 24-hour strip is 100%; grouping pieces loses their time-of-day sequence, but retains each activity’s total duration.
Reflections after the body-state exercise
1. A different teacher: The bar may be similar, but need not match. Travel, teaching style, health, timetable and habits can change time spent in each state.
2. All seven days: It may change because weekend routines differ. Find the new mean from all seven daily totals for each state; do not simply average the five-day mean and the two-day mean equally.
3. Other occupations: A teacher and shop worker might have similar standing time; an office worker and student might have similar sitting time. Matching bars do not prove matching occupations.
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