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

Page 8 — Why does a simple average fail?

There are 8 senior trainees with a combined height of 1324 cm and 3 junior trainees totalling 448 cm. The correct combined average is (1324 + 448) ÷ (8 + 3) = 1772/11 ≈ 161.09 cm. The two group averages cannot be treated equally because the groups have different sizes; their averages must be weighted by 8 and 3.

Page 26 — Stacked bars and 100% stacked bars

1. A stacked bar chart shows actual component values and their total. A 100% stacked bar chart shows the proportion each component contributes to its own total.

2. Yes, convert a stacked bar to 100% by dividing every segment by that bar’s total and multiplying by 100.

3. A percentage bar cannot be converted back to actual quantities unless the actual total for each bar is also known.

4. Stacked bars compare actual amounts and totals; 100% stacked bars compare shares. A greater percentage does not necessarily mean a greater actual count.

Page 27 — Daily time-use chart

1. Sleep takes roughly 38% of the day for children, 34% for youth, 32% for adults and 37% for older people. Learning takes a larger share for children (about 22%) than youth (about 11%). Adults spend about 14% on paid work and 15% on unpaid work and care; the elderly group spends about one-third on leisure, social activity and travel.

2. Sleep does not decrease steadily across all age groups. The approximate hours are 9.12, 8.16, 7.68 and 8.88 respectively.

3. The 15–24 age group includes people who may have left full-time school and begun work or other responsibilities. The graph suggests the difference but does not establish its cause.

4. For adults, 14% of 24 hours is 3.36 hours and 15% is 3.60 hours, a difference of about 14.4 minutes. Thus the averages are nearly equal.

Page 28 — Grouping activity strips

Yes. If each strip represents a full day (24 hours = 100%), group all small pieces of each activity into one segment. Keep the order and colours consistent. The chart shows each activity’s share but loses information about when during the day the activity took place.

Page 30 — Interpreting body-state data

1. Another teacher’s chart may show similar patterns but need not be identical because routines differ.

2. Including all seven days may change the chart if weekends differ. For each state, weekly daily average = (5 × weekday average + Saturday hours + Sunday hours) ÷ 7.

3. Similar amounts of sitting, lying down and standing/moving can occur in different occupations, so the chart alone cannot identify someone’s job.

Question 1 — Combined class average

Section A: 30 students, mean 72. Section B: 25 students, mean 76.

Total marks = 30×72 + 25×76 = 2160 + 1900 = 4060. Combined mean = 4060/(30+25) = 4060/55 = 73.82 approximately.

Question 2 — Nitrogen in equal quantities

The three nitrogen fractions are 1/10, 9/50 and 3/60. Equal quantities mean equal weights.

Mean = (1/10 + 9/50 + 3/60)/3 = (10/100 + 18/100 + 5/100)/3 = 11/100.

Question 3 — Combined gold purity

Mix 9 units at 12 varṇa, 5 units at 10 varṇa and 17 units at 11 varṇa.

Combined purity = (9×12 + 5×10 + 17×11)/(9+5+17) = 345/31 ≈ 11.13 varṇa.

Question 4 — Combined rainfall average

May, June and July have 31, 30 and 31 days; average daily rainfall is 3.5, 10 and 8.7 mm.

Combined mean = (3.5×31 + 10×30 + 8.7×31)/(31+30+31) = 678.2/92 ≈ 7.37 mm per day.

Question 5 — Spice concentration

(i) Combine 100 mL at 5%, 200 mL at 10% and 300 mL at 15%. Mean concentration = (100×5+200×10+300×15)/600 = 7000/600 ≈ 11.67%.

(ii) Combine 300 mL at 5%, 200 mL at 10% and 100 mL at 15%. Mean = (300×5+200×10+100×15)/600 = 5000/600 ≈ 8.33%. The first mixture is stronger because more of it comes from the 15% solution.

Question 1 — Savitri’s annual score

Internal tests: 35/50×100 = 70%. Project: 44/60×100 = 73⅓%. Final exam: 80%. Apply weights 3:4:5.

Weighted percentage = [70×3 + (73⅓)×4 + 80×5]/12 = 1355/18 ≈ 75.28%.

Correct expressions: (iii) and (iv), because the marks must first be put on the same percentage scale before applying the weights.

Question 1 — Total profit percentage

Profit on books = 30% of ₹8000 = ₹2400. Profit on covers = 50% of ₹1000 = ₹500. Total profit = ₹2900 and sales = ₹9000.

Profit percentage = (2900/9000)×100 ≈ 32.22%.

Question 2 — Average daily distance of a stork

Distance over the first 20 days = 20×44.5 = 890 km. Add 55 km: total = 945 km in 21 days. New average = 945/21 = 45 km per day.

Question 3 — Salt and sugar solutions

Mix 600 mL at 5% salt with 300 mL at 8% sugar. Total = 900 mL.

Salt = (600×5 + 300×0)/900 = 3⅓%. Sugar = (600×0 + 300×8)/900 = 2⅔%. Correct option: (v).

Question 4 — Diluting panipuri water

Initial mixture: 10 L at 8%; target is three-quarters of 8%, or 6%. If x litres of plain water are added, spice quantity remains unchanged.

80/(10+x)=6 ⇒ 80=60+6x ⇒ x=10/3. Add 3⅓ litres of water.

Question 5 — Weighted fitness marks

Weights for strength, flexibility, agility are 4:5:6.

Keerthi is 5 marks lower in strength but 5 higher in agility, which has the larger weight, so Keerthi’s score is higher.

Rashi = (60×4+65×5+70×6)/15 = 985/15 ≈ 65.67.

Keerthi = (55×4+65×5+75×6)/15 = 995/15 ≈ 66.33.

Question 6 — Restaurant rating

Food average = (5×5+3×4+2×3)/10 = 4.3. Ambience = (4×4+5×3+1×2)/10 = 3.3. Service = (1×5+2×4+2×3+4×2+1×1)/10 = 2.8.

Use weights food:ambience:service = 6:5:4. Overall = (4.3×6+3.3×5+2.8×4)/15 = 107/30 ≈ 3.57 out of 5.

Question 7 — Langurs in an animal facility

There are 60 langurs. Let x be males and y be females. The combined mean is 14.925 kg; average male weight is 16.5 kg and average female weight is 13.8 kg.

(16.5x+13.8y)/(x+y)=14.925 and x+y=60. Substitute y=60−x: 16.5x+13.8(60−x)=14.925×60 ⇒ 2.7x=67.5 ⇒ 25 males and 35 females. The valid given expression is (a), with denominator x+y.

(iii) New female mean = (35×13.8+15.2)/36 = 498.2/36 ≈ 13.84 kg.

(iv) Removing males weighing 16.9 kg and 16.1 kg leaves 412.5−33=379.5 kg across 23 males; mean = 16.5 kg.

(v) After one of these males loses 1 kg, mean = (379.5−1)/23 ≈ 16.46 kg.

Question 8 — Salinity of water mixtures

Take Dead Sea water as 34% salinity, purified water as 0.001%, and groundwater as 0.01%.

(i) One litre of Dead Sea water plus 2 litres purified water gives (34×1+0.001×2)/3 ≈ 11.334%.

(ii) To obtain groundwater salinity, set (34+0.001x)/(1+x)=0.01. Then 33.99=0.009x, so x≈3776.67 L purified water.

(iii) It is impossible to reach 0.001% by mixing Dead Sea water with groundwater at 0.01%, because both concentrations exceed the target. Mixing cannot produce a value below both starting concentrations.

Question 1 — IUCN Red List

(i) The 14,234 written above the 2019 column is the total number of listed animal species across the classes shown for 2019.

(ii) The reptile segment for 2016 is approximately 1,000–1,200 species; this is a visual estimate from the chart.

(iii) Mammals and birds show relatively small increases between 2007 and 2019 compared with some other classes.

Question 2 — Bowler’s wickets at home and overseas

(i) Home vs overseas: Home wickets in Test, ODI and T20 are 62, 100 and 32; total = 194, cumulative segment boundaries 0, 62, 162, 194. Overseas: 172, 49, 71; total = 292, boundaries 0, 172, 221, 292. More wickets were taken overseas.

(ii) By format: Test = 62+172 = 234; ODI = 100+49 = 149; T20 = 32+71 = 103. Test has the greatest total. If each bar stacks Home then Overseas, the segment boundaries are 0–62–234, 0–100–149 and 0–32–103.

Question 1 — Smartwatch data for five people

(i) A reasonable guess for Sahana is 8 hours lying down, 10 hours sitting, and 6 hours standing/moving. Total = 24 hours. This is an estimate, so other reasonable rows that total 24 hours are possible.

(ii) Julie’s pattern could fit a person who spends much of the day sitting, such as an office worker, student or tailor. The chart alone cannot establish her actual occupation.

(iii) Convert hours into percentages using (hours/24)×100.

For Sahana, the stacked segments end at 33⅓%, 75% and 100%.

Question 1 — Cricket run rate

(i) Total after two overs = 6+12 = 18, so run rate = 18/2 = 9 runs per over.

(ii) Runs after 19 overs = 19×6 = 114; after the final 12-run over, total = 126. Run rate = 126/20 = 6.3 runs per over.

Question 2 — Average price of shuttlecocks

The five friends collected 6, 4, 5, 4 and 4 prices: 23 prices in total. If their averages are aᵧ, aₛ, aₖ, aₚ and a𝗀, combined mean = (6aᵧ+4aₛ+5aₖ+4aₚ+4a𝗀)/23.

The list totals are ₹917, ₹493, ₹766, ₹897 and ₹990. Total = ₹4063, so mean = 4063/23 ≈ ₹176.65.

By type, 13 nylon prices total ₹1429 and 10 feather prices total ₹2634, giving (1429+2634)/23 ≈ ₹176.65 as well. Both approaches use all the same observations.

Question 3 — Average share price

(i) Shreyas’s existing cost = 25×₹150 = ₹3750. New cost = 10×₹30 = ₹300. New average = ₹4050/35 ≈ ₹115.71 per share.

(ii) Let Vaishnavi buy x shares. (750+30x)/(5+x)=70 ⇒ 750+30x=350+70x ⇒ x=10. She bought 10 shares.

Question 4 — Mean depth of a pool

Weight each depth by the length of its section: (4×9 + 5×7 + 6×7 + 7×3 + 8×2)/30 = 150/30 = 5 hastas.

Question 5 — Suvarna’s mean gold price

Initially she owns 1 g bought at ₹15,000.

(i) Buy 1 g at ₹30,000: mean = (15+30)/2 = ₹22,500/g.

(ii) Buy 2 g at ₹30,000: mean = (15+2×30)/3 = ₹25,000/g.

(iii) Buy 1 g at ₹10,000: mean = (15+10)/2 = ₹12,500/g.

(iv) Buy 10 g at ₹10,000: mean = (15+100)/11 ≈ ₹10,454.55/g.

(v) Buy 0.5 g at ₹30,000: mean = (15+15)/1.5 = ₹20,000/g.

(vi) Buy 0.5 g at ₹15,000: price rate is unchanged, so the mean remains ₹15,000/g.

Question 6 — Double every weight

If M=(w₁x₁+…+wₙxₙ)/(w₁+…+wₙ), then after doubling all weights M′=2(w₁x₁+…+wₙxₙ)/[2(w₁+…+wₙ)]=M. The weighted average remains unchanged.

Question 7 — Objects orbiting Earth

(i) The total in 2003 is about 10,000; it doubles to about 20,000 around 2021.

(ii) The 2025 total is about 33,000: approximately 17,000 payload objects and 16,000 other objects. Payload share ≈ 17,000/33,000×100 ≈ 52%; other-object share ≈ 48%.

(iii) Both totals rise overall, with a sharp recent rise in payload objects. The payload share grows towards about half of the total, while the other-object share falls proportionally despite its count remaining high. Values are approximate graph readings.

Question 8 — Schools with playgrounds

(i) Correct inferences: (b) and (d). (a) cannot be inferred because state totals are unknown. (b) is reasonable because Arunachal Pradesh is about 68%, close to two-thirds. (c) cannot be inferred from percentages alone. (d) follows because Maharashtra’s percentage (93%) is higher than Telangana’s (74%) and Maharashtra is stated to have more schools.

(ii) Nationwide percentage requires the total number of schools in every state/UT (or actual playground-school counts), then total schools with playgrounds divided by total schools.

Question 9 — Choosing a play from rating charts

(i) One reasonable initial choice is Play A because its combined share of 4- and 5-star ratings is strong. Play C has a large 5-star share but also a relatively large 1-star share, indicating divided opinions.

(ii) After considering the counts of ratings, one might choose Play B because its favourable reviews represent a much larger number of responses. The conclusion depends on whether we compare proportions or the evidence represented by actual counts.

Question 10 — Triathlon times

A stacked bar chart of actual time is better for comparing who finished first.

Athlete 1: 68+300+195 = 563 min. Athlete 2: 65+310+215 = 590 min. Athlete 3: 82+355+230 = 667 min. Athlete 1 finishes first.

From a 100% stacked chart alone, only (b) can be answered: Athlete 1’s cycling share = 300/563 ≈ 53.3%. Equal-length bars conceal total time, and percentages alone do not show which athlete spent the longest actual time running.

Question 11 — Disability data by age

Notice: Each bar totals 100%, so the graph compares disability types within each age group. Shares for seeing and movement difficulties are larger in the oldest group, while speech difficulties take a smaller share.

Wonder: How many people are represented by each bar, and how might the distribution have changed since the data were collected?

Inference: The distribution differs by age group, but the bars alone cannot show absolute counts, the percentage of the entire age group affected, or the reasons for the differences.

Question 12 — Individual project

(i) Time-use chart: Record daily hours for lying down, sitting and standing/moving for yourself and two family members. Ensure every row totals 24. Example rows could be 8/10/6 hours, 8/8/8 hours and 7/5/12 hours. Plot all bars on the same 0–24 scale.

(ii) Budget chart: Sort monthly expenditure into categories such as housing, food, education, transport and healthcare, record three months, convert each category to a percentage of its month’s total and draw equal-length 100% bars. Discuss both percentage shares and actual totals.

Question 13 — Create a weighted rating scheme

This is an open project. Example: bus travel with weights safety 4, punctuality 3, comfort 2, cleanliness 1; total weight = 10. If ratings are S, P, C and L, overall passenger rating R=(4S+3P+2C+L)/10.

An example data set gives ten overall ratings 4.3, 3.8, 4.7, 3.3, 3.6, 4.6, 4.4, 3.1, 4.0 and 4.3. Their average is 40.1/10 = 4.01/5. Convert raw 1–5 star counts into percentages to make the 100% stacked chart. Mention limitations such as not including fare or accessibility.

Question 14 — Average age across families

Use each family’s size as the weight. Example: Family A has 4 members averaging 25 years; B has 6 averaging 30; C has 5 averaging 28.

Combined average = (4×25+6×30+5×28)/(4+6+5) = 420/15 = 28 years.

Question 15 — Add 1 to every weight

Adding the same constant to weights does not generally leave the weighted mean unchanged; it moves the mean towards the ordinary mean of the values.

Values 10 and 20 with weights 3 and 1 give 12.5. Increasing weights to 4 and 2 gives (40+40)/6 = 13⅓. With starting weights 1 and 3, the mean falls from 17.5 to 16⅔.

Let W be total weight, M the weighted mean, n the number of values and A their ordinary mean. After adding 1 to each weight, M′=(WM+nA)/(W+n), so M′−M=n(A−M)/(W+n). Thus it increases if A>M, decreases if A<M and remains unchanged if A=M.

Question 16 — Changes in animal population shares

Possible options are (iii), (iv) and (v); options (i) and (ii) are impossible because the three shares must sum to 100%.

Starting counts 120 cows, 50 sheep and 30 chickens give 60%, 25%, 15%. New counts 60, 25, 15 leave shares unchanged (iii). Counts 55, 30, 15 give 55%, 30%, 15% (iv). Counts 65, 26, 9 give 65%, 26%, 9% (v). In all examples each actual animal count falls, while percentage shares respond differently.