Class 9 · MathsNCERT 2026 Revised Syllabus

Chapter 10: How Quantities Combine: Understanding Data (0904CH10)

Ganita Manjari Part II · 39 questions · 7 exercises

Chapter code: 0904ch10 · Book: Ganita Manjari Part II · Class 9 Maths

Find complete NCERT solutions for Class 9 Maths, Ganita Manjari Part II Chapter 10. Every exercise has a clear, step-by-step answer for chapter code 0904CH10.

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Exercise Set 10.1(5 questions)

Section A has 30 students with average score 72; Section B has 25 students with average score 76. Find the combined average.

School Level Solution

Combined total of marks = 30×72 + 25×76 = 2160 + 1900 = 4060. Total students = 30 + 25 = 55.Combined average = 4060/55 = 73.818… ≈ 73.82 marks.

Three equal quantities of fertiliser contain nitrogen fractions 1/10, 9/50 and 3/60 respectively. Find the fraction of nitrogen in the mixture.

School Level Solution

The quantities are equal, so take their weighted mean with equal weights.Nitrogen fraction = [(1/10) + (9/50) + (3/60)]/3 = (0.10 + 0.18 + 0.05)/3 = 0.33/3 = 0.11 = 11/100.

A goldsmith combines 9 units of gold at 12 varṇa, 5 units at 10 varṇa and 17 units at 11 varṇa. Find the combined purity.

School Level Solution

Total pure-gold equivalent = 9×12 + 5×10 + 17×11 = 108 + 50 + 187 = 345. Total alloy = 9 + 5 + 17 = 31 units.Combined purity = 345/31 = 11.129… ≈ 11.13 varṇa.

Average daily rainfall in May, June and July is 3.5 mm, 10 mm and 8.7 mm, respectively. Write an expression for their combined average.

School Level Solution

May, June and July have 31, 30 and 31 days. The combined average must weight each monthly average by the number of days.Combined mean = (31×3.5 + 30×10 + 31×8.7)/(31+30+31) = 678.2/92 ≈ 7.37 mm per day.

Find the spice concentration when (i) 100 mL at 5%, 200 mL at 10%, and 300 mL at 15% are mixed; (ii) 300 mL at 5%, 200 mL at 10%, and 100 mL at 15% are mixed.

School Level Solution

(i) Spice amount = 100(0.05)+200(0.10)+300(0.15)=5+20+45=70 mL. Total volume = 600 mL. Concentration = 70/600×100% = 11.67%.(ii) Spice amount = 300(0.05)+200(0.10)+100(0.15)=15+20+15=50 mL. Concentration = 50/600×100% = 8.33%.

Exercise Set 10.2(1 questions)

Savitri scores 35 out of 50 in internal tests, 44 out of 60 in a project and 80 out of 100 in the final exam. The three components are weighted in the ratio 3:4:5. Which of the given expressions correctly represents her annual percentage? Calculate the annual percentage using the weighted mean.

School Level Solution

Step 1: Convert each component to a percentage. Internal tests: (35/50)×100 = 70%. Project: (44/60)×100 = 73⅓%. Final exam: (80/100)×100 = 80%.Step 2: Apply the weights 3:4:5. The sum of the weights is 3+4+5=12.Weighted percentage = [3(70)+4(73⅓)+5(80)]/12 = [210+293⅓+400]/12 = 903⅓/12 = 75.277…% ≈ 75.28%.The ordinary total-marks percentage (159/210×100) is not the correct method because it ignores the stated weights.

Exercise Set 10.3(8 questions)

A shop makes ₹8,000 from book sales with 30% profit and ₹1,000 from book-cover sales with 50% profit. Find the profit percentage on total sales.

School Level Solution

Profit from books = 30% of ₹8,000 = ₹2,400. Profit from covers = 50% of ₹1,000 = ₹500. Total sales = ₹9,000; total profit = ₹2,900.Profit percentage = (2900/9000)×100% = 32.22%.

A white stork travels an average of 44.5 km daily over 20 days and flies 55 km on day 21. Find its average daily distance over all 21 days.

School Level Solution

Distance in first 20 days = 20×44.5 = 890 km. Total for 21 days = 890+55 = 945 km. Average = 945/21 = 45 km per day. The new average rises slightly because 55 km exceeds the previous average.

Mix 600 mL of solution containing 5% salt with 300 mL of solution containing 8% sugar. Find the salt and sugar concentrations in the mixture.

School Level Solution

Salt amount = 5% of 600 = 30 mL. Sugar amount = 8% of 300 = 24 mL. Total mixture = 900 mL. Salt concentration = 30/900×100% = 3.33%. Sugar concentration = 24/900×100% = 2.67%. Therefore option (v) is correct.

Ten litres of pani has 8% spice. How much regular water must be added to reduce spice concentration to 3/4 of its original value?

School Level Solution

Spice initially = 8% of 10 = 0.8 L-equivalent. The target concentration is (3/4)×8% = 6%. Let x L of water be added.0.8/(10+x)=0.06 ⇒ 0.8=0.6+0.06x ⇒ x=0.2/0.06=10/3 L = 3⅓ L.

Final fitness marks combine strength, flexibility and agility in ratio 4:5:6. Rashi scores 60,65,70 and Keerthi scores 55,65,75. (i) Whose score is higher without full calculation? (ii) Find both final marks.

School Level Solution

(i) Compared with Rashi, Keerthi has 5 fewer strength marks (weight 4), equal flexibility marks, and 5 more agility marks (weight 6). The gain 5×6 = 30 exceeds the loss 5×4 = 20, so Keerthi scores higher.(ii) Rashi = [4(60)+5(65)+6(70)]/15 = 985/15 = 65⅔. Keerthi = [4(55)+5(65)+6(75)]/15 = 995/15 = 66⅓.

Ten customers rated food, ambience and service on a 1–5 scale. Counts for ratings 5,4,3,2,1 are Food (5,3,2,0,0), Ambience (0,4,5,1,0), Service (1,2,2,4,1). Find the combined average for weights food:ambience:service = 6:5:4.

School Level Solution

For each aspect, multiply every rating by its frequency, add these products and divide by the 10 customers.Food: (5×5+4×3+3×2)/10 = (25+12+6)/10 = 4.3.Ambience: (5×0+4×4+3×5+2×1+1×0)/10 = (16+15+2)/10 = 3.3.Service: (5×1+4×2+3×2+2×4+1×1)/10 = (5+8+6+8+1)/10 = 2.8.Now apply weights 6, 5 and 4 to food, ambience and service: overall rating = [6(4.3)+5(3.3)+4(2.8)]/(6+5+4) = (25.8+16.5+11.2)/15 = 53.5/15 = 3.566… ≈ 3.57 out of 5.

There are 60 male and female langurs together. Their average weights are 16.5 kg (males), 13.8 kg (females), and 14.925 kg overall. (i) Select the correct equation. (ii) Find the number of males. (iii) Add a 15.2 kg female and find the new female mean. (iv) Release males weighing 16.9 kg and 16.1 kg and find the new male mean. (v) One remaining male loses 1 kg; find the average male weight.

School Level Solution

Let x be the number of male langurs and y the number of female langurs. The total number is x+y=60.(i) Correct expression(s). The mean formula gives (16.5x+13.8y)/(x+y)=14.925. Since x+y=60, this is also (16.5x+13.8y)/60=14.925. Therefore both (a) and (b) are correct.(ii) Number of males. Solve x+y=60 and 16.5x+13.8y=14.925(60)=895.5. Put y=60−x: 16.5x+13.8(60−x)=895.5 ⇒ 2.7x=67.5 ⇒ x=25. Hence y=60−25=35; there are 25 males and 35 females.(iii) New female mean. Original total female weight = 35×13.8=483 kg. After adding a 15.2 kg female, total weight = 483+15.2=498.2 kg and number =36. Mean =498.2/36=13.838… ≈ 13.84 kg.(iv) New male mean after releasing two males. Original male total =25×16.5=412.5 kg. Released weight =16.9+16.1=33 kg, so remaining total =412.5−33=379.5 kg for 23 males. Mean =379.5/23=16.5 kg. This remains unchanged because the two released males’ average, (16.9+16.1)/2, is exactly 16.5 kg.(v) One remaining male loses 1 kg. The male total becomes 379.5−1=378.5 kg, while the number remains 23. New mean =378.5/23=16.4565… ≈ 16.46 kg.

Dorjee has 1 litre of Dead Sea water (salinity ≈34%). (i) Find the salinity after mixing it with 2 litres of purified drinking water (≈0.001%). (ii) Can he obtain groundwater salinity (≈0.01%) by mixing Dead Sea water and purified water? If so, how much purified water must be mixed with 1 litre of Dead Sea water? (iii) Can he obtain purified-drinking-water salinity (≈0.001%) by mixing Dead Sea water and groundwater (≈0.01%)? If so, how much groundwater should be mixed with 1 litre of Dead Sea water?

School Level Solution

Treat the listed salinities as percentages of the mixture.(i) Dead Sea + purified water. Salt-equivalent amount = 1(0.34)+2(0.00001)=0.34002 L. Total volume = 3 L. Salinity = 0.34002/3×100% = 11.334% approximately.(ii) Reach groundwater salinity 0.01%. Let x litres of purified water be added to the 1 L of Dead Sea water. Using fractions,(0.34+0.00001x)/(1+x)=0.0001.Thus 0.34+0.00001x=0.0001+0.0001x ⇒ 0.3399=0.00009x ⇒ x≈3,776.67 L.(iii) Reach purified-water salinity 0.001% using groundwater. The two starting salinities are 34% and 0.01%. Any mixture containing positive amounts of both has salinity between 0.01% and 34%, so it cannot be as low as 0.001%. Even adding unlimited groundwater only approaches 0.01% from above. Therefore it is impossible, and no nonnegative quantity of groundwater satisfies the request.

Exercise Set 10.4(2 questions)

A stacked column chart gives the number of animal species on the IUCN Red List by class in 2007, 2010, 2013, 2016 and 2019. (i) What does 14,234 represent? (ii) Approximately how many reptile species were listed in 2016? (iii) Which classes show a relatively small increase from 2007 to 2019?

School Level Solution

(i) The number 14,234 is the total height of the full 2019 stacked column, so it represents all animal species included across the classes shown on the IUCN Red List in 2019.(ii) Read the reptile segment (the red segment in the legend) in the 2016 column against the vertical scale. Its height is about 1,500 species.(iii) Compare each coloured segment in the 2007 and 2019 columns. The green insects segment and the small top segment for other invertebrates (“Others”) show relatively small increases compared with the larger increases in several other classes.

A bowler took wickets at home/overseas: Test 62/172, ODI 100/49 and T20 32/71. (i) Describe stacked bars comparing total home versus overseas wickets. (ii) Describe stacked bars comparing totals for Test, ODI and T20.

School Level Solution

(i) Home total = 62+100+32 = 194 wickets; overseas total = 172+49+71 = 292 wickets. Draw two bars: Home, subdivided into Test 62, ODI 100 and T20 32; Overseas, subdivided into Test 172, ODI 49 and T20 71. Use a consistent scale.(ii) Totals by format: Test = 62+172 = 234; ODI = 100+49 = 149; T20 = 32+71 = 103. Draw three bars of those total lengths, split into Home and Overseas portions using the given counts.

Exercise Set 10.5(1 questions)

The table records daily average hours for five people spent lying down, sitting, and standing/moving: Pavani (20,3,1), Raghu (7,4,13), Zakir (8,6,10), Sahana (student, to be estimated), Julie (8,10,6). (i) Make a reasonable 24-hour estimate for Sahana. (ii) Guess Julie’s work based on the data. (iii) Complete a 100% stacked bar chart.

School Level Solution

(i) Sahana’s estimate. One reasonable illustrative row is 8 h lying down, 6 h sitting and 10 h standing/moving. The total is 8+6+10=24 hours. Many other estimates are possible, but the three entries should add to 24.(ii) Julie’s possible work. She spends 10 hours sitting, compared with 8 hours lying down and 6 hours standing/moving. A plausible guess is a desk-based or seated job with substantial rest; this is only an inference from the table, not a definite identification.(iii) 100% stacked bars. Divide each time by 24 and multiply by 100%. The segment percentages (lying down, sitting, standing/moving) are:- Pavani: 20/24=83.33%, 3/24=12.50%, 1/24=4.17%. - Raghu: 7/24=29.17%, 4/24=16.67%, 13/24=54.17%. - Zakir: 8/24=33.33%, 6/24=25.00%, 10/24=41.67%. - Sahana (sample row): 8/24=33.33%, 6/24=25.00%, 10/24=41.67%. - Julie: 8/24=33.33%, 10/24=41.67%, 6/24=25.00%.Draw five equal-length bars from 0% to 100%, and split each bar in this order using the percentages above. Rounded percentages may total 99.99% or 100.01%; use the exact hour fractions to draw the segments.

End-of-Chapter Exercises(16 questions)

In a T20, a team scores 6 runs in the first over. (i) It scores 12 in the second over; find the run rate now. (ii) After 19 overs its run rate is 6; it scores 12 in the last over. Find its final run rate.

School Level Solution

(i) Total after two overs = 6+12=18. Run rate = 18/2 = 9 runs per over.(ii) Runs after 19 overs = 19×6=114. After scoring 12, total =126 in 20 overs. Final run rate =126/20=6.3 runs per over.

Five friends report shuttlecock prices. (i) If their respective averages are ay, as, ak, ap, ag, write the combined average. (ii) If an and af are the average prices of nylon and feather shuttlecocks, give the overall average and compare methods.

School Level Solution

Let the five friends have collected 6, 4, 5, 4 and 4 prices, respectively, for 23 prices in total.(i) If their means are aᵧ, aₛ, aₖ, aₚ and aɢ, the combined mean must weight each mean by how many prices contributed:$$(6a_y+4a_s+5a_k+4a_p+4a_g)/23.$$Using the actual prices listed in the question, the total is ₹4,063, so the combined mean is 4063/23 ≈ ₹176.65.(ii) The 13 nylon prices have mean aₙ=1429/13≈₹109.92 and the 10 feather prices have mean aբ=2634/10=₹263.40. Therefore the combined mean is (13aₙ+10aբ)/23 = 4063/23 ≈ ₹176.65. Both approaches agree because they give each individual price equal weight.

Shreyas owns 25 shares at an average ₹150. (i) He buys 10 more at ₹30 each; find his new average. (ii) Vaishnavi owns 5 shares averaging ₹150 and buys x shares at ₹30 each; her new average is ₹70. Find x.

School Level Solution

(i) Initial cost =25×150=₹3,750. New purchase =10×30=₹300. Total cost =₹4,050 for 35 shares. New average =4050/35≈₹115.71 per share.(ii) Vaishnavi’s initial cost =5×150=₹750. Let x be the number bought at ₹30. (750+30x)/(5+x)=70 ⇒ 750+30x=350+70x ⇒ 400=40x ⇒ x=10 shares.

A 30-hasta pool is divided into lengths 4,5,6,7,8 hastas with depths 9,7,7,3,2 hastas. Find the mean depth.

School Level Solution

Mean depth is weighted by the lengths of the sections. Total length =4+5+6+7+8=30 hastas. Weighted depth sum =4×9+5×7+6×7+7×3+8×2=36+35+42+21+16=150. Mean depth =150/30=5 hastas.

Suvarna owns 1 g gold at an average price ₹15,000/g. Estimate the new average after: (i) 1 g at ₹30,000; (ii) 2 g at ₹30,000; (iii) 1 g at ₹10,000; (iv) 10 g at ₹10,000; (v) 0.5 g at ₹30,000; (vi) 0.5 g at ₹15,000.

School Level Solution

Treat the original gram as costing ₹15,000. The new average is total cost divided by total grams. (i) (15,000+30,000)/2=₹22,500/g. (ii) (15,000+2×30,000)/3=₹25,000/g. (iii) (15,000+10,000)/2=₹12,500/g. (iv) (15,000+10×10,000)/11=115,000/11≈₹10,454.55/g. (v) (15,000+0.5×30,000)/1.5=30,000/1.5=₹20,000/g. (vi) (15,000+0.5×15,000)/1.5=22,500/1.5=₹15,000/g.

If every weight in a weighted mean is doubled, how does the weighted mean change? Justify algebraically.

School Level Solution

For values xᵢ with weights wᵢ, the mean is μ=Σwᵢxᵢ/Σwᵢ. After doubling every weight, μ′=Σ(2wᵢ)xᵢ/Σ(2wᵢ)=2Σwᵢxᵢ/(2Σwᵢ)=Σwᵢxᵢ/Σwᵢ=μ. Therefore doubling all weights does not change the weighted mean.

Use the cumulative graph of objects orbiting Earth by launch/separation year. (i) Estimate the total in 2003 and when it doubled. (ii) Estimate payload and other objects and their shares in 2025. (iii) Describe number and share trends.

School Level Solution

(i) Doubling the total. Read the stacked graph at 2003: the total is approximately 10,000 objects. Twice this is 2 × 10,000 = 20,000. The cumulative total reaches about 20,000 near 2021 (read approximately from the graph).(ii) Split in 2025. The total height is approximately 33,000. The payload portion is about 17,000, so $\frac{17,000}{33,000}\times100\%\approx51.5\%\approx52\%$ . The other-object count is about 33,000 − 17,000 = 16,000, or $\frac{16,000}{33,000}\times100\%\approx48.5\%\approx48\%$ . These are estimates because the graph is read visually.(iii) Trend. Both the payload count and other-object count increase over time. Payloads increase more sharply in the later years; consequently, payloads make up about 52% of the 2025 total and other objects about 48%, even though the number of other objects also continues to grow.

Observe the school-playground infographic. (i) Decide which statements can be inferred: (a) more schools have a playground in Haryana than in Uttarakhand; (b) approximately 2 out of 3 schools in Arunachal Pradesh have a playground; (c) Punjab has the highest number of schools with a playground; (d) if Maharashtra has more schools than Telangana, then Maharashtra has more schools with a playground. (ii) Can the nationwide percentage of schools with playgrounds be found from the information given? If not, what additional information is needed?

School Level Solution

(i)(a) The statement compares absolute counts, not percentages. The chart gives the share of schools with playgrounds; without knowing how many schools each state has, the count cannot be compared. Not inferable.(i)(b) Two out of three is about 66.7%, matching the displayed share for Arunachal Pradesh. Inferable.(i)(c) Punjab may have the highest displayed percentage, but that does not prove it has the largest count; state totals may differ. Not inferable.(i)(d) Given that Maharashtra has more schools than Telangana and the chart shows Maharashtra’s playground percentage is higher, the number with playgrounds must also be higher. If M and T are total schools and pM>pT while M>T, then MpM>TpT. Inferable.(ii) A national percentage is a weighted percentage: total schools with playgrounds divided by all schools. We need the number of schools in every state/UT (and then the corresponding playground counts), or equivalent nationwide totals. An unweighted average of state percentages would generally be incorrect.

A 100% stacked bar chart shows the shares of ratings for Plays A, B and C. (i) Which play would you choose from this chart? (ii) Would the corresponding absolute-count stacked bar chart change your choice?

School Level Solution

(i) Percentage chart. All bars total 100%, so compare the distribution of ratings. Combining the 4-star (red) and 5-star (blue) portions, Play B has the largest favourable share. Play C has the largest 5-star portion alone, but also a sizeable 1-star (grey) portion. On an overall balance of very positive ratings and fewest 1-star ratings, Play B is the most defensible choice. A different personal preference (for example, valuing only 5-star ratings) could lead to a different choice.(ii) Absolute-count chart. This chart shows counts rather than shares. Play B has far more ratings than the other plays and therefore also the largest absolute number of 4- and 5-star ratings. Thus the second chart supports choosing Play B too, but it answers a different question: total counts versus proportion of each play’s ratings.

Triathlon times: Athlete 1 swimming 1:08, cycling 5:00, running 3:15; Athlete 2 1:05, 5:10, 3:35; Athlete 3 1:22, 5:55, 3:50. (i) Choose stacked or 100% stacked bars. (ii) Which questions can a 100% chart answer: first finisher, fraction of Athlete 1’s time cycling, longest running time?

School Level Solution

(i) Use a stacked bar chart, because the total elapsed time matters and must be compared in absolute units. Convert to minutes if needed: Athlete 1 =68+300+195=563 min (9 h 23 min); Athlete 2 =65+310+215=590 min (9 h 50 min); Athlete 3 =82+355+230=667 min (11 h 7 min). Thus Athlete 1 finishes first.(ii) A 100% chart shows proportions of each athlete’s time, so it answers (b): Athlete 1’s cycling share is 300/563≈53.3%. It does not establish (a), since total durations are lost; nor (c), because a segment’s share cannot by itself tell which athlete had the greatest absolute running time.

Observe the 100% stacked bar chart of disability type by age group in India (Census 2011). State observations, questions and inferences.

School Level Solution

Each bar represents 100% of the disabled persons in one age group, so compare segment shares, not absolute numbers. A suitable observation is that the mix of disability types differs across age groups; movement-related disability tends to occupy a larger share among older groups, while some other types take larger relative shares among younger groups. The chart alone does not tell how many disabled persons are in each age group because absolute group totals are not shown. Useful follow-up questions include the counts by age group, causes of disability, and how trends differ by gender or region.

Individual project: (i) estimate time spent lying, sitting and standing/moving for yourself and at least two family members, then draw a stacked bar chart; or (ii) show monthly family expenditure with a 100% stacked bar chart for at least three months.

School Level Solution

Sample response for (i): Record 24-hour totals (lying/sitting/moving): Student = (8,7,9), parent 1 = (7,8,9), parent 2 = (7,7,10). Each row sums to 24 hours. Percentages are each hours÷24×100; for the student, 33.3%, 29.2%, 37.5%. Draw equal-length 100% bars split into these three categories.Sample response for (ii): Suppose monthly expenditure in ₹ is Housing/Food/Education/Transport/Health/Other: January 8,000/6,000/2,000/1,500/1,000/1,500; February 8,000/6,200/2,000/1,600/800/1,400; March 8,000/5,800/2,000/1,700/1,200/1,300. Convert each category to a percentage of that month’s total and draw one 100% bar per month. Compare changes in shares, not just rupee amounts; note that actual family data should replace these illustrative figures.

Small-group project: design a weighted rating scheme for a shop, bus trip, tourist spot or clinic. Choose aspects, weights, collect ratings from ten people, calculate means and make a 100% stacked chart.

School Level Solution

Example: local tourist spot. Rate cleanliness, accessibility, safety, information/signage and attractions on a 1–5 scale, with weights 3:2:3:1:4 respectively. The weights sum to 13. For a visitor who gives 4, 3, 5, 4, 5, the score is [3(4)+2(3)+3(5)+1(4)+4(5)]/13=53/13≈4.08.Use the same weighted-mean formula for each of ten visitors, then average their ten scores. For the 100% stacked chart, show each visitor’s ratings as shares of their weighted total or show the proportion of respondents in each rating band per aspect. Report which aspects score highest/lowest and acknowledge that a 1–5 rating may not capture every aspect. This is a complete model; replace sample ratings with the group’s ten responses.

Each student shares their family’s average age and family size. How can the class calculate the average age of all people in all families?

School Level Solution

For family i, let nᵢ be the number of members and aᵢ their average age. Their combined ages total nᵢaᵢ. Add all family totals and divide by the total number of people: class average age = (Σ nᵢaᵢ)/(Σ nᵢ). A simple average of the family averages would be wrong unless every family had the same size.

If each weight wᵢ is increased by a constant such as 1, what happens to a weighted average? Justify algebraically.

School Level Solution

Original weighted mean: μ=Σwᵢxᵢ/Σwᵢ. If each of n weights is increased by c, then μ′=(Σwᵢxᵢ+cΣxᵢ)/(Σwᵢ+nc). This is generally different from μ and can be written as a weighted combination of the old weighted mean and the unweighted mean.Example: values 10 and 20 with weights 1 and 3 give μ=(10+60)/4=17.5. Increasing both weights by 1 gives μ′=(2×10+4×20)/6=100/6≈16.67. Thus increasing all weights by the same additive constant is not the same as multiplying all weights by a common factor.

Last year a farm population was 60% cows, 25% sheep and 15% chickens. All three populations decrease. Which share-change scenarios are possible: (i) all shares decrease, (ii) all increase, (iii) all unchanged, (iv) cows share decreases, sheep share increases, chickens unchanged, (v) cows and sheep shares increase, chickens decrease?

School Level Solution

The three shares must always add to 100%, so they cannot all decrease or all increase; (i) and (ii) are impossible.(iii) is possible if all three populations shrink by the same percentage, preserving their ratios.(iv) is possible if the chicken population shrinks by the same factor as the total, sheep shrink less rapidly, and cows shrink more rapidly. For example, initial counts (60,25,15) can become (24,18.5,7.5), with all populations smaller but shares 48%, 37%, 15%.(v) is possible if cows and sheep shrink less rapidly than the chicken population, so their shares rise while the chicken share falls.

In-text and Think and Reflect(6 questions)

Eight senior students have average height 165.5 cm and three junior students have average height 149⅓ cm. Find the combined average and explain why simply averaging the two group averages is wrong.

School Level Solution

Total height of seniors = 8 × 165.5 = 1324 cm. Total height of juniors = 3 × 149⅓ = 448 cm. Combined total = 1772 cm for 11 students. Combined average = 1772/11 = 161.0909… ≈ 161.09 cm. The simple mean of 165.5 and 149⅓ gives equal weight to the two group averages even though the group sizes are 8 and 3; use a weighted mean.

A 500 mL bowl of lemonade contains 10% jaggery, and a 200 mL glass contains 20% jaggery. What percentage of jaggery is in the mixture?

School Level Solution

Jaggery in the bowl = 10% of 500 = 50 mL. Jaggery in the glass = 20% of 200 = 40 mL. Total jaggery = 90 mL and total mixture = 700 mL. Concentration = (90/700) × 100% = 12.857…% ≈ 12.86%. It is closer to 10% because the larger quantity (500 mL) has the lower concentration.

Answer the following about stacked bar charts and 100% stacked bar charts: (i) What does each show? (ii) Can a 100% stacked chart be made from a stacked chart? (iii) Can the original stacked chart generally be recovered from a 100% stacked chart? (iv) What comparisons suit each chart?

School Level Solution

(i) A stacked bar preserves each category’s count/quantity and the overall total. A 100% stacked bar normalizes each total to 100% and shows the share of each category. (ii) Yes: divide each segment by that bar’s total and multiply by 100%. (iii) Not without the original totals: the same percentages can describe bars of very different sizes. (iv) Use stacked bars to compare totals and absolute amounts; use 100% stacked bars to compare compositions or shares.

From a chart comparing how people of different age groups spend a typical day, state useful observations and explain why such observations should be treated as estimates rather than exact universal facts.

School Level Solution

Read the displayed segments and compare the age groups rather than assuming every person follows the average. School-age children generally devote more time to learning than adults, while adults may devote more time to paid and unpaid work. Sleep time may also vary with age. The graph describes aggregated data, so it can suggest patterns but cannot prove causes or describe every individual. Any numerical readings from the chart are approximate.

In the “What Can a Strip Say?” activity, each strip represents a full day divided into activities. If the pieces for each activity are grouped, can the information be displayed as a 100% stacked bar chart?

School Level Solution

Treat one full day (24 hours) as 100%. For each person/day, calculate each activity’s share as (hours in that activity ÷ 24) × 100%. Place the activity segments consecutively in a bar of equal total length. The total remains 100%, so the result is a 100% stacked bar chart.

For a chart based on one teacher’s five weekdays of activity, would the bars necessarily look the same for another teacher or when all seven days are included? Give a reasoned answer.

School Level Solution

Different teachers have different routines, travel, duties and home responsibilities, so their category shares can differ. Adding Saturday and Sunday may change the averages or proportions because weekend schedules are often unlike weekdays. Some patterns could resemble those of other jobs, but the graph alone cannot uniquely identify a person’s occupation.

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