NCERT Solutions Class 9 Mathematics Ganita Manjari 2 Chapter 10 How Quantities Combine Understanding Data Examples

Examples and In-text Questions

Example 1. A badminton academy has 11 trainees: 8 seniors and 3 juniors. Heights (cm) of the seniors: 165, 169, 164, 167, 170, 159, 164, 166 (average 165.5). Heights of the juniors: 146, 149, 153 (average 149.33). Find the average height of the whole group.
Two students work it out differently:
Method 1: \( \frac{165.5 + 149.33}{2} = \frac{314.83}{2} = 157.415 \)
Method 2: \( \frac{165 + 169 + 164 + 167 + 170 + 159 + 164 + 166 + 146 + 149 + 153}{11} = \frac{1772}{11} = 161.09 \)
Whose calculation gives the correct average height of the whole group?
Answer:
The group average must be (total of all 11 heights) ÷ 11.
Step 1: Seniors' total = 165 + 169 + 164 + 167 + 170 + 159 + 164 + 166 = 1324 cm (and 1324 ÷ 8 = 165.5 ✓).
Step 2: Juniors' total = 146 + 149 + 153 = 448 cm (and 448 ÷ 3 ≈ 149.33 ✓).
Step 3: Group average = \( \frac{1324 + 448}{8 + 3} = \frac{1772}{11} \approx 161.09 \) cm.
Method 1 only averages the two averages and gives 157.415 cm, which is not the group's average.
Method 2 is correct: the average height is about 161.09 cm.

Example 1 (Think and Reflect). Why did Method 1 not work?
Answer:
Method 1 computes \( \frac{165.5 + 149.33}{2} \), so it treats the seniors' average and the juniors' average as equally important.
But the seniors' average stands for 8 heights while the juniors' average stands for only 3. The seniors' average deserves more importance.
Giving the smaller (and shorter) junior group equal say drags the answer down to 157.415 cm, below the true 161.09 cm.
Method 1 works only when both groups have the same number of members.

Teacher's Note:
1. Every average carries a hidden 'how many' behind it.
2. Averaging averages is safe only when group sizes are equal.

Example 1 (In-text Question). Shreyas calculated it as \( \frac{(165.5 \times 8) + (149.33 \times 3)}{8 + 3} = \frac{1324 + 448}{11} = \frac{1772}{11} = 161.09 \). Why does this also work? Why does 165.5 × 8 give the total height of the seniors and 149.33 × 3 the total height of the juniors?
Answer:
Since average = \( \frac{\text{sum of values}}{\text{number of values}} \), multiplying both sides by the number of values gives: sum of values = average × number of values.
Seniors: 165.5 × 8 = 1324 cm. Juniors: 149.33 × 3 = 447.99 ≈ 448 cm (the tiny gap is only because 149.33 is a rounded form of 448 ÷ 3).
So Shreyas first rebuilds each group's total, adds them, and divides by the total number 11, which is exactly what Method 2 does.
Hence his method also gives 161.09 cm.
Key idea: if one collection has n values with average a and another has m values with average b, the combined average is \( \frac{an + bm}{n + m} \).

Example 2. Jaspreet tracked the minutes she cycled on weekdays for 3 weeks (Table 10.1).
Week 1: 10, 13, 10, 15, 16 (weekly average 12.8)
Week 2: 14, 10, 20, 17, 18 (weekly average 15.8)
Week 3: 15, 18, 14, 20, 23 (weekly average 18)
Find the mean time spent cycling per weekday over the 3 weeks. Two methods are suggested:
Method 1: \( \frac{12.8 + 15.8 + 18}{3} \); Method 2: \( \frac{(12.8 \times 5) + (15.8 \times 5) + (18 \times 5)}{5 + 5 + 5} = \frac{64 + 79 + 90}{15} \).
Answer:
Step 1: Weekly totals: Week 1 = 64 min, Week 2 = 79 min, Week 3 = 90 min (these match 5 × 12.8, 5 × 15.8 and 5 × 18).
Step 2: Mean over all 15 weekdays = \( \frac{64 + 79 + 90}{15} = \frac{233}{15} \approx 15.53 \) minutes.
Step 3: Method 1: \( \frac{12.8 + 15.8 + 18}{3} = \frac{46.6}{3} \approx 15.53 \), the same value. (The figure 46.63 printed in the book should read 46.6; the final value 15.53 is right.)
Mean cycling time ≈ 15.53 minutes per weekday. Both methods agree here.

Example 2 (In-text Question). In the badminton example Method 1 gave a wrong value. Why does Method 1 give the correct answer here? When does it work and when does it not?
Answer:
For three collections with averages a, b, c and sizes p, q, r, the combined average is \( \frac{ap + bq + cr}{p + q + r} \).
Here every week has 5 days, so p = q = r. Then \( \frac{ap + bp + cp}{3p} = \frac{(a + b + c)p}{3p} = \frac{a + b + c}{3} \), which is just Method 1.
Method 1 works when all collections are the same size; it fails when the sizes differ (8 seniors and 3 juniors), because the bigger collection must count more.

Teacher's Note:
1. Equal sizes make the common factor cancel.
2. Unequal sizes need the weighted formula.

Example 3. Two glasses hold equal quantities of lemonade. One has 10% jaggery and the other 20% jaggery. If the two are mixed, what is the concentration of jaggery in the mixture?
Answer:
Let each glass hold y units of lemonade.
Jaggery: 0.1y in the first glass and 0.2y in the second.
Concentration = \( \frac{\text{jaggery in mixture}}{\text{quantity of mixture}} = \frac{0.1y + 0.2y}{y + y} = \frac{0.3y}{2y} = 0.15 \).
The mixture is 15% jaggery, exactly midway between 10% and 20%, because the quantities are equal.

Example 4. A bowl of 500 mL lemonade has 10% jaggery and a glass of 200 mL lemonade has 20% jaggery. If they are mixed, what is the concentration of jaggery in the mixture? Estimate first: is it 15%, more, or less? Why?
Answer:
Estimate: the bowl supplies far more lemonade (500 mL against 200 mL), so the mixture should lie nearer the bowl's 10%, i.e. below 15%.
Jaggery in the bowl = 500 × 0.1 = 50 mL; jaggery in the glass = 200 × 0.2 = 40 mL.
Concentration = \( \frac{50 + 40}{500 + 200} = \frac{90}{700} \approx 0.1286 \).
The mixture is about 12.9% (≈ 13%) jaggery, less than 15%, as estimated.

Example 5. Brass is an alloy of copper and zinc. Batch A weighs 200 kg and is 70% copper. Batch B weighs 120 kg and is 50% copper. Batch C is 45% copper. All three batches together give an alloy that is 55% copper. Find the weight of Batch C.
Answer:
Let Batch C weigh y kg.
Copper: A = 200 × 0.7 = 140 kg; B = 120 × 0.5 = 60 kg; C = 0.45y kg.
\( \frac{140 + 60 + 0.45y}{200 + 120 + y} = 0.55 \Rightarrow 200 + 0.45y = 0.55(320 + y) \)
\( 200 + 0.45y = 176 + 0.55y \Rightarrow 0.1y = 24 \Rightarrow y = 240 \)
Check: copper = 140 + 60 + 108 = 308 kg in 560 kg; 308 ÷ 560 = 0.55 ✓
Batch C weighs 240 kg.
Key idea (weighted mean): mixing things of concentrations x₁, x₂, …, x_n in amounts w₁, w₂, …, w_n gives concentration \( \frac{w_1x_1 + w_2x_2 + \cdots + w_nx_n}{w_1 + w_2 + \cdots + w_n} \). Brahmagupta wrote this formula (c. 628 CE) for the mean depth of a pit.

Teacher's Note:
1. Write copper ÷ total = required fraction, then clear the denominator.
2. Always check by recomputing the final concentration.

Example 6. Rehmat scored 60% in internal tests, 64% in the project and 73% in the final exam in Maths. The annual score combines internals, project and final exam in the ratio 3 : 2 : 5. Find Rehmat's annual score.
Answer:
With equal importance the score would be \( \frac{60 + 64 + 73}{3} = \frac{197}{3} \approx 65.67 \)%.
But the ratio 3 : 2 : 5 sets the weights, so use the weighted mean:
\( \frac{60 \times 3 + 64 \times 2 + 73 \times 5}{3 + 2 + 5} = \frac{180 + 128 + 365}{10} = \frac{673}{10} = 67.3 \)
Rehmat's annual score is 67.3%.
This equals the plain mean of 60, 60, 60, 64, 64, 73, 73, 73, 73, 73 (each value written as many times as its weight). Weights show how much importance each value gets.

In-text Question (Table 10.2). The table shows the average weekly expenditure (₹) of three families.
Family A: Housing 1940, Food 1700, Education 1280, Transportation 1200, Healthcare 1770, Recreation 535
Family B: Housing 1750, Food 1546, Education 1500, Transportation 1280, Healthcare 1210, Recreation 0
Family C: Housing 950, Food 1700, Education 1540, Transportation 1400, Healthcare 1300, Recreation 150
What would a cluster-column chart of this data look like?
Answer:
There are two natural choices for what forms a cluster.
Choice 1: each cluster is a category (Housing, Food, …), with one column for each family.

Choice 2: each cluster is a family, with one column for each category.

Fig. 10.1: Two cluster-column charts of Table 10.2.

In-text Questions 1–5. Use the charts above (try Choice 1 first, then Choice 2). Which was easier?
1. Which family spent the least on housing?
2. Which family spent the most on healthcare?
3. Approximately how much did Family A spend on education?
4. Which category did Family B spend most on?
5. Did Family A spend more on food or on healthcare?
Answer:
1. In the Housing cluster, Family C's column is shortest (₹950): Family C.
2. In the Healthcare cluster, Family A's column is tallest (₹1770): Family A.
3. Family A's education column reaches roughly ₹1300 (exactly ₹1280).
4. In Family B's cluster the tallest column is Housing (₹1750).
5. Healthcare (₹1770) is just above Food (₹1700): more on healthcare.
Easier chart: questions 1 and 2 compare families inside one category, so Choice 1 suits them. Questions 4 and 5 compare categories inside one family, so Choice 2 suits them. Question 3 is easy on either.

In-text Questions 6–8. 6. Which family spent the least amount overall?
7. Approximately how much did Family C spend overall?
8. Approximately what percentage of Family C's expenses were on housing?
Was it as easy as before? Is there a chart that makes such questions simpler? Study the chart below (Fig. 10.2) and answer again.
Answer:
Totals: Family A = ₹8425, Family B = ₹7286, Family C = ₹7040.
6. Family C spent the least (₹7040).
7. About ₹7000 (exactly ₹7040).
8. \( \frac{950}{7040} \times 100\% \approx 13.5\% \), i.e. about 13.5% (roughly one-seventh).
From cluster columns this is slow, since six columns must be added for each family. A stacked bar chart joins each family's bars end to end, so the full bar length is the total:

In Fig. 10.2 Family C's bar is shortest and ends a little above 7000; its housing piece (0 to about 950) is roughly one-seventh of the bar. So all three answers can be read almost directly.
Key idea: a stacked bar chart is used when we want to compare totals while still seeing the parts that make them up.

Teacher's Note:
1. Totals are hidden in cluster charts but visible in stacked charts.
2. Percentage share = part ÷ total × 100.

In-text Question. Now use the stacked bar chart (Fig. 10.2) to answer Questions 1–5. Is it possible? Is it easy? If not, what makes it harder?
Answer:
Possible, but not always easy.
1. Housing is the first piece and starts at 0 for every family, so Family C (≈ ₹950) is easy to spot.
2. Healthcare pieces start at different places (about 6100, 6100 and 5600), so their lengths (≈ 1770, 1210, 1300) must be measured: Family A. Harder.
3. Family A's education piece runs from about 3640 to 4920, so it is about ₹1300. Needs subtraction.
4. Family B's longest piece is Housing (≈ ₹1750).
5. Family A's healthcare piece (≈ 1770) is only slightly longer than food (≈ 1700), which is hard to judge by eye.
Why harder: except the first piece, pieces do not start at 0, and the same category starts at different points in different bars. Totals, however, are much easier to see.

In-text Question. Using the stacked bar chart, which family spends the smallest share of its total on healthcare? Was it easy? What chart would make such questions easy?
Answer:
A share is part ÷ total:
Family A: \( \frac{1770}{8425} \times 100\% \approx 21.0\% \); Family B: \( \frac{1210}{7286} \times 100\% \approx 16.6\% \); Family C: \( \frac{1300}{7040} \times 100\% \approx 18.5\% \).
Family B has the smallest healthcare share.
It was not easy from Fig. 10.2 because the bars have different lengths; a short piece in a short bar can still be a big share. Charts built on shares help: pie charts (Fig. 10.3) or a 100% stacked bar chart (Fig. 10.4).


Shares used (A, B, C): Housing 23.0%, 24.0%, 13.5%; Food 20.2%, 21.2%, 24.1%; Education 15.2%, 20.6%, 21.9%; Transportation 14.2%, 17.6%, 19.9%; Healthcare 21.0%, 16.6%, 18.5%; Recreation 6.4%, 0%, 2.1%.
Key idea: in a 100% stacked bar chart every bar has the same length (the whole, 100%) and each piece is proportional to its part, just as a pie splits 360° in proportion.

Teacher's Note:
1. Same length bars make shares comparable across groups.
2. Pies and 100% bars carry the same information; bars are easier to compare side by side.

Example 7. The average daily electricity use of a house (kWh) by category is:
2005: Lighting 360, Cooling 300, Kitchen appliances 120, Other appliances 220
2025: Lighting 220, Cooling 800, Kitchen appliances 460, Other appliances 520
The stacked bar chart of this data is shown. Make the corresponding 100% stacked bar chart.

Answer:
2005: total = 360 + 300 + 120 + 220 = 1000 kWh.
Shares: Lighting 360/1000 = 36%, Cooling 30%, Kitchen 12%, Other 22%.
2025: total = 220 + 800 + 460 + 520 = 2000 kWh.
Shares: Lighting 220/2000 = 11%, Cooling 40%, Kitchen 23%, Other 26%.
Check: 36 + 30 + 12 + 22 = 100 and 11 + 40 + 23 + 26 = 100 ✓

The 100% stacked bar chart is shown above.

Example 7 (In-text Question). What do you observe? What is interesting in the data and the charts?
Answer:
1. Total daily use doubled, from 1000 to 2000 kWh.
2. Cooling went from 300 to 800 kWh and from 30% to 40%; it is now the biggest category (more ACs and coolers, hotter summers).
3. Lighting dropped from 360 to 220 kWh and its share fell from 36% to 11%, probably thanks to LED bulbs.
4. Kitchen appliances almost quadrupled (120 to 460 kWh), share 12% to 23%.
5. Other appliances more than doubled (220 to 520 kWh), yet their share rose only a little (22% to 26%).
6. The 100% chart shows only shares and hides the doubling of the total; the stacked chart shows both.

Example 8. The 100% stacked bar chart compares the share of flowers that bloomed in each season in Fatima's and Naveen's gardens.
Which of the following can be inferred from the chart?
(i) In Fatima's garden there were more blooms in summer than in the monsoon.
(ii) In summer, Fatima's garden had more blooms than Naveen's.
(iii) In winter, Fatima's garden had fewer blooms than Naveen's.
(iv) Both gardens had the same total number of flowers across the seasons, but it varied in each season.
(Hint: work out the blooms per season if both gardens had the same yearly total, and again if the totals were very different.)
Answer:
From the chart: Fatima 50% summer, 30% monsoon, 20% winter; Naveen 40% summer, 25% monsoon, 35% winter.
(i) Can be inferred. Both shares belong to the same total (Fatima's), and 50% > 30%, so summer had more blooms.
(ii) and (iii) cannot be inferred. They compare different gardens whose totals are unknown. Try two cases:
Case 1: Fatima 210 flowers, Naveen 120. Fatima: 105, 63, 42; Naveen: 48, 30, 42.

Summer: Fatima 105 > Naveen 48; winter: both 42, so (iii) fails.
Case 2: Fatima 180 flowers, Naveen 240. Fatima: 90, 54, 36; Naveen: 96, 60, 84.

Summer: Fatima 90 < Naveen 96, so (ii) fails. Both cases produce the same 100% chart, so the chart cannot decide (ii) or (iii).
(iv) Cannot be inferred. A 100% chart draws every total as 100% whatever its real size.
Only statement (i) can be inferred.

Teacher's Note:
1. Within one bar, shares compare fairly; across bars, they need the totals.
2. Inventing two sets of totals is a quick way to test any cross-bar claim.

Think and Reflect. 1. What does this say about the scope of stacked bars and 100% stacked bars?
2. Given a stacked bar chart, can we make the corresponding 100% stacked bar chart?
3. Given a 100% stacked bar chart, can we make the corresponding stacked bar chart?
4. What inferences or comparisons can be made from each kind of chart?
Answer:
1. A stacked bar chart shows actual amounts (totals and parts). A 100% stacked bar chart shows only proportions and hides the totals, so it is useful only for comparing shares.
2. Yes. Read each part and the total, and work out \( \frac{\text{part}}{\text{total}} \times 100\% \) for every part.
3. Not in general. Percentages alone do not give the actual values; many stacked charts (like Cases 1 and 2) lead to one 100% chart. It becomes possible only if the totals are also given.
4. Stacked bar: compare totals, compare the first pieces (common base), estimate each part and roughly its fraction of the bar. 100% stacked bar: compare shares within a bar and the share of one category across bars, but not amounts or totals.

Example 9. The chart shows how an average Indian spends the 24 hours of a day. What do you notice? What do you wonder about?

Answer:
Converting the shares to hours (share × 24):
Sleep 34% ≈ 8.2 h; Personal care 13% ≈ 3.1 h; Paid work 10% = 2.4 h; Unpaid work & care 11% ≈ 2.6 h; Learning 5% = 1.2 h; Leisure, social & travel 27% ≈ 6.5 h. Total 100% = 24 h.
Noticing:
1. Sleep is the largest part, about one-third of the day (≈ 8 h).
2. Leisure, social life and travel together take more than a quarter (≈ 6.5 h).
3. Paid work is only about 2.4 h and learning about 1.2 h.
4. Unpaid work and care (housework, looking after children and elders) is a little more than paid work.
Wondering:
1. Paid work looks small because the average includes children, students, the elderly and people without paid jobs; learning is small because most adults do not study.
2. How would the strip look for different age groups, for men and women, or for villages and cities?
The strip mixes people of all ages and backgrounds, so it may not match the day of any single real person.

Think and Reflect. Grouping the same data by age gives the stacked column chart below (left axis: share of the day; right axis: hours).
1. What do you find interesting? What can you infer?
2. Does the sleep-with-age trend you studied last year match this chart?
3. Why is the learning time of the 15–24 group much lower than that of the 6–14 group?
4. Do adults spend about equal time on paid and unpaid work? What do you think?
Answer:
1. Children (6–14) spend about 22% of the day (≈ 5.3 h) learning and almost nothing on paid work. Paid work appears among youth (7%) and peaks for adults (14% ≈ 3.4 h). Leisure is about 26% for the first three groups but about 33% (≈ 8 h) for the elderly. Personal care stays nearly constant (13–14%). The single average day of Fig. 10.5 hides these big differences.
2. Sleep: children 38% ≈ 9.1 h, youth 34% ≈ 8.2 h, adults 32% ≈ 7.7 h, elderly 37% ≈ 8.9 h. Sleep falls from childhood to adulthood, matching the usual trend; it rises again for the elderly, who have more free time after retiring.
3. Many 15–24 year-olds have finished school or college and have begun paid work (7%) or household/care work (9%). Only some are still students, so the group's average learning falls to 11%, against 22% for 6–14 year-olds, nearly all of whom attend school.
4. On average adults spend 14% (≈ 3.4 h) on paid work and 15% (≈ 3.6 h) on unpaid work, which is nearly equal. But this is an average; often some people (frequently women) do most of the unpaid work while others do mostly paid work. Data split by gender would be needed to be sure.

Teacher's Note:
1. An average over a mixed group can hide large differences between sub-groups.
2. Convert shares to hours (share × 24) to make them concrete.

In-text Question. What type of chart is Fig. 10.6: a stacked bar chart or a 100% stacked bar chart?
Answer:
It is both.
It is a stacked chart because each piece shows an actual amount of time (hours, right-hand axis).
It is also a 100% stacked chart because every column has the same total, 24 hours = 100% (left-hand axis), so the pieces also show shares.
(Whenever every total is the same, a stacked chart is automatically a 100% stacked chart as well.)

Think and Reflect. Recall the 'What Can A Strip Say?' activity from last year. If, in each strip, we club together all the tiny pieces of the same activity, do we get a 100% stacked bar like those in Fig. 10.6?
Answer:
Yes. Each strip stands for one full day of 24 hours.
Joining all the pieces of the same activity (all sleep periods together, all learning periods together, and so on) leaves one strip of fixed length divided into one piece per activity.
That is exactly one bar of a 100% stacked bar chart, like each column of Fig. 10.6 (which shows such a bar for the average person of each age group).