Chapter 10: How Quantities Combine: Understanding Data

Class 9 Ganita Manjari Part 2 · notes, exercises and mini test.

Learning map

What this chapter is teaching

This chapter connects weighted averages, mixtures, percentages, tables and charts. The central question is: when quantities are combined, which values should count more?

CombineFind a mean using every quantity and its size.
WeightGive larger groups or more important parts their correct influence.
UpdateRevise an average when new data enters or leaves.
VisualiseChoose charts that show totals, parts or percentages clearly.
Study tip: Before calculating, ask what the weights are: number of items, volume, marks ratio, frequency or total size.
Coverage check: Notes 1–9 cover the methods needed for Exercises 10.1–10.5 and the end-of-chapter questions. For chart-based questions, read the original figure for exact approximate values; the supplied text did not include those images.

1Ordinary mean and combined mean (Average of Averages)

In earlier grades, we saw how the average (arithmetic mean) conveys the ‘centre’ of given data:

Mean = total of all values / number of values

When combining multiple data sets, a fundamental question arises: can we simply average the individual averages? Let us examine the textbook examples.

Example 1: Trainees at a Badminton Academy

In a badminton academy, there is a group of 11 trainees — 8 seniors and 3 juniors. Their individual heights and average heights are given below:

GroupHeights (in cm)CountAverage Height
Seniors165, 169, 164, 167, 170, 159, 164, 1668165.5 cm
Juniors146, 149, 1533149.33 cm

Two students calculate the overall average height of the whole group in two different ways:

Method 1 (Simple average of averages):
(165.5 + 149.33) / 2 = 314.83 / 2 = 157.415 cm

Method 2 (Sum of all individual heights ÷ 11):
(165 + 169 + 164 + 167 + 170 + 159 + 164 + 166 + 146 + 149 + 153) / 11 = 1772 / 11 ≈ 161.09 cm

Whose calculation is correct? Method 2 gives the correct average height of the whole group because the mean is the sum of all 11 heights divided by 11.

Think and Reflect — Why did Method 1 not work?
Method 1 does not work because it treats the senior and junior groups equally (giving each a weight of 50%), but there are 8 seniors and only 3 juniors! The seniors heavily outnumber the juniors, so the group average must be pulled much closer to the seniors' average (165.5 cm).
Shreyas's Calculation:
[(165.5 × 8) + (149.33 × 3)] / (8 + 3) = (1324 + 448) / 11 = 1772 / 11 ≈ 161.09 cm
Notice why this works: 165.5 × 8 = 1324 cm recovers the sum of heights of all 8 seniors, and 149.33 × 3 ≈ 448 cm recovers the sum of heights of all 3 juniors. Dividing the combined total by (8 + 3) gives the exact combined mean!

Generalisation & Algebraic Derivation: Combined Mean of Two Collections

Suppose we have two collections of data:

  1. Collection 1 has n values: x1, x2, …, xn with average a.
    By definition: (x1 + x2 + … + xn) / n = a &implies; Sum1 = a × n.
  2. Collection 2 has m values: y1, y2, …, ym with average b.
    By definition: (y1 + y2 + … + ym) / m = b &implies; Sum2 = b × m.
  3. Combined Average:
    Combined Mean = (Sum1 + Sum2) / (n + m) = (a × n + b × m) / (n + m)

Example 2: Jaspreet's Cycling Times (Table 10.1)

Jaspreet tracked how much time she spent cycling on weekdays over 3 weeks (5 days each week):

WeekDay 1Day 2Day 3Day 4Day 5Weekly Average
Week 1101310151612.8 min
Week 2141020171815.8 min
Week 3151814202318.0 min
Method 1 (Average of weekly averages):
(12.8 + 15.8 + 18.0) / 3 = 46.6 / 3 ≈ 15.53 min

Method 2 (Sum of all daily times ÷ 15 days):
[(12.8 × 5) + (15.8 × 5) + (18.0 × 5)] / (5 + 5 + 5) = (64 + 79 + 90) / 15 = 233 / 15 ≈ 15.53 min

Why does Method 1 work in this case?

Theorem — When does simple average of averages work?
For 3 collections with averages a, b, c and sizes p, q, r:
Combined Mean = (a×p + b×q + c×r) / (p + q + r)
When all three collections have equal size (i.e. p = q = r = 5 weekdays):
(a×p + b×p + c×p) / (p + p + p) = [(a + b + c)×p] / (3p) = (a + b + c) / 3
Rule: When the sizes of all collections are equal, we can simply add their averages and divide by the number of collections. If collection sizes differ, we MUST use the weighted mean.
PracticeExercise 10.1 Q1, Q4End Q1, Q3, Q14

2Weighted average: the main tool

The weighted average (or weighted mean) is a generalisation of the arithmetic mean where each value is multiplied by a measure of how important, frequent, or large it is:

Weighted Mean x̄ = (w1x1 + w2x2 + … + wnxn) / (w1 + w2 + … + wn) = Σ(wi × xi) / Σwi

Here, the prefix ‘weighted’ does not necessarily refer to physical weight, but abstract weights (sizes, volumes, counts, credits, frequencies, or priority ratios).

Historical Origins in Ancient India:
  • Brahmagupta in his seminal treatise Brāhmasphuṭasiddhānta (c. 628 CE) formulated this exact principle to find the mean depth x̄ of an irregularly shaped pit: if the pit has depth x1 for length w1, depth x2 for length w2, …, then the average depth is given by the weighted mean formula.
  • Śrīdharācārya (c. 750 CE) in Pāṭīga logic;ita used the weighted arithmetic mean to calculate the purity of mixed gold-alloys (the ancient Indian varṇa system, identical to modern karat calculations).
  • By the 1700s, the concept gained widespread prominence across Europe, becoming the cornerstone of modern statistics.
Equal weightsWhen w1 = w2 = … = wn = w, the weights cancel out: wΣxi / (n×w) = Σxi / n, giving the ordinary arithmetic mean.
Doubling weightsMultiplying all weights by a constant k > 0 leaves the weighted average completely unchanged: Σ(kwixi) / Σ(kwi) = Σ(wixi) / Σwi.
Adding a constantAdding a constant (e.g. +1) to every weight alters their relative ratios, changing the weighted average (unless all xi are already equal to the mean).
Bounding propertyThe weighted average must always lie strictly between the smallest and largest values: min(xi) ≤ x̄ ≤ max(xi).
Practice10.110.210.310.5End exercises

3Mixtures, concentration and purity

When substances with different concentrations are blended, the mixture concentration is determined by the total quantity of solute divided by the total volume of mixture:

Concentration of Mixture = quantity of active ingredient in mixture / total quantity of mixture

Example 3: Mixing Equal Quantities of Lemonade

Two glasses of equal quantities of lemonade are prepared. One glass has 10% jaggery and the other has 20% jaggery. If we mix both glasses, what is the concentration of jaggery in the mixture?

Mathematical Derivation:
Let y be the quantity of lemonade in each glass.
Amount of jaggery in first glass = (10/100) × y = 0.1y
Amount of jaggery in second glass = (20/100) × y = 0.2y
Total mixture volume = y + y = 2y

Concentration = (0.1y + 0.2y) / (y + y) = 0.3y / 2y = 0.15 = 15%
Insight: Because both glasses had equal quantities, the concentration of jaggery in the combined mixture is exactly midway (15%) between 10% and 20%.

Example 4: Mixing Unequal Quantities of Lemonade

A bowl of 500 mL lemonade has 10% jaggery. A glass of 200 mL lemonade has 20% jaggery. If we mix the bowl and the glass, what is the concentration of jaggery in the mixture?

Can you estimate beforehand? Is it 15%, more, or less?
Since the bowl has a larger volume (500 mL vs 200 mL), the final concentration must be pulled closer to 10% (i.e. less than midway).

Concentration = (500 × 0.10 + 200 × 0.20) / (500 + 200) = (50 + 40) / 700 = 90 / 700 ≈ 0.1286 ≈ 12.86% (≈ 13%)
The final concentration is not the simple average (15%), but closer to 10% because the 10% mixture contributed over 71% of the total volume.

Example 5: Brass Alloy — Finding the Weight of Batch C

Brass is an alloy of copper and zinc. Batch A weighs 200 kg (70% copper). Batch B weighs 120 kg (50% copper). Batch C is 45% copper. When all three batches are combined, the resulting alloy is 55% copper. What is the weight of Batch C?

Step-by-step Solution:
  1. Let y be the weight (in kg) of Batch C.
  2. Quantity of copper in Batch A = 200 × 0.70 = 140 kg
  3. Quantity of copper in Batch B = 120 × 0.50 = 60 kg
  4. Quantity of copper in Batch C = y × 0.45 = 0.45y kg
  5. Total weight of mixture = 200 + 120 + y = (320 + y) kg
  6. Set up the concentration equation:
    0.55 = (140 + 60 + 0.45y) / (320 + y) = (200 + 0.45y) / (320 + y)
  7. Cross-multiply and solve for y:
    0.55 × (320 + y) = 200 + 0.45y
    176 + 0.55y = 200 + 0.45y
    0.55y − 0.45y = 200 − 176
    0.10y = 24 &implies; y = 24 / 0.10 = 240 kg
Answer: Batch C weighs 240 kg.
PracticeExercise 10.1 Q2, Q3, Q5Exercise 10.3 Q3, Q4, Q8End Q4, Q5

4Marks, profit and custom weights (Rating Systems)

The weighted mean is indispensable whenever different parts of data require different levels of importance or weightage (prescribed assessment ratios, store margins, or customer rating rubrics).

Example 6: Rehmat's Annual Score in Mathematics

Rehmat’s marks in Maths are as follows: 60% in internal tests, 64% in the project, and 73% in the final exam. The annual percentage score is calculated by combining internals, project, and final exam in the ratio 3 : 2 : 5. What is Rehmat’s annual score in Maths?

Equal Weightage vs. Prescribed Ratio:
  • If all modes had equal weightage, the arithmetic mean would be:
    (60 + 64 + 73) / 3 = 197 / 3 ≈ 65.67%
  • The ratio 3 : 2 : 5 indicates that internals get 3 parts, project gets 2 parts, and the final exam gets 5 parts of importance out of a total of (3 + 2 + 5 = 10) parts. This is equivalent to having 10 test scores: 60, 60, 60, 64, 64, 73, 73, 73, 73, 73.
  • Arithmetic Mean with relative weights:
    Annual Score = (60 × 3 + 64 × 2 + 73 × 5) / (3 + 2 + 5) = (180 + 128 + 365) / 10 = 673 / 10 = 67.3%
  • Expressed as weighted percentages:
    [(60/100)×3 + (64/100)×2 + (73/100)×5] / (3 + 2 + 5) = (673/100) / 10 = 0.673 = 67.3%

Real-Life Situations for Custom Weights

1. Hotel / Restaurant RatingsA customer experience depends on Food, Ambience, and Service. If assigned weights are Food : Service : Ambience = 4 : 3 : 2, and a customer rates Food = 5, Service = 3, Ambience = 4, then:
Combined rating = (5×4 + 3×3 + 4×2) / (4 + 3 + 2) = (20 + 9 + 8) / 9 = 37 / 9 ≈ 4.11 / 5.
2. Product Review RecencyA store selling portable furniture frequently upgrades its designs. To calculate the overall store rating, recent reviews (e.g. last 3 months) can be assigned twice the weightage of older reviews so current quality is accurately reflected.
Crucial Reminder for Exercise 10.2: When marks are given out of different maximum scores (e.g. 35/50, 44/60, 80/100), you must first convert each score into a percentage (or fraction of 1) before applying the weights. Combining raw marks directly without normalisation is incorrect!
PracticeExercise 10.2 Q1Exercise 10.3 Q1, Q5, Q6End Q13

5How an average changes

Keep the old total visible instead of starting from the old average alone.

New mean = (old number × old mean + added value) / (old number + 1)
  • If the new value is above the old mean, the mean rises.
  • If the new value is below the old mean, the mean falls.
  • If the new value equals the old mean, the mean stays unchanged.
  • When values are removed, subtract their total and subtract their count.

This method applies to the stork distance, run rate, gold purchases, share prices and changing langur groups.

Practice10.3 Q2, Q710.5End Q1, Q3–Q6

6Dilution and possible target values

When a solution is diluted with pure water, the amount of spice or salt stays the same while the total volume increases.

New concentration = amount of solute / new total volume
  • To reduce 8% spice to three-fourths of its original concentration, the target is 6%.
  • A mixture of two concentrations must have a concentration between them, including the endpoints.
  • It cannot have a concentration lower than both sources unless a still-purer source is added.
  • Set up an equation using solute amount before and after mixing; do not subtract percentages directly.
Practice10.1 Q510.3 Q3–Q4, Q8

7Unknown counts and frequency tables

Use a variable for each unknown group size and let the total count provide a second equation.

Overall mean = Σ(value × frequency) / Σfrequency
  • For the langur table, let x and y be the male and female counts. Then x + y = 60.
  • The overall-weight equation is (16.5x + 13.8y) / 60 = 14.925.
  • For ratings, multiply each rating by its frequency before adding.
  • After a person or animal enters or leaves, update both the total and the count.
Reasoning before calculation: If a group’s mean is above the overall mean, that group must pull the combined mean upward; this helps check whether an answer is sensible.
Practice10.3 Q6–Q8End Q2, Q14–Q15

8Tables, stacked bars and 100% stacked bars

Data containing multiple categories across several groups can be visualised in different formats depending on whether our goal is comparing individual categories, total quantities, or relative percentage shares.

Textbook Case Study: Weekly Household Expenditure (Table 10.2)

FamilyHousingFoodEducationTransportationHealthcareRecreationTotal
Family A₹1,940₹1,700₹1,280₹1,200₹1,770₹535₹8,425
Family B₹1,750₹1,546₹1,500₹1,280₹1,210₹0₹7,286
Family C₹950₹1,700₹1,540₹1,400₹1,300₹150₹7,040
Cluster Column ChartsBest for comparing specific categories across groups (e.g. comparing Housing across Families A, B, and C). However, it is very hard to see the total expenditure of each family.
Stacked Bar ChartsBars are stacked one after another. The overall length immediately reveals the grand total for each family. However, categories beyond the first do not start at zero, making subtle comparisons between individual middle segments harder.
100% Stacked Bar ChartsEvery bar has the exact same fixed length (100%). The segments show the relative proportions (shares) of each category within that family. Ideal alternative to pie charts.

Example 7: Powering Up! Household Electricity Consumption

Average category-wise daily electricity consumption of a house (in kWh) in 2005 and 2025:

YearLightingCoolingKitchen appliancesOther appliancesTotal Consumption
2005360 kWh300 kWh120 kWh220 kWh1000 kWh
2025220 kWh800 kWh460 kWh520 kWh2000 kWh

Converting to a 100% Stacked Bar Chart:

  1. For Year 2005 (Total = 1000 kWh):
    • Lighting: (360 / 1000) × 100 = 36%
    • Cooling: (300 / 1000) × 100 = 30%
    • Kitchen: (120 / 1000) × 100 = 12%
    • Other: (220 / 1000) × 100 = 22%
    Sum = 36% + 30% + 12% + 22% = 100%.
  2. For Year 2025 (Total = 2000 kWh):
    • Lighting: (220 / 2000) × 100 = 11%
    • Cooling: (800 / 2000) × 100 = 40%
    • Kitchen: (460 / 2000) × 100 = 23%
    • Other: (520 / 2000) × 100 = 26%
    Sum = 11% + 40% + 23% + 26% = 100%.
Key Observation: Although lighting consumption in kWh decreased only moderately (from 360 to 220 kWh), its share of total household energy plummeted from 36% down to 11% due to energy-efficient LED adoption and the massive rise of cooling and electric kitchen appliances.
PracticeExercise 10.4 Q1, Q2Exercise 10.5 Q1End Q7, Q9, Q10

9Percentages, shares and inferences (What Can We Infer?)

A critical statistical skill is knowing what can and cannot be inferred from percentage-based visualisations.

Example 8: Flower Blooms across Seasons (Fatima vs. Naveen)

A 100% stacked bar chart shows the proportion of flowers blooming across seasons:

  • Fatima's Garden: Summer = 50%, Monsoon = 30%, Winter = 20% (Total = 100%)
  • Naveen's Garden: Summer = 40%, Monsoon = 35%, Winter = 25% (Total = 100%)

Evaluating Statements:

Statement (i): In Fatima’s garden, there were more blooms in the summer than in the monsoon.
True — This compares seasons within the same garden. Since both percentages refer to the same total, 50% of the total is definitely greater than 30% of the total.

Statement (ii): In summer, Fatima’s garden had more blooms than Naveen’s garden.
Cannot be inferred — This compares across two different gardens. A 100% chart does not state total flowers! If Naveen had 1,000 flowers (40% = 400) and Fatima had 100 flowers (50% = 50), Naveen had far more flowers.

Statement (iii): In winter, Fatima’s garden had fewer blooms than Naveen’s garden.
Cannot be inferred — Same logic as (ii). Percentage share cannot determine absolute count without knowing group totals.

Statement (iv): The total number of flowers, across seasons, is the same in both gardens.
False / Cannot be inferred — Any two totals can be normalised to a 100% bar.
Concrete Proof: Case 1 vs. Case 2
• Case 1: Fatima has 210 blooms (Summer: 105) and Naveen has 120 blooms (Summer: 48). Here Fatima has more summer blooms.
• Case 2: Fatima has 180 blooms (Summer: 90) and Naveen has 240 blooms (Summer: 96). Here Fatima has fewer summer blooms!
Both cases produce the exact same 100% stacked bar chart, proving absolute comparisons across groups are impossible.
Think and Reflect — Conversions between Charts:
  • Stacked Bar → 100% Stacked Bar: Always possible. Calculate the total length of each bar and find each segment’s percentage.
  • 100% Stacked Bar → Stacked Bar: Impossible without external data. You cannot reconstruct absolute numbers without knowing the grand total.

Example 9: A Typical Day — National Time Use Survey 2024

The time spent on daily activities by the average Indian (out of 24 hours):

ActivitySleepPersonal CarePaid WorkUnpaid Work & CareLearningLeisure & Social
National Average34% (8.2 h)13% (3.1 h)10% (2.4 h)11% (2.6 h)5% (1.2 h)27% (6.5 h)
Children (6–14)38%13%0%7%22%26%
Youth (15–24)37%14%13%9%9%26%
Adults (25–59)34%11%26%14%1%13%
Elderly (60+)32%15%7%9%1%33%
Is this a stacked bar chart or a 100% stacked bar chart?
It is BOTH! Because every human’s day consists of exactly 24 hours, the total length of each bar is constant (24h = 100%). Thus, the lengths of segments represent both absolute hours and percentage proportions simultaneously.
Practice10.410.5End Q7, Q8, Q11, Q16

10Chapter Summary — Key Takeaways

  • The weighted average (weighted mean) of x1, x2, …, xn with weights w1, w2, …, wn is given by:
    x̄ = (w1x1 + w2x2 + … + wnxn) / (w1 + w2 + … + wn)
  • An average of averages can be computed accurately using the weighted mean. The weighted mean is also used to calculate the new mean when values are added to or removed from data.
  • The weighted average is essential in calculating the concentration of mixtures and alloys.
  • Whenever different parts of data have different group sizes or levels of importance, the weighted mean must be used by assigning appropriate weights (e.g. rating systems, multi-component exam scoring).
  • Stacked bar charts are preferred over cluster bar charts when the total of each cluster is also a matter of primary interest.
  • 100% stacked bar charts, like pie charts, compare proportions/shares and not absolute values.
  • Given a stacked bar chart, a corresponding 100% stacked bar chart can always be made, but not vice versa.

Exercise Set 10.1 — Solutions

Combined averages, equal quantities, purity and concentration.

1The average score of students on a test in Section A is 72 and that of students in Section B is 76. What is the combined average of both the sections given that Section A has 30 students and Section B has 25 students?
Answer / method

Step-by-step Solution:

  1. Given:
    • Section A: Number of students n = 30, Average score a = 72
    • Section B: Number of students m = 25, Average score b = 76
  2. Calculate total scores for each section:
    • Total score in Section A = 30 × 72 = 2,160
    • Total score in Section B = 25 × 76 = 1,900
  3. Combined total score:
    Total score = 2,160 + 1,900 = 4,060
  4. Total number of students:
    Total students = 30 + 25 = 55
  5. Combined Average:
    Combined Mean = (n×a + m×b) / (n + m) = 4060 / 55 = 812 / 11 ≈ 73.82
Answer: The combined average score of both sections is 73.82 (or 73 9/11).
2A farmer mixes three equal quantities of fertilisers. The first one contains 1/10 nitrogen, the second contains 9/50 nitrogen, and the third contains 3/60 nitrogen. What is the fraction of nitrogen in the mixture?
Answer / method

Step-by-step Solution:

  1. Simplify the given nitrogen fractions:
    • First fertiliser = 1/10
    • Second fertiliser = 9/50
    • Third fertiliser = 3/60 = 1/20
  2. Use the Equal Quantities Principle:
    Since the farmer mixes three equal quantities, the combined concentration is the ordinary arithmetic mean of the three fractions:
    Combined Fraction = (1/10 + 9/50 + 1/20) / 3
  3. Find a common denominator (LCM of 10, 50, 20 is 100):
    • 1/10 = 10/100
    • 9/50 = 18/100
    • 1/20 = 5/100
    Sum of fractions = (10 + 18 + 5) / 100 = 33/100
  4. Divide by 3:
    (33 / 100) ÷ 3 = 33 / (100 × 3) = 11 / 100
Answer: The fraction of nitrogen in the mixture is 11/100 (or 11%).
3(Śrīdharācārya, Pāṭīgaṇita, c. 750 CE) In ancient India, Varṇa was the measure of gold purity. A purity of 16 varṇa meant pure gold; in general, a purity of k varṇa meant that the gold-alloy was k/16 gold and the rest impurities. (Now the term used is karat; 16 Varna = 24 karat.) Suppose a goldsmith melts together three pieces of gold: 9 units at 12 varṇa, 5 units at 10 varṇa, and 17 units at 11 varṇa. Find the purity in varṇa of the combined gold.
Answer / method

Step-by-step Solution:

  1. Given:
    • Piece 1: Weight w1 = 9 units, Purity x1 = 12 varṇa
    • Piece 2: Weight w2 = 5 units, Purity x2 = 10 varṇa
    • Piece 3: Weight w3 = 17 units, Purity x3 = 11 varṇa
  2. Total weight of melted gold:
    W = 9 + 5 + 17 = 31 units
  3. Weighted sum of purity:
    Σ(wi × xi) = (9 × 12) + (5 × 10) + (17 × 11) = 108 + 50 + 187 = 345
  4. Combined Purity in varṇa:
    Combined Purity = Σ(wi × xi) / Σwi = 345 / 31 = 11 4/31 ≈ 11.13 varṇa
Answer: The purity of the combined gold is 345/31 ≈ 11.13 varṇa (equivalent to ≈ 16.69 karat).
4The average rainfall per day in the months of May, June, and July in a certain location are 3.5 mm, 10 mm and 8.7 mm respectively. Write an expression that gives their combined average.
Answer / method

Step-by-step Solution:

  1. Method A — Accounting for exact number of days in each month:
    • May has 31 days (Daily average = 3.5 mm &implies; Total rain = 3.5 × 31 mm)
    • June has 30 days (Daily average = 10 mm &implies; Total rain = 10 × 30 mm)
    • July has 31 days (Daily average = 8.7 mm &implies; Total rain = 8.7 × 31 mm)
    • Total days = 31 + 30 + 31 = 92 days.
    Combined Daily Average = (3.5 × 31 + 10 × 30 + 8.7 × 31) / (31 + 30 + 31) = (108.5 + 300 + 269.7) / 92 = 678.2 / 92 ≈ 7.37 mm/day
  2. Method B — If each month is treated as an equal unit:
    Combined Average = (3.5 + 10 + 8.7) / 3 = 22.2 / 3 = 7.4 mm/day
Answer Expression: (3.5 × 31 + 10 × 30 + 8.7 × 31) / 92 (or simplified monthly average (3.5 + 10 + 8.7) / 3).
5Calculate the concentration of spice mix in these two scenarios. (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 / method

Step-by-step Solution:

  1. Scenario (i):
    • Spice from 100 mL @ 5% = 100 × 0.05 = 5 mL
    • Spice from 200 mL @ 10% = 200 × 0.10 = 20 mL
    • Spice from 300 mL @ 15% = 300 × 0.15 = 45 mL
    • Total spice = 5 + 20 + 45 = 70 mL
    • Total volume = 100 + 200 + 300 = 600 mL
    Concentration = (70 / 600) × 100% = 70 / 6 % = 35 / 3 % ≈ 11.67%
  2. Scenario (ii):
    • Spice from 300 mL @ 5% = 300 × 0.05 = 15 mL
    • Spice from 200 mL @ 10% = 200 × 0.10 = 20 mL
    • Spice from 100 mL @ 15% = 100 × 0.15 = 15 mL
    • Total spice = 15 + 20 + 15 = 50 mL
    • Total volume = 300 + 200 + 100 = 600 mL
    Concentration = (50 / 600) × 100% = 50 / 6 % = 25 / 3 % ≈ 8.33%
Key Takeaway: In scenario (i), the largest volume (300 mL) had 15% spice, pulling the concentration up to 11.67%. In scenario (ii), the largest volume (300 mL) had 5% spice, pulling the concentration down to 8.33%.

Exercise Set 10.2 — Solutions

Weighted marks and converting scores to percentages.

1Savitri's marks in Kashmiri are as follows: 35 out of 50 in internal tests, 44 out of 60 in the project, and 80 out of 100 in the final exam. The annual percentage score is calculated by combining the internals, project, and final exam in the ratio 3:4:5. Which of the following expression(s) gives her annual score (as a percentage) in Kashmiri?
  • (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 / method

Step-by-step Analysis:

  1. Normalise raw marks to percentages:
    • Internal tests: 35 out of 50 = (35/50) × 100 = 70%
    • Project: 44 out of 60 = (44/60) × 100 = 73.33%
    • Final exam: 80 out of 100 = (80/100) × 100 = 80%
  2. Evaluate each expression:
    • (i) is INCORRECT: It averages raw marks (35, 44, 80) directly. Because they are out of different maximums (50, 60, 100), combining raw scores creates false weightage.
    • (ii) is INCORRECT: It divides 35 and 44 by 100 instead of their actual maximum totals of 50 and 60.
    • (iii) is CORRECT: It calculates the fractional scores (35/50, 44/60, 80/100), weights them by 3, 4, 5, divides by 12, and multiplies by 100 at the end to get the percentage.
    • (iv) is CORRECT: It converts each component directly into a percentage first [(35/50)×100, (44/60)×100, (80/100)×100], and then computes their weighted average with weights 3, 4, 5.
Correct Expressions: (iii) and (iv) are both mathematically correct and yield approximately 75.28%.

Exercise Set 10.3 — Solutions

Updating means, mixtures, dilution, weighted ratings and frequency tables.

1A stationery shop owner made ₹8000 selling books, of which 30% is the profit amount, and ₹1000 selling book covers, of which 50% is the profit amount. What is the percentage of profit on the total sales?
Answer / method

Step-by-step Solution:

  1. Find profit from selling books:
    Sales = ₹8,000, Profit% = 30%
    Profit from books = 30% of 8,000 = (30/100) × 8,000 = ₹2,400
  2. Find profit from selling book covers:
    Sales = ₹1,000, Profit% = 50%
    Profit from covers = 50% of 1,000 = (50/100) × 1,000 = ₹500
  3. Calculate combined total sales and total profit:
    • Total Sales = ₹8,000 + ₹1,000 = ₹9,000
    • Total Profit = ₹2,400 + ₹500 = ₹2,900
  4. Calculate overall profit percentage on total sales:
    Profit % = (Total Profit / Total Sales) × 100 = (2,900 / 9,000) × 100 = 290 / 9 % = 32 2/9 % ≈ 32.22%
Answer: The percentage of profit on the total sales is 32.22% (or 32 2/9 %).
2A white stork's migration is tracked. The average daily distance travelled, calculated over 20 days, is 44.5 km. On the 21st day, it flew 55 km. What is the average daily distance travelled over these 21 days? Make a guess before you calculate.
Answer / method

Step-by-step Solution:

  1. Guess before calculating:
    Since the distance flown on day 21 (55 km) is greater than the 20-day average (44.5 km), the new average must increase above 44.5 km (a reasonable guess is around 45 km).
  2. Total distance flown in the first 20 days:
    Total Distance20 = 20 × 44.5 = 890 km
  3. Add distance on the 21st day:
    Total Distance21 = 890 + 55 = 945 km
  4. Compute new average daily distance over 21 days:
    New Mean = 945 / 21 = 45 km
Answer: The average daily distance travelled over the 21 days is exactly 45 km.
3A 600 mL solution with 5% salt is mixed with a 300 mL solution with 8% sugar. What are the concentrations of salt and sugar in the mixture?
  • (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 / method

Step-by-step Solution:

  1. Total volume of the combined mixture:
    Total Volume = 600 mL + 300 mL = 900 mL
  2. Amount of salt in the mixture:
    Salt comes only from the first solution: 5% of 600 mL = 0.05 × 600 = 30 mL
    Salt Concentration = (30 / 900) × 100% = 30 / 9 % = 10 / 3 % ≈ 3.33%
  3. Amount of sugar in the mixture:
    Sugar comes only from the second solution: 8% of 300 mL = 0.08 × 300 = 24 mL
    Sugar Concentration = (24 / 900) × 100% = 24 / 9 % = 8 / 3 % ≈ 2.67%
Correct Option: (v) Salt: 3.33%, Sugar: 2.67%. (Note: salt and sugar are separate solutes; their concentrations are computed individually against the new 900 mL total).
4At a panipuri (golgappa) stall, the concentration of spice in the pani (spiced water) was 8%. Many customers complained that it was too spicy. What quantity of regular water should be mixed into the 10 litres of pani so that the spice level is reduced to (3/4)th of the original concentration?
Answer / method

Step-by-step Solution:

  1. Find target spice concentration:
    Target concentration = (3/4) × 8% = 6% (or 0.06)
  2. Quantity of pure spice present in original 10 L:
    Amount of spice = 8% of 10 L = 0.08 × 10 = 0.8 litres
  3. Set up equation with added water w:
    When regular water (0% spice) is added, the amount of spice remains 0.8 L while the total volume becomes (10 + w) L:
    0.8 / (10 + w) = 0.06
  4. Solve for w:
    10 + w = 0.8 / 0.06 = 80 / 6 = 40 / 3 = 13 1/3 litres
    w = 13 1/3 − 10 = 3 1/3 litres ≈ 3.33 litres
Answer: 3 1/3 litres (or 3.33 L) of regular water must be added.
5A physical fitness evaluation is being undertaken. The final marks are calculated by combining the marks for strength, flexibility, and agility in the ratio 4:5:6. Rashi has scored 60, 65, and 70. Keerthi has scored 55, 65, and 75 respectively. (i) Find out whose total is more without calculating. (ii) What are their final marks?
Answer / method

Step-by-step Solution:

  1. (i) Comparing without calculating:
    • Both scored identical marks in flexibility (65).
    • In strength (weight 4), Rashi scored 60 and Keerthi scored 55. Rashi is ahead by 5 marks: advantage = 5 × 4 = +20.
    • In agility (weight 6), Keerthi scored 75 and Rashi scored 70. Keerthi is ahead by 5 marks: advantage = 5 × 6 = +30.
    • Since agility carries a higher weight (6 > 4), Keerthi's lead of 5 marks in agility contributes more points (+30) than Rashi's lead in strength (+20).
    Therefore, Keerthi's weighted total is higher!
  2. (ii) Computing final marks:
    Total weight = 4 + 5 + 6 = 15
    • Rashi's Final Marks:
    (60 × 4 + 65 × 5 + 70 × 6) / 15 = (240 + 325 + 420) / 15 = 985 / 15 = 197 / 3 = 65 2/3 ≈ 65.67
    • Keerthi's Final Marks:
    (55 × 4 + 65 × 5 + 75 × 6) / 15 = (220 + 325 + 450) / 15 = 995 / 15 = 199 / 3 = 66 1/3 ≈ 66.33
Answer: (i) Keerthi's total is more; (ii) Rashi scored 65.67 (65 2/3) and Keerthi scored 66.33 (66 1/3).
6A restaurant collected ratings from 10 customers on a scale of 1 to 5. The resulting data is shown in the table below. What is the average rating if the metrics are combined with the weights food : ambience : service = 6:5:4?
Metric5★4★3★2★1★Total Customers
Food5320010
Ambience0451010
Service1224110
Answer / method

Step-by-step Solution:

  1. Find average rating for Food (out of 10 customers):
    Sum = (5 × 5) + (4 × 3) + (3 × 2) + (2 × 0) + (1 × 0) = 25 + 12 + 6 = 43
    Average Food Rating = 43 / 10 = 4.3
  2. Find average rating for Ambience (out of 10 customers):
    Sum = (5 × 0) + (4 × 4) + (3 × 5) + (2 × 1) + (1 × 0) = 0 + 16 + 15 + 2 = 33
    Average Ambience Rating = 33 / 10 = 3.3
  3. Find average rating for Service (out of 10 customers):
    Sum = (5 × 1) + (4 × 2) + (3 × 2) + (2 × 4) + (1 × 1) = 5 + 8 + 6 + 8 + 1 = 28
    Average Service Rating = 28 / 10 = 2.8
  4. Combine with weights Food : Ambience : Service = 6 : 5 : 4 (Total weight = 15):
    Combined Rating = (4.3 × 6 + 3.3 × 5 + 2.8 × 4) / (6 + 5 + 4) = (25.8 + 16.5 + 11.2) / 15 = 53.5 / 15 = 107 / 30 ≈ 3.57
Answer: The combined average restaurant rating is 3.57 out of 5.
7The following table shows the weight data of langurs in an animal facility. Without doing any computations, can you tell whether there are more male langurs or more female langurs? Or are they equal in number?
Male langursFemale langursAll langurs
Average weight16.5 kg13.8 kg14.925 kg
Number of langursxy60
  1. (i) Which expression(s) describe 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
  2. (ii) Find out how many male langurs are present.
  3. (iii) A female langur weighing 15.2 kg is admitted to the facility. What is the average weight of the female langurs after this?
  4. (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?
  5. (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 / method

Step-by-step Solution:

  1. Intuition without computing:
    The midpoint of male (16.5 kg) and female (13.8 kg) averages is (16.5 + 13.8) / 2 = 15.15 kg. Since the actual combined average (14.925 kg) is closer to the female average (13.8 kg) than the male average, there must be more female langurs than male langurs!
  2. (i) Correct Expressions:
    Both (a) and (b) are correct, because x + y = 60.
  3. (ii) Finding male count x:
    x + y = 60 &implies; y = 60 − x.
    16.5x + 13.8(60 − x) = 14.925 × 60 = 895.5
    16.5x + 828 − 13.8x = 895.5
    2.7x = 895.5 − 828 = 67.5 &implies; x = 67.5 / 2.7 = 25 male langurs (and y = 35 female langurs).
  4. (iii) Female admitted (15.2 kg):
    Old female sum = 35 × 13.8 = 483 kg.
    New female sum = 483 + 15.2 = 498.2 kg, New count = 36.
    New Female Average = 498.2 / 36 = 13.838 ≈ 13.84 kg
  5. (iv) Two males released (16.9 kg and 16.1 kg):
    Sum of released males = 16.9 + 16.1 = 33.0 kg.
    Old male sum = 25 × 16.5 = 412.5 kg.
    New male sum = 412.5 − 33.0 = 379.5 kg, New count = 25 − 2 = 23.
    New Male Average = 379.5 / 23 = 16.5 kg
    Notice: Since (16.9 + 16.1) / 2 = 16.5 kg (equal to the group average), removing them leaves the average unchanged!
  6. (v) One male loses 1 kg:
    New male sum = 412.5 − 1 = 411.5 kg, Count = 25.
    New Male Average = 411.5 / 25 = 16.46 kg
Answers: There are more females; (i) (a) & (b); (ii) 25 male langurs; (iii) 13.84 kg; (iv) 16.5 kg; (v) 16.46 kg.
8Dorjee has collected 1 litre of water from the Dead Sea! Using the information in the table, answer the following questions. A calculator can be used if necessary.
Water SourceSalinity
Dead Sea≈ 34%
Other seas and oceans≈ 3.5%
Ground water≈ 0.01%
Purified drinking water≈ 0.001%
  1. (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?
  2. (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?
  3. (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 / method

Step-by-step Solution:

  1. (i) Salinity of 1 L Dead Sea + 2 L purified water:
    Salt from Dead Sea = 1 × 34% = 0.34 L
    Salt from Purified water = 2 × 0.001% = 0.00002 L
    Total volume = 1 + 2 = 3 L
    Salinity = (0.34 + 0.00002) / 3 × 100% = 34.002% / 3 ≈ 11.334%
  2. (ii) Mixing Dead Sea & Purified water to achieve Groundwater salinity (0.01%):
    • Is it possible? Yes, because the target salinity (0.01%) lies strictly between purified water (0.001%) and Dead Sea water (34%).
    • Let x be the litres of purified drinking water mixed with 1 L of Dead Sea water:
    (1 × 34 + x × 0.001) / (1 + x) = 0.01
    34 + 0.001x = 0.01(1 + x) = 0.01 + 0.01x
    34 − 0.01 = 0.01x − 0.001x &implies; 0.009x = 33.99
    x = 33.99 / 0.009 ≈ 3,776.67 litres
  3. (iii) Mixing Dead Sea & Groundwater to achieve Purified drinking water salinity (0.001%):
    • Is it possible? NO!
    • Reason: Any mixture of two sources must always have a concentration lying between the two sources (min ≤ mixture ≤ max). Here, both sources (34% and 0.01%) have salinity strictly higher than 0.001%. You cannot dilute a solution below the salinity of its least saline component!
Answers: (i) 11.334%; (ii) Yes, mix approx. 3,776.67 litres; (iii) No, impossible because 0.001% is lower than both component salinities.

Exercise Set 10.4 — Solutions

Reading and constructing stacked bar charts.

1The following stacked column chart shows the number of animal species in the IUCN Red List, by class, over time.
Chart Context — “The Number of Endangered Species is Rising”:
Stacked column chart representing animal species of the IUCN Red List by class: Mammals, Reptiles, Birds, Insects, Amphibians, Molluscs (e.g. snails), Fish, and Others.
Cumulative Column Totals: 2007: 7,851 • 2010: 9,618 • 2013: 11,212 • 2016: 12,630 • 2019: 14,234.
  1. (i) What does the number 14,234 on the top of the column 2019 represent?
  2. (ii) Approximately how many reptile species were in the list in 2016?
  3. (iii) Which class(es) of species have seen a relatively small increase in count between 2007 and 2019?
Answer / method

Step-by-step Solution:

  1. (i) Meaning of 14,234 on the 2019 column:
    In a stacked column chart, the top of the bar marks the cumulative total of all stacked slices. Therefore, 14,234 represents the total number of endangered animal species across all classes listed in the IUCN Red List for the year 2019.
  2. (ii) Number of reptile species in 2016:
    To find the count of an individual middle slice in a stacked column:
    Slice Count = Top boundary of slice − Bottom boundary of slice
    Reading the 2016 bar: the bottom of the reptile segment starts at approximately 1,800 and the top reaches approximately 3,150 on the vertical scale.
    Reptile Species in 2016 ≈ 3,150 − 1,800 ≈ 1,350 species (approx. 1,300 to 1,400)
  3. (iii) Classes with relatively small increases between 2007 and 2019:
    Comparing the vertical heights of each class segment between 2007 and 2019:
    • Mammals and Birds show relatively minor increases in their segment heights.
    • In contrast, classes such as Amphibians, Molluscs, and Fish / Insects show massive vertical expansions in their endangered counts over the same 12-year period.
Key Principle: Reading a middle segment in a stacked bar chart always requires finding the difference between its upper and lower endpoints.
2The 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).
LocationTestODIT20Total Wickets
Home6210032194
Overseas1724971292
Format Total234149103486
Two representations to complete:
Chart (i): Comparing total wickets taken at Home vs. Overseas (Axis: 0 to 300 wickets; Bars for Home and Overseas, stacked by match format Test, ODI, T20).
Chart (ii): Total wickets taken in each format (Axis: 0 to 250 wickets; Bars for Test, ODI, and T20, stacked by match location Home and Overseas).
Answer / method

Step-by-step Construction & Segment Calculations:

  1. Chart (i) — Wickets at Home vs. Overseas:
    • Home Bar (Total = 194 wickets):
      – Test segment: 0 to 62 (height = 62)
      – ODI segment: 62 to 162 (height = 100)
      – T20 segment: 162 to 194 (height = 32)
    • Overseas Bar (Total = 292 wickets):
      – Test segment: 0 to 172 (height = 172)
      – ODI segment: 172 to 221 (height = 49)
      – T20 segment: 221 to 292 (height = 71)
    Visual inference: Overseas wickets (292) exceed Home wickets (194) by nearly 100 wickets.
  2. Chart (ii) — Wickets in Different Formats:
    • Test Bar (Total = 234 wickets):
      – Home segment: 0 to 62 (height = 62)
      – Overseas segment: 62 to 234 (height = 172)
    • ODI Bar (Total = 149 wickets):
      – Home segment: 0 to 100 (height = 100)
      – Overseas segment: 100 to 149 (height = 49)
    • T20 Bar (Total = 103 wickets):
      – Home segment: 0 to 32 (height = 32)
      – Overseas segment: 32 to 103 (height = 71)
  3. Cross-verification:
    Grand Total = 194 + 292 = 486 wickets.
    Format Total = 234 + 149 + 103 = 486 wickets. Both representations reconcile perfectly.
Answer: Chart (i) has 2 bars of lengths 194 and 292; Chart (ii) has 3 bars of lengths 234, 149, and 103.

Exercise Set 10.5 — Solutions

Tables and 100% stacked bar charts for body-state data.

1The smart watch data of 5 people was tracked from Monday to Friday. The recorded data tells us how many hours each person spent lying down, sitting, and standing or moving around, etc., as shown in Table 10.3.
Table 10.3: Average number of hours spent per day (Monday–Friday) in each body-state (Total = 24 h)
Person (Occupation)Lying downSittingStanding / Moving around etc.Total Hours
Pavani (Patient)20 h3 h1 h24 h
Raghu (Nurse)7 h4 h13 h24 h
Zakir (Teacher)8 h6 h10 h24 h
Sahana (Student)[ ? ][ ? ][ ? ]24 h
Julie (________)8 h10 h6 h24 h
  1. (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.
  2. (ii) Guess what activity or work Julie could be engaged in based on the data. Find out what your classmates' guesses are.
  3. (iii) Complete the 100% stacked horizontal bar chart based on the tabular data.
Answer / method

Step-by-step Solution:

  1. (i) Reasonable Guess for Sahana (Student) — Sum must equal 24 hours:
    A typical school student's daily weekday breakdown:
    • Lying down: 8 hours (night sleep + rest)
    • Sitting: 10 hours (school classroom, study, homework, transport, and meals)
    • Standing or Moving around: 6 hours (sports, recess, walking, helping at home)
    Check: 8 + 10 + 6 = 24 hours.
  2. (ii) Julie's Probable Work / Occupation:
    • Julie spends 10 hours sitting and only 6 hours moving (with 8 hours of sleep).
    • Spending 10 hours seated per day indicates a desk-bound, screen, or office profession, such as a software developer, accountant, call centre executive, writer, bank cashier, or data analyst.
  3. (iii) Converting Hours into 100% Stacked Bar Percentages (out of 24 h):
    • Pavani (Patient):
      – Lying down: (20/24) × 100 ≈ 83.33%
      – Sitting: (3/24) × 100 = 12.50%
      – Moving: (1/24) × 100 ≈ 4.17%
    • Raghu (Nurse):
      – Lying down: (7/24) × 100 ≈ 29.17%
      – Sitting: (4/24) × 100 ≈ 16.67%
      – Moving: (13/24) × 100 ≈ 54.17% (busy hospital ward shift)
    • Zakir (Teacher):
      – Lying down: (8/24) × 100 ≈ 33.33%
      – Sitting: (6/24) × 100 = 25.00%
      – Moving: (10/24) × 100 ≈ 41.67% (standing while teaching, blackboard work)
    • Sahana (Student):
      – Lying down: (8/24) × 100 ≈ 33.33%
      – Sitting: (10/24) × 100 ≈ 41.67%
      – Moving: (6/24) × 100 = 25.00%
    • Julie (Office Worker):
      – Lying down: (8/24) × 100 ≈ 33.33%
      – Sitting: (10/24) × 100 ≈ 41.67%
      – Moving: (6/24) × 100 = 25.00%
Key Mathematical Fact: Because every single person has an identical total time of 24 hours, this chart is both an ordinary stacked bar chart (hours) and a 100% stacked bar chart (percentages) simultaneously!

End-of-Chapter Exercises — Solutions

Comprehensive step-by-step solutions for mixed applications, proofs, and real-life projects.

1In cricket, the run rate is the average number of runs scored per over. In a T20 match, a team scored 6 runs in the first over making the run rate 6. (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 / method

Step-by-step Solution:

  1. (i) Run rate after 2 overs:
    • Runs in 1st over = 6
    • Runs in 2nd over = 12
    • Total runs after 2 overs = 6 + 12 = 18 runs
    Run Rate = Total Runs / Total Overs = 18 / 2 = 9.0 runs per over
  2. (ii) Run rate after 20 overs:
    • Run rate after 19 overs = 6
    • Total runs after 19 overs = 19 × 6 = 114 runs
    • Runs scored in 20th over = 12
    • Total runs after 20 overs = 114 + 12 = 126 runs
    New Run Rate = Total Runs / 20 = 126 / 20 = 63 / 10 = 6.3 runs per over
Answers: (i) 9.0 runs/over; (ii) 6.3 runs/over. Notice how scoring 12 runs in Over 2 surged the run rate by +3.0 (from 6 to 9), but scoring 12 runs in Over 20 only lifted the run rate by +0.3 (from 6.0 to 6.3) due to the much larger base (weight of 20 overs).
2Five friends who play badminton surveyed the city and collected the price of shuttlecocks across brands and types (Nylon and Feather). The compiled list of prices (in ₹) is shown below:
• Yusuf: 40(N), 105(N), 383(F), 108(N), 165(F), 116(F) — 6 prices (3 N, 3 F)
• Srikanth: 194(N), 85(N), 93(N), 121(N) — 4 prices (4 N)
• Kashvi: 49(N), 297(F), 105(N), 275(F), 40(N) — 5 prices (3 N, 2 F)
• Prasanna: 124(N), 333(F), 182(N), 258(F) — 4 prices (2 N, 2 F)
• Gracy: 220(F), 458(F), 129(F), 183(N) — 4 prices (1 N, 3 F)

Questions:
(i) Each one calculated the average of the prices they gathered: ay, as, ak, ap, ag. Write an expression that gives their combined average.
(ii) Suppose an and af are the average prices of nylon and feather shuttlecocks, respectively. Write an expression that gives the average price of a shuttlecock. Will this equal the answer from part (i)?
Answer / method

Step-by-step Solution:

  1. Total observations gathered:
    Total prices = 6 (Yusuf) + 4 (Srikanth) + 5 (Kashvi) + 4 (Prasanna) + 4 (Gracy) = 23 prices.
    By type: Nylon prices = 3 + 4 + 3 + 2 + 1 = 13; Feather prices = 3 + 0 + 2 + 2 + 3 = 10 (13 + 10 = 23).
  2. (i) Combined average using friends' individual averages:
    Each friend's sum of prices is (count × average):
    Total Sum = 6ay + 4as + 5ak + 4ap + 4ag.
    Combined Average = (6ay + 4as + 5ak + 4ap + 4ag) / 23
  3. (ii) Combined average using shuttlecock types:
    Sum of nylon prices = 13 × an; Sum of feather prices = 10 × af.
    Overall Average = (13an + 10af) / 23
  4. Will the two expressions be equal?
    YES, absolutely! Both expressions compute the exact total sum of all 23 individual shuttlecock prices divided by 23. Grouping by person or grouping by material does not change the grand sum.
Answers: (i) (6ay + 4as + 5ak + 4ap + 4ag) / 23; (ii) (13an + 10af) / 23; they are equal.
3Shreyas holds 25 shares of a company at an average price of ₹150 and Vaishnavi holds 5 shares of the same company at an average price of ₹150. (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 / method

Step-by-step Solution:

  1. (i) Shreyas's new average share price:
    • Existing shares: 25 @ ₹150 &implies; Total cost = 25 × 150 = ₹3,750
    • New purchase: 10 @ ₹30 &implies; Cost = 10 × 30 = ₹300
    • Combined total cost = 3,750 + 300 = ₹4,050
    • Total shares = 25 + 10 = 35 shares
    New Average Price = 4050 / 35 = 810 / 7 ≈ ₹115.71
  2. (ii) Number of shares Vaishnavi bought (let count = x):
    • Existing shares: 5 @ ₹150 &implies; Cost = 5 × 150 = ₹750
    • New purchase: x shares @ ₹30 &implies; Cost = 30x
    • Total shares = 5 + x
    • Target average price = ₹70
    (750 + 30x) / (5 + x) = 70
    750 + 30x = 70(5 + x)
    750 + 30x = 350 + 70x
    750 − 350 = 70x − 30x
    400 = 40x &implies; x = 400 / 40 = 10 shares
Answers: (i) ₹115.71 per share; (ii) Vaishnavi bought 10 shares.
4(Pṛthūdakasvāmī, commentary on Brahmagupta's Brāhmasphuṭasiddhānta, c. 864 CE) A “hasta” (meaning “forearm”) refers to a unit of length measuring around 18 inches. A pool 30 hastas long is dug to different depths along its length. It is divided into 5 sections having lengths 4, 5, 6, 7, and 8 hastas, and is dug to depths of 9, 7, 7, 3, and 2 hastas, respectively. Find the mean depth of the pool.
Answer / method

Step-by-step Solution:

  1. Identify weights and values:
    • The lengths of the sections serve as the weights (wi): 4, 5, 6, 7, 8 hastas.
    • Total length = 4 + 5 + 6 + 7 + 8 = 30 hastas.
    • The depths of the sections are the values (xi): 9, 7, 7, 3, 2 hastas.
  2. Compute total longitudinal cross-sectional area Σ(wi × xi):
    • Section 1: 4 × 9 = 36
    • Section 2: 5 × 7 = 35
    • Section 3: 6 × 7 = 42
    • Section 4: 7 × 3 = 21
    • Section 5: 8 × 2 = 16
    Sum = 36 + 35 + 42 + 21 + 16 = 150 square hastas
  3. Calculate the weighted mean depth:
    Mean Depth = Σ(wi × xi) / Σwi = 150 / 30 = 5 hastas
Answer: The mean depth of the pool is exactly 5 hastas.
5Suvarna had purchased 1g gold at ₹15k (short for ₹15,000). This is shown by point O denoting the average price of gold that she possesses. Six different scenarios are given below for her next transaction. For each scenario estimate and mark the average price of gold she will have after the transaction.
Number line from 0 to 30 (in ₹k), with O marked at 15.
(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 / method

Step-by-step Solution for All Six Scenarios:

  1. (i) Buy 1g gold at ₹30k:
    Equal quantities (1g and 1g) &implies; exact midpoint:
    (1 × 15 + 1 × 30) / (1 + 1) = 45 / 2 = ₹22.5k
  2. (ii) Buy 2g gold at ₹30k:
    (1 × 15 + 2 × 30) / (1 + 2) = (15 + 60) / 3 = 75 / 3 = ₹25.0k
  3. (iii) Buy 1g gold at ₹10k:
    Equal quantities (1g and 1g) &implies; exact midpoint:
    (1 × 15 + 1 × 10) / (1 + 1) = 25 / 2 = ₹12.5k
  4. (iv) Buy 10g gold at ₹10k:
    Heavily weighted towards ₹10k (10g vs 1g):
    (1 × 15 + 10 × 10) / (1 + 10) = 115 / 11 ≈ ₹10.45k
  5. (v) Buy 0.5g gold at ₹30k:
    (1 × 15 + 0.5 × 30) / (1 + 0.5) = (15 + 15) / 1.5 = 30 / 1.5 = ₹20.0k
  6. (vi) Buy 0.5g gold at ₹15k:
    Since the new purchase is at the existing average price (₹15k):
    (1 × 15 + 0.5 × 15) / (1 + 0.5) = 22.5 / 1.5 = ₹15.0k (Unchanged!)
Key Principle: The quantity purchased acts as the weight. A large purchase (10g) pulls the average almost all the way to ₹10k, whereas buying at the existing mean leaves the average unchanged.
6Given some data with corresponding weights, how would the weighted average change if all the weights are doubled? If required, experiment with some data. What do you observe? Justify your answer using algebra.
Answer / method

Observation & Algebraic Proof:

  1. Observation: The weighted average remains completely unchanged.
  2. Algebraic Justification:
    Let the data values be x1, x2, …, xn with corresponding weights w1, w2, …, wn.
    The original weighted average is:
    x̄ = (w1x1 + w2x2 + … + wnxn) / (w1 + w2 + … + wn) = Σ(wixi) / Σwi
  3. Now double every weight: new weights are w'⊂i⊃ = 2wi.
    The new weighted average is:
    x̄' = Σ(2wi × xi) / Σ(2wi) = [2 × Σ(wixi)] / [2 × Σwi] = Σ(wixi) / Σwi = x̄
  4. Generalisation: In fact, multiplying all weights by any non-zero constant k > 0 cancels out completely from numerator and denominator, leaving the weighted average identical.
Answer: The weighted average does not change because the factor of 2 factors out and cancels: Σ(2wixi) / Σ(2wi) = Σ(wixi) / Σwi.
7Answer the following questions based on the space objects graph.
Graph Context: “Cumulative number of objects currently orbiting Earth, by year of launch/separation (as of Apr. 24, 2026)”. Categories: Payload objects (satellites, space stations) and Other objects (spent rocket stages, mission debris). Y-axis: 0 to 35,000 objects. Source: J. McDowell, General Catalogue of Artificial Space Objects.
  1. (i) In 2003, approximately how many total objects were found orbiting Earth in space? In what year did this number double?
  2. (ii) Find the approximate number and share of payload objects and other objects in the year 2025.
  3. (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 / method

Step-by-step Solution:

  1. (i) Orbiting objects in 2003 and doubling year:
    • Reading the 2003 bar: Total objects ≈ 9,000 to 9,500 objects.
    • Doubling this count yields ≈ 18,000 to 19,000 objects.
    • Reading across the graph, the bar reaches 18,000–19,000 around 2018–2020. Thus, it took about 15–17 years to double.
  2. (ii) Counts and percentage shares in 2025:
    • Total objects in 2025 ≈ 34,000 objects.
    • Payload objects ≈ 12,000 objects &implies; Share = (12,000 / 34,000) × 100% ≈ 35.3%.
    • Other objects (debris/stages) ≈ 22,000 objects &implies; Share = (22,000 / 34,000) × 100% ≈ 64.7%.
  3. (iii) Temporal Trends (Absolute Counts vs. Percentage Shares):
    • Absolute counts: Both payload objects and debris objects have grown substantially year-on-year.
    • Relative shares: For decades (1970–2015), debris (“Other objects”) dominated space (> 75% share). However, post-2019, the share of payload objects surged dramatically (from ≈ 15% up to > 35%) due to commercial mega-constellations of low-Earth orbit satellites (e.g. Starlink, OneWeb).
Key Finding: While debris still forms the majority, active satellite launches have grown at an exponential pace in the 2020s.
8Observe the infographic showing the percentage of schools having a playground across Indian states (source: statsofindia.in).
  1. (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.
  2. (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 / method

Step-by-step Evaluation:

  1. (i) Evaluating each statement:
    • Statement (a) is UNCERTAIN / CANNOT BE INFERRED: The infographic shows percentages, not total school counts. Even if Haryana has a higher playground percentage, Uttarakhand might have more total schools, making the absolute count unpredictable without state school totals.
    • Statement (b) is CORRECT: "2 out of 3" equals 2/3 ≈ 66.7%. The map shows Arunachal Pradesh with approximately 66%–67% playground coverage.
    • Statement (c) is INCORRECT / CANNOT BE INFERRED: A high percentage (e.g. 90% in Punjab) does not mean the highest count. Larger states like Uttar Pradesh or Maharashtra have vastly more total schools.
    • Statement (d) is INCORRECT / CANNOT BE INFERRED: Knowing Maharashtra has more schools than Telangana is not sufficient unless we know both the exact number of schools and the specific playground percentages of each state.
  2. (ii) Can we find the nation-wide percentage?
    NO. We cannot simply calculate the unweighted average of the state percentages. Because states have drastically different school populations (e.g., UP has >250,000 schools while Sikkim has ~1,200), smaller states would have unfair influence.
    Required information: The total number of schools in every state/UT (Ni). Then the nation-wide percentage is computed via the weighted mean:
    Nation-wide Percentage = Σ(Ni × Pi) / ΣNi
Answers: (i) Only (b) is a valid inference; (ii) No; the total number of schools in each state is required.
9Decision Dilemma: (i) Which of the plays would you choose to watch based on a 100% stacked bar chart showing the share of ratings (5★ to 1★) for Play A, Play B, and Play C? (ii) The corresponding stacked bar chart showing absolute number of ratings (axis 0 to 750) is shown. Would you change your decision after looking at this chart? Why/Why not? Discuss.
Answer / method

Step-by-step Statistical Dilemma Analysis:

  1. (i) Decision based only on the 100% Stacked Bar Chart:
    In a 100% chart, all bars have equal length. A viewer naturally chooses the play with the highest combined percentage of 5★ and 4★ reviews (suppose Play A shows 90% positive ratings, while Play B shows 80%). Based purely on percentages, Play A appears to be the superior choice.
  2. (ii) Looking at the Ordinary Stacked Bar Chart (Total Volume Revealed):
    • The absolute stacked bar chart reveals sample size: Play A received only 10 total ratings (9 positive reviews out of 10), whereas Play B received 700 total ratings (560 positive reviews out of 700)!
    • Would you change your decision? YES!
    • Reason: A 90% approval rating from just 10 people could be biased (e.g., cast members' friends and family). In contrast, an 80% approval rating across 700 independent theatregoers is statistically robust and reliable. High sample size provides far greater confidence.
Key Takeaway: 100% stacked bars can create misleading impressions when sample sizes across categories differ by orders of magnitude. Always inspect absolute totals before concluding!
10A triathlon is an endurance race consisting of swimming 3.8 km, cycling 180 km, and running 42.2 km. Athletes compete for the fastest overall time, completing each segment in that order. Table 10.4 shows the finish times of 3 athletes in each segment.
Table 10.4: Finish time of three athletes in each segment of a triathlon race (hh : mm)
AthleteSwimming (3.8 km)Cycling (180 km)Running (42.2 km)Total Finish Time
Athlete 101:08 (68 min)05:00 (300 min)03:15 (195 min)09:23 (563 min)
Athlete 201:05 (65 min)05:10 (310 min)03:35 (215 min)09:50 (590 min)
Athlete 301:22 (82 min)05:55 (355 min)03:50 (230 min)11:07 (667 min)
  1. (i) What is a suitable representation of this data — a stacked bar chart or a 100% stacked bar chart? Why do you think so?
  2. (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 / method

Step-by-step Solution:

  1. (i) Suitable representation: Stacked Bar Chart
    In an athletic race, the total finish time is the primary metric of interest because the shortest total bar wins the race. A 100% stacked bar chart forces all three bars to equal length (100%), completely masking who finished first and how long each segment took in real hours. Hence, an ordinary stacked bar chart is far more suitable.
  2. (ii) Questions answerable from a 100% Stacked Bar Chart:
    • Question (a) — Who finished first? CANNOT be answered. All bars are 100% long, hiding the winner.
    • Question (b) — Fraction of time Athlete 1 spent cycling? CAN BE ANSWERED. A 100% chart displays proportions directly: cycling fraction = 300 / 563 ≈ 53.3%.
    • Question (c) — Which athlete took the longest for running? CANNOT be answered. A 100% chart shows only running percentage share of that athlete's own total, not absolute time. For example, Athlete 1's running share is 195/563 ≈ 34.6%, whereas Athlete 3's is 230/667 ≈ 34.5%. Even though Athlete 1 has a slightly higher percentage share, Athlete 3 took 35 minutes longer in real time!
Answers: (i) Ordinary stacked bar chart; (ii) Only (b) can be answered from a 100% stacked bar chart.
11Look at the 100% stacked bar chart showing the “Distribution of disabled persons by age group and type of disability in India (Census 2011)” across age groups (0–19, 20–39, 40–59, 60 & above). Categories: seeing, hearing, speech, movement, mental retardation, mental illness, any other, multiple disability. What do you notice? What do you wonder? Write your inferences.
Answer / method

Notice, Wonder, and Statistical Inferences:

  1. What I Notice (Observations):
    • Children & Youth (0–19): Speech and hearing disabilities account for a notably larger proportion of total disabilities compared to older brackets.
    • Working Age (20–39 & 40–59): Mental illness, mental retardation, and movement disabilities make up significant segments.
    • Elderly (60 & above): Movement disability and seeing / hearing disabilities expand massively, occupying the overwhelming majority of the 100% bar.
  2. What I Wonder (Critical Inquiries):
    • I wonder how much of the surge in seeing and movement disability in the 60+ group is caused by age-related biological degeneration (cataracts, arthritis) versus preventable workplace injuries.
    • I wonder whether early medical interventions and childhood screenings are reducing speech and hearing impairments over time.
  3. Crucial Statistical Inference:
    Remember that this is a 100% stacked bar chart: it illustrates the composition within each age group, NOT the absolute number of disabled persons. A smaller percentage in the 60+ bracket could still represent millions more individuals than a larger percentage in the 0–19 cohort.
Summary: Disability types shift systematically from congenital/developmental conditions in youth towards degenerative sensory and physical conditions in old age.
12Individual project: (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) Identify expense categories: Housing, Food, Education, Transport, Healthcare, Recreation, Savings.
• (b) Collect expense data for at least 3 months.
• (c) Represent the data in a 100% stacked bar chart.
• (d) Write observations and inferences based on the chart.
Answer / method

Project Guide & Worked Example:

  1. Part (i) — Tracking Daily Body-States (24-Hour Rule):
    Create a table ensuring every row sums to 24 hours:
    • Student (Weekday): 8 h sleeping + 10 h sitting (school/study) + 6 h moving = 24 h.
    • Student (Sunday): 9 h sleeping + 6 h sitting + 9 h moving/playing = 24 h.
    • Parent (Working): 7 h sleeping + 9 h sitting (office) + 8 h moving/commute = 24 h.
    Plot these as horizontal stacked bars of length 24 on a common scale (0 to 24 hours).
  2. Part (ii) — Monthly Family Expenditure 100% Stacked Bar Chart:
    Track 3 consecutive months across standard categories:
    • Month 1 (Total = ₹40,000): Food (₹16,000 = 40%), Housing (₹10,000 = 25%), Education (₹6,000 = 15%), Transport (₹4,000 = 10%), Health & Misc (₹4,000 = 10%).
    • Month 2 (Total = ₹45,000): Food (₹18,000 = 40%), Housing (₹10,000 = 22.2%), Education (₹6,000 = 13.3%), Transport (₹5,000 = 11.1%), Health & Misc (₹6,000 = 13.3%).
    • Month 3 (Total = ₹50,000 — festival month): Food (₹22,000 = 44%), Housing (₹10,000 = 20%), Education (₹6,000 = 12%), Transport (₹5,000 = 10%), Health & Gifts (₹7,000 = 14%).
  3. Sample Inference: Fixed costs like housing rent remain constant in rupees, but their percentage share shrinks during months with higher total expenditure (festival months). Food remains the single largest budget component.
Deliverable: 2 completed tables, 2 clean charts (one ordinary 24h stacked bar, one 100% expenditure stacked bar), and a short written paragraph of insights.
13Small-group project: Make a group of 3–4 students. Choose one scenario to design a custom rating scheme by assigning appropriate weights: (a) shopping experience at a cloth store, (b) travel experience in a bus, (c) tourism experience of a nearby tourist spot, (d) clinic/hospital experience. (i) Discuss and arrive at 4–6 aspects to rate with justifications. (ii) Decide relative weights. (iii) Collect ratings from at least 10 people on a 1–5 scale. (iv) Compute individual and overall weighted ratings. (v) Make a 100% stacked bar chart. (vi) Write a note on what the scheme captures and what it misses.
Answer / method

Worked Framework — Scenario (d): Clinic / Hospital Experience:

  1. (i) & (ii) Aspects and Weights Selection:
    • Doctor competence & consultation time (Weight = 5): The primary reason for visiting.
    • Nursing care & politeness (Weight = 4): Directly impacts patient comfort and well-being.
    • Cleanliness & sanitisation (Weight = 4): Crucial to prevent secondary infections.
    • Waiting time & punctuality (Weight = 3): Respects patient time.
    • Billing clarity & medicine availability (Weight = 2): Administrative convenience.
    Total weight = 5 + 4 + 4 + 3 + 2 = 18.
  2. (iv) Computing Individual Weighted Ratings:
    For each respondent, calculate:
    Patient Rating = (5×Doctor + 4×Nursing + 4×Cleanliness + 3×Wait + 2×Billing) / 18
    Overall average rating across N respondents is the mean of individual ratings.
  3. (v) Visualisation:
    Plot the ratings distribution (5★, 4★, 3★, 2★, 1★) for each aspect as a 100% stacked bar chart to identify which department has the highest dissatisfaction.
  4. (vi) Critical Evaluation:
    • What it captures well: Prioritises medical quality over administrative convenience through higher weights.
    • What it misses: It does not differentiate between emergency acute care vs routine checkups, where waiting time priority may differ significantly.
Key Lesson: A thoughtful weighting rubric prevents minor conveniences (like free parking) from overshadowing fundamental core qualities (like diagnostic accuracy).
14Whole class project: Each student shares the average age of their family and the number of family members. Discuss among the class and come up with a way to find out the average age of all the families of the class.
Answer / method

Mathematical Formulation & Discussion:

  1. Why the simple average of family averages FAILS:
    If Student 1 has a family of 3 with average age 25, and Student 2 has a joint family of 10 with average age 45, simply averaging (25 + 45)/2 = 35 gives a misleading result. The 10 individuals in the larger family have been given the same combined weight as the 3 individuals in the smaller family!
  2. Correct Formulation: Weighted Mean by Family Size:
    Let student i's family have ni members with average age ai.
    • Total age of family i = ni × ai.
    • Grand total age of all people across all families = Σ(ni × ai).
    • Grand total number of individuals = Σni.
    True Average Age of All People = Σ(ni × ai) / Σni
  3. Class Procedure:
    • Create a 3-column table on the blackboard: [Student Name | Family Members (ni) | Family Average Age (ai)].
    • Add a 4th column for Product [ni × ai].
    • Sum column 2 to get total people, sum column 4 to get total age, and divide sum(4) by sum(2).
Core Statistical Takeaway: When combining subgroup averages, subgroups MUST be weighted by their sample sizes (ni) to obtain the true population mean.
*15Given some data with corresponding weights, what would happen to the weighted average if all the weights are increased by a constant value, say 1? If required, experiment with some data. What do you observe? Justify your answer using algebra.
Answer / method

Comprehensive Algebraic Proof & Behavior Analysis:

  1. Definitions:
    Let the n data values be x1, x2, …, xn with original weights w1, w2, …, wn.
    • Original weighted mean: x̄w = Σ(wixi) / Σwi
    • Ordinary arithmetic mean: x̄ord = Σxi / n &implies; Σxi = n × x̄ord
  2. Increase all weights by +1 (new weights w'⊂i⊃ = wi + 1):
    • New numerator = Σ[(wi + 1)xi] = Σ(wixi + xi) = Σ(wixi) + Σxi
    • New denominator = Σ(wi + 1) = Σwi + Σ1 = Σwi + n
    New Weighted Mean x̄' = [Σ(wixi) + Σxi] / [Σwi + n]
  3. Proving the Effect:
    Notice that x̄' is the combined mean of the original weighted mean x̄w (with weight Σwi) and the unweighted mean x̄ord (with weight n):
    x̄' = [(Σwi) × x̄w + n × x̄ord] / [Σwi + n]
    • Case 1: If x̄w > x̄ord, adding +1 pulls the new average down towards x̄ord.
    • Case 2: If x̄w < x̄ord, adding +1 pulls the new average up towards x̄ord.
    • Case 3: The weighted average remains unchanged if and only if x̄w = x̄ord (for example, when all original weights were already equal, or by symmetrical coincidence).
Conclusion: Adding 1 to all weights pulls the weighted average closer to the simple unweighted arithmetic mean: x̄' = [Σ(wixi) + Σxi] / [Σwi + n].
*16A farm has some cows, sheep, and chickens. Last year the cows made up 60%, the sheep 25%, and the chickens 15%. There was a decrease in the number of all three animals' population over the year. Choose the possibilities for the change in their respective shares of the population —
(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 / method

Logical & Mathematical Analysis of Population Shares:

  1. The 100% Constant Sum Constraint:
    The percentage shares of all groups in a population must ALWAYS sum to 100%:
    % Cows + % Sheep + % Chickens = 100%
  2. Evaluating Scenarios (i) and (ii):
    • (i) is IMPOSSIBLE: If all three percentages decrease, their new sum would be strictly less than 100%, which violates the mathematical definition of shares.
    • (ii) is IMPOSSIBLE: If all three percentages increase, their new sum would exceed 100%, which is equally impossible.
  3. Evaluating Scenarios (iii), (iv), and (v):
    • (iii) is POSSIBLE: If the populations of cows, sheep, and chickens all decrease by the exact same percentage rate (e.g., all drop by 20%), their relative proportions remain strictly 60%, 25%, and 15%.
    • (iv) is POSSIBLE: If cows decrease at a faster rate than the overall farm, sheep decrease at a slower rate (or barely decrease), and chickens decrease at exactly the farm's average rate, then cow share falls, sheep share rises, and chicken share stays at 15%.
    • (v) is POSSIBLE: If chickens suffer a massive epidemic (e.g. dropping by 80%) while cows and sheep suffer only minor reductions (e.g. dropping by 2%), chickens' share plummets, causing both cows' and sheep's shares to increase.
Correct Choices: (iii), (iv), and (v) are all mathematically possible. Options (i) and (ii) are strictly impossible.
Source note: All exercise questions retain authentic problem tables and visual chart specifications from the Grade 9 Ganita Manjari Part 2 textbook.

Mini Test - 10 Questions

Select one option for each question.

Q1. Which formula combines two groups with means a and b and sizes n and m?

Q2. When three mixtures have equal quantities, their combined concentration is:

Q3. A weighted average is calculated as:

Q4. If every weight is doubled, the weighted average:

Q5. To combine concentrations with different volumes, use volume as:

Q6. A new value above the old mean will usually make the new mean:

Q7. A 100% stacked bar chart is best for comparing:

Q8. To combine marks out of different totals, first convert them to:

Q9. A combined concentration of two sources must lie:

Q10. If all weights are increased by 1, the weighted average:

Score: 0 / 10