Class 9 Mathematics Chapter 4: Data Handling – Complete Important Questions Bank with Solutions

Data Handling carries 8-10 marks in the Class 9 final exam, making it a high-value chapter. The CBSE focuses on application-based questions that test conceptual understanding rather than mere formula memorization. Recent board trends show increased emphasis on: (1) Reading and interpreting grouped frequency distributions; (2) Comparing datasets using multiple graph types; (3) Real-world probability problems involving equally likely outcomes; (4) Converting raw data into visual forms with proper labels and scales. Students who master frequency distribution tables and histogram construction gain an edge, as these concepts directly link to Class 10 Statistics and Class 11 Probability. The chapter also integrates problem-solving skills—you won't just calculate, you'll analyze and interpret. This landing page provides strategically designed questions mirroring actual board difficulty, ensuring you're prepared for any variation the examiner may ask.

**Question 1:** The class mark of the interval 10–20 is: (A) 10 (B) 15 (C) 20 (D) 25 **Answer: (B) 15** *Explanation: Class mark = (Lower limit + Upper limit) ÷ 2 = (10 + 20) ÷ 2 = 15* **Question 2:** A pie chart shows angles of 90°, 120°, and 150° for three categories. What is the angle for the fourth category? (A) 360° (B) 240° (C) 0° (D) 45° **Answer: (C) 0°** *Explanation: Sum of all angles in a pie chart = 360°. Fourth angle = 360° – (90° + 120° + 150°) = 0°. This means the fourth category has no data.* **Question 3:** When a fair die is thrown, the probability of getting an even number is: (A) 1/6 (B) 1/3 (C) 1/2 (D) 2/3 **Answer: (C) 1/2** *Explanation: Favorable outcomes (even) = {2, 4, 6} = 3. Total outcomes = 6. Probability = 3/6 = 1/2* **Question 4:** The width of class interval 25–35 is: (A) 10 (B) 25 (C) 35 (D) 60 **Answer: (A) 10** *Explanation: Class width = Upper limit – Lower limit = 35 – 25 = 10* **Question 5:** In a frequency distribution table, if the frequency of a class is 12 and the total frequency is 60, the relative frequency is: (A) 0.2 (B) 0.5 (C) 0.8 (D) 1.2 **Answer: (A) 0.2** *Explanation: Relative frequency = Class frequency ÷ Total frequency = 12 ÷ 60 = 0.2*

**Question 1:** The marks obtained by 10 students in a test are: 35, 42, 50, 50, 65, 70, 72, 80, 85, 90. Construct a frequency distribution table using intervals of width 10 (starting from 30). **Solution:** Class Interval | Frequency | Tally Marks 30–40 | 1 | | 40–50 | 2 | || 50–60 | 2 | || 60–70 | 1 | | 70–80 | 2 | || 80–90 | 2 | || Total | 10 | *Each class shows the count of marks falling within that range, providing a clear summary of the data distribution.* **Question 2:** In a survey of 200 families, 80 own a car, 120 own a motorcycle, and 10 own both. Draw a pie chart representation (find the angles). **Solution:** Car only = 80 – 10 = 70 Motorcycle only = 120 – 10 = 110 Both = 10 Neither = 200 – (70 + 110 + 10) = 10 Angle for car only = (70/200) × 360° = 126° Angle for motorcycle only = (110/200) × 360° = 198° Angle for both = (10/200) × 360° = 18° Angle for neither = (10/200) × 360° = 18° *The pie chart would show these four slices proportional to their angles.* **Question 3:** Two coins are tossed simultaneously. List all possible outcomes and find the probability of getting at least one head. **Solution:** Possible outcomes = {HH, HT, TH, TT} (4 total outcomes) Favorable outcomes (at least one head) = {HH, HT, TH} (3 outcomes) Probability = 3/4 = 0.75 *This demonstrates the equally likely outcome principle—each outcome has equal chance.* **Question 4:** The following data shows the height (in cm) of 15 students: 150, 152, 155, 158, 160, 162, 160, 158, 155, 165, 167, 170, 172, 170, 168. Group the data into classes of width 5, starting from 150, and find the class mark of the median class. **Solution:** Class | Frequency | Cumulative Frequency 150–155 | 3 | 3 155–160 | 4 | 7 160–165 | 2 | 9 165–170 | 4 | 13 170–175 | 2 | 15 Median class (n/2 = 7.5) = 155–160 Class mark = (155 + 160) ÷ 2 = 157.5 cm **Question 5:** A bag contains 5 red balls, 3 blue balls, and 2 green balls. If a ball is drawn at random, find the probability of drawing (i) a red ball, (ii) a non-blue ball. **Solution:** Total balls = 5 + 3 + 2 = 10 (i) Probability of red = 5/10 = 1/2 = 0.5 (ii) Non-blue balls = 5 + 2 = 7 Probability of non-blue = 7/10 = 0.7 *This demonstrates the application of probability to practical scenarios with equally likely outcomes.*

**Question 1:** The following histogram shows the distribution of weights (in kg) of 50 workers. Interpret the histogram and answer: (a) Which weight class has the maximum frequency? (b) How many workers have weight between 60–70 kg? (c) What is the class width? [Assume histogram shows: 40–50 (5), 50–60 (15), 60–70 (20), 70–80 (8), 80–90 (2)] **Solution:** (a) The class 60–70 kg has the maximum frequency of 20 workers. This represents the modal class. (b) Workers between 60–70 kg = 20 (c) Class width = 50 – 40 = 10 kg (or any consecutive class difference) *Histograms allow quick visual identification of data concentration and patterns.* **Question 2:** A survey of 360 students' favorite sports shows: Cricket 120°, Football 90°, Basketball 80°, Tennis 70°. Find the number of students who prefer: (a) Cricket (b) Basketball and Tennis combined (c) Sports other than Football **Solution:** (a) Students preferring Cricket = (120/360) × 360 = 120 students (b) Basketball angle = 80°, Tennis angle = 70°, Combined = 150° Students = (150/360) × 360 = 150 students (c) Football angle = 90°, Remaining angle = 360° – 90° = 270° Students = (270/360) × 360 = 270 students *Pie charts enable conversion of angles directly into frequencies.* **Question 3:** Three cards bearing numbers 2, 4, 6 are placed in a box. Two cards are drawn simultaneously without replacement. Find: (a) Total number of possible outcomes (b) Probability of getting both even numbers (c) Probability of getting sum ≥ 8 **Solution:** (a) Possible outcomes = {(2,4), (2,6), (4,6)} = 3 outcomes (b) All outcomes have both even numbers, so P = 3/3 = 1 (c) Favorable outcomes: (2,6) sum=8, (4,6) sum=10 → 2 outcomes P = 2/3 *Systematic listing ensures no outcome is missed in equally likely scenarios.* **Question 4:** The ages (in years) of 20 employees are: 25, 28, 30, 25, 32, 35, 28, 30, 32, 35, 38, 25, 28, 30, 35, 38, 40, 42, 40, 45. Construct a frequency table and find: (a) Modal age (b) Number of employees below 35 years (c) Percentage of employees aged 40 or above **Solution:** Age | Frequency 25 | 3 28 | 3 30 | 3 32 | 2 35 | 3 38 | 2 40 | 2 42 | 1 45 | 1 (a) Modal age = 25, 28, 30, 35 (all appear 3 times, multimodal) (b) Employees below 35 = 3 + 3 + 3 + 2 = 11 (c) Employees aged ≥40 = 2 + 1 + 1 = 4 Percentage = (4/20) × 100 = 20% *Frequency tables organize ungrouped data for quick statistical analysis.*

**Question 1:** A school conducted a survey of 100 students' daily study hours. The data is given below: Study Hours | 1–2 | 2–3 | 3–4 | 4–5 | 5–6 Number of Students | 10 | 20 | 35 | 25 | 10 Prepare a frequency distribution table with class marks, class width, cumulative frequency, and relative frequency. Draw a histogram and answer: (a) What is the class width? (b) Which class has the maximum number of students? (c) How many students study for more than 3 hours daily? (d) What percentage of students study between 2–4 hours? **Complete Solution:** Frequency Distribution Table: Class | Class Mark | Frequency | Cumulative Frequency | Relative Frequency 1–2 | 1.5 | 10 | 10 | 0.10 2–3 | 2.5 | 20 | 30 | 0.20 3–4 | 3.5 | 35 | 65 | 0.35 4–5 | 4.5 | 25 | 90 | 0.25 5–6 | 5.5 | 10 | 100 | 0.10 (a) **Class width** = Upper limit – Lower limit = 2 – 1 = 1 hour (b) **Maximum frequency class** = 3–4 hours with 35 students (c) **Students studying more than 3 hours** = 35 + 25 + 10 = 70 students (Alternative method: 100 – (10 + 20) = 70) (d) **Students studying 2–4 hours** = 20 + 35 = 55 students **Percentage** = (55/100) × 100 = 55% **Histogram Construction Notes:** - X-axis: Study Hours (1–2, 2–3, 3–4, 4–5, 5–6) - Y-axis: Number of Students (scale: 0 to 40, intervals of 5) - Height of each bar = Frequency of that class - Bars are adjacent with no gaps (characteristic of histograms for continuous data) - The 3–4 hour bar reaches height 35, clearly showing the mode --- **Question 2:** A factory produces light bulbs. The quality control team tests 200 bulbs and records their lifespan (in hours): Lifespan (Hours) | 800–1000 | 1000–1200 | 1200–1400 | 1400–1600 | 1600–1800 Number of Bulbs | 20 | 50 | 70 | 40 | 20 Represent this data using: (a) A bar graph (b) A pie chart (with angles) (c) Find the probability that a randomly selected bulb has lifespan ≥1200 hours (d) Calculate the percentage of bulbs with lifespan less than 1200 hours **Complete Solution:** (a) **Bar Graph Construction:** - X-axis: Lifespan intervals - Y-axis: Number of Bulbs (0–80, intervals of 10) - Bars for each interval with heights: 20, 50, 70, 40, 20 - Label each bar clearly (b) **Pie Chart Angles:** 800–1000: (20/200) × 360° = 36° 1000–1200: (50/200) × 360° = 90° 1200–1400: (70/200) × 360° = 126° 1400–1600: (40/200) × 360° = 72° 1600–1800: (20/200) × 360° = 36° *Verification: 36° + 90° + 126° + 72° + 36° = 360° ✓* (c) **Probability of lifespan ≥1200 hours:** Favorable bulbs = 70 + 40 + 20 = 130 Total bulbs = 200 Probability = 130/200 = 13/20 = 0.65 (d) **Percentage with lifespan <1200 hours:** Bulbs <1200 hours = 20 + 50 = 70 Percentage = (70/200) × 100 = 35% **Key Insight:** The modal class (1200–1400) represents the most common lifespan, suggesting the manufacturing process typically produces bulbs in this range. --- **Question 3:** Two spinners are used in a game. Spinner A has sections labeled 1, 2, 3, 4 (equal size). Spinner B has sections labeled Red, Blue, Green (equal size). Both spinners are spun simultaneously. (a) List all possible outcomes (b) Find the probability of getting an even number and Red (c) Find the probability of getting 3 and any color (d) Find the probability of getting an odd number and not Green (e) How many outcomes are equally likely? **Complete Solution:** (a) **All Possible Outcomes (equally likely):** (1,R), (1,B), (1,G), (2,R), (2,B), (2,G), (3,R), (3,B), (3,G), (4,R), (4,B), (4,G) **Total = 4 × 3 = 12 outcomes** (b) **Probability of even number and Red:** Even numbers on Spinner A = {2, 4} Favorable outcomes = {(2,R), (4,R)} = 2 Probability = 2/12 = 1/6 ≈ 0.167 (c) **Probability of 3 and any color:** Outcomes with 3 = {(3,R), (3,B), (3,G)} = 3 Probability = 3/12 = 1/4 = 0.25 (d) **Probability of odd number and not Green:** Odd numbers = {1, 3} Not Green = {Red, Blue} Favorable outcomes = {(1,R), (1,B), (3,R), (3,B)} = 4 Probability = 4/12 = 1/3 ≈ 0.333 (e) **Equally likely outcomes = 12** *Each outcome has probability 1/12, confirming the equally likely principle.* **Verification Table:** Total probability = 2/12 + 3/12 + 4/12 + 3/12 = 12/12 = 1 ✓ *This demonstrates the multiplication principle: when two independent events occur, total outcomes = outcomes from event 1 × outcomes from event 2.*

**Case Study:** Environmental Survey – Plastic Waste Management A municipality surveyed 360 households to understand plastic waste disposal habits. The pie chart shows the distribution: - Burning (40°) - Landfill (120°) - Recycling (100°) - Composting (60°) - Other methods (40°) Additionally, 20 households were randomly selected, and their weekly plastic waste (in kg) was recorded: 0.5, 0.8, 1.0, 1.2, 1.5, 1.8, 2.0, 2.5, 0.6, 0.9, 1.1, 1.3, 1.6, 1.9, 2.1, 2.3, 0.7, 1.4, 2.2, 2.4 **Questions:** (A) From the pie chart data: (i) How many households use recycling? (ii) What is the ratio of households using landfill to those using composting? (iii) If a household is randomly chosen, what is the probability it uses either burning or "other methods"? (B) From the waste data: (i) Construct a frequency distribution table with class width 0.5, starting from 0.5 (ii) Draw a histogram (iii) Find the probability that a household produces ≥1.5 kg weekly plastic waste (iv) What percentage of households produce between 1.0–2.0 kg waste? **Step-by-Step Solution:** **Part A:** (i) **Recycling households:** Recycling angle = 100° Number of households = (100/360) × 360 = 100 households (ii) **Ratio of landfill to composting:** Landfill angle = 120°, households = (120/360) × 360 = 120 Composting angle = 60°, households = (60/360) × 360 = 60 Ratio = 120 : 60 = 2 : 1 (iii) **Probability of burning OR other methods:** Burning angle = 40°, Other angle = 40° Combined angle = 80° Probability = 80/360 = 2/9 ≈ 0.222 **Part B:** (i) **Frequency Distribution Table:** Class (kg) | Tally | Frequency | Cumulative Frequency 0.5–1.0 | |||| | 5 | 5 1.0–1.5 | |||| | 4 | 9 1.5–2.0 | |||| | 6 | 15 2.0–2.5 | ||||| | 5 | 20 Total: 20 households (ii) **Histogram:** - X-axis: Weekly plastic waste (kg) from 0.5 to 2.5 - Y-axis: Frequency (0 to 7, intervals of 1) - Bar heights: 5, 4, 6, 5 (adjacent bars, no gaps) - The 1.5–2.0 kg class shows maximum frequency (iii) **Probability of ≥1.5 kg waste:** Households with ≥1.5 kg = 6 + 5 = 11 Probability = 11/20 = 0.55 (iv) **Percentage producing 1.0–2.0 kg:** Households in 1.0–1.5 kg class = 4 Households in 1.5–2.0 kg class = 6 Total = 10 Percentage = (10/20) × 100 = 50% **Critical Thinking Extension:** If the municipality wants to promote recycling, which waste-producing segment (low, medium, high) should it target first? Based on the data, 55% of households produce ≥1.5 kg weekly—a significant volume. Targeting households in the 1.5–2.0 kg range (30% of total) with recycling incentives could yield maximum environmental benefit. **Numerical verification:** 5 + 4 + 6 + 5 = 20 ✓, all angles sum to 360° ✓

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**Mistake 1: Confusing Class Interval Notation** Students often treat class 10–20 as having limits 10 and 20, forgetting that in a grouped frequency distribution, a value of exactly 20 belongs to the *next* class (20–30). Always clarify whether intervals are inclusive: does 20–30 mean 20 ≤ x < 30 or 20 < x ≤ 30? NCERT Class 9 uses the convention that the lower limit is included and the upper is excluded, *except* for the final class. This distinction affects cumulative frequency calculations. **Mistake 2: Incorrect Histogram Construction** Many students draw histograms with gaps between bars or with unequal bar widths. Remember: histograms are for **continuous grouped data** and must have adjacent bars (no gaps). Bar height = Frequency ÷ Class width (if widths differ). Failing to adjust heights when class widths vary leads to visual misinterpretation. **Mistake 3: Pie Chart Angle Errors** Forget the formula, not the principle. Angle = (Frequency ÷ Total frequency) × 360°. Many students miscalculate totals or use percentages (0–100) instead of angles (0–360°). Always verify: sum of all angles must equal 360°. **Mistake 4: Probability ≠ Frequency** Probability = (Favorable outcomes) ÷ (Total equally likely outcomes), *not* the frequency from a histogram. If a die shows 6 once in 10 rolls, probability of 6 is 1/6 (theory), not 1/10 (observed frequency). The sample experiment approaches 1/6 as rolls increase (Law of Large Numbers). **Mistake 5: Forgetting to List All Outcomes** In equally likely outcome problems (e.g., two coins), students mentally shortcut and miss branches. Always use systematic methods: lists, tree diagrams, or tables. For two coins: {HH, HT, TH, TT}—not just {H, T}. Incomplete listing leads to wrong probabilities. **Mistake 6: Misinterpreting Cumulative Frequency** Cumulative frequency at a class = sum of frequencies *up to and including* that class. It answers "how many scored up to this value?" not "how many scored in this range?" Plotting cumulative frequency points at upper class limits (not midpoints) creates an *ogive* (cumulative curve).