Class 9 Mathematics Chapter 14: Data Handling and Probability – Important Questions with Full Solutions

The CBSE Class 9 Mathematics board exam weights Data Handling and Probability at 6–8 marks, typically distributed across multiple choice (1 mark), short-answer (2–3 marks), and extended-response questions (5 marks). The rationalized 2024-25 syllabus emphasizes conceptual clarity over rote learning: examiners expect you to not only *calculate* mean, median, and mode but also to *justify which measure* best represents a dataset, interpret real-world graphs critically, and reason about probability using sample spaces and counting principles. Questions often blend topics — for example, a 5-mark question might ask you to compute mean from a frequency table, then represent it as a pie chart, then compare probability outcomes. The question patterns tested are predictable: mean calculations from ungrouped/grouped data, median from cumulative frequency, mode from frequency tables, reading/drawing bar and pie charts, identifying sample spaces, calculating theoretical and experimental probability, and applying tree diagrams. By practicing these 18 variations, you'll recognize the 4–5 core templates the examiner cycles through, build speed, and eliminate careless errors. Many students lose marks not from lack of knowledge but from inconsistent formula application or misreading graph scales — these questions are designed to expose and remedy those gaps.

Multiple-choice questions test your recall of definitions, formulas, and quick conceptual checks. These are often gateway questions — getting them right builds momentum and saves time for longer problems. **Q1:** The mean of 4, 8, 6, and 10 is: A) 7 B) 8 C) 7.5 D) 6 **Answer: A) 7** Mean = (4 + 8 + 6 + 10) ÷ 4 = 28 ÷ 4 = 7 **Q2:** In a frequency distribution, the value that appears most often is called: A) Mean B) Median C) Mode D) Range **Answer: C) Mode** Mode is the observation with the highest frequency. **Q3:** If a die is rolled once, the total number of possible outcomes (sample space) is: A) 1 B) 3 C) 6 D) 12 **Answer: C) 6** A standard die has faces numbered 1 to 6; each is equally likely. **Q4:** The probability of an impossible event is: A) 0 B) 0.5 C) 1 D) –1 **Answer: A) 0** Probability ranges from 0 (impossible) to 1 (certain). **Q5:** A pie chart is divided into three sectors with angles 90°, 120°, and 150°. What is the central angle for the remaining sector? A) 30° B) 40° C) 50° D) 60° **Answer: A) 30°** Total angle in a pie chart = 360°. Remaining = 360° – (90° + 120° + 150°) = 30°

These questions require brief calculations and one-line justification. Expect questions on mean, median identification, mode from frequency tables, and simple probability. **Q1:** Find the median of the following data: 3, 7, 2, 9, 5, 4, 8 **Solution:** Arrange in order: 2, 3, 4, 5, 7, 8, 9 Number of values (n) = 7 (odd) Median = value at position (n+1)/2 = (7+1)/2 = 4th position **Median = 5** **Q2:** A frequency table shows: Value 10 (freq 3), Value 15 (freq 7), Value 20 (freq 5). Find the mode. **Solution:** The highest frequency is 7, corresponding to value 15. **Mode = 15** **Q3:** Two coins are tossed together. Write the sample space and find the probability of getting at least one tail. **Solution:** Sample space = {HH, HT, TH, TT} Total outcomes = 4 Favourable outcomes (at least one T) = {HT, TH, TT} = 3 **Probability = 3/4** **Q4:** In a pie chart representing 100 students' favourite subjects, if 25 students prefer Mathematics, what is the central angle for the Mathematics sector? **Solution:** Central angle = (Frequency / Total) × 360° = (25 / 100) × 360° = 0.25 × 360° = **90°** **Q5:** Find the mean of the data: 12, 18, 14, 20, 11 **Solution:** Mean = (12 + 18 + 14 + 20 + 11) / 5 = 75 / 5 = **15**

Three-mark questions typically combine two concepts or require multi-step reasoning: mean from a frequency table, median class identification, probability with multiple conditions, and interpretation of graphs. **Q1:** Calculate the mean from the following frequency table: Class: 10–20, 20–30, 30–40, 40–50 Frequency: 5, 8, 4, 3 **Solution:** Find class midpoints: 15, 25, 35, 45 Create column: Midpoint × Frequency 15 × 5 = 75 25 × 8 = 200 35 × 4 = 140 45 × 3 = 135 Sum of (f × x) = 75 + 200 + 140 + 135 = 550 Total frequency = 5 + 8 + 4 + 3 = 20 **Mean = 550 / 20 = 27.5** **Q2:** A bar graph shows that in a survey of 200 people, 60 prefer Tea, 80 prefer Coffee, 40 prefer Juice, and 20 prefer Water. Find the probability that a randomly selected person prefers either Tea or Coffee. **Solution:** People preferring Tea = 60 People preferring Coffee = 80 Favourable outcomes = 60 + 80 = 140 Total outcomes = 200 **Probability = 140 / 200 = 7/10 = 0.7** **Q3:** A student scores 85, 92, 78, and 88 in four tests. If a fifth test is added, what score must the student achieve to bring the mean to 85? **Solution:** Sum of first four scores = 85 + 92 + 78 + 88 = 343 For mean of 5 tests to be 85: Total needed = 85 × 5 = 425 Fifth test score = 425 – 343 = **82** **Q4:** In a box, there are 3 red balls, 5 blue balls, and 2 green balls. A ball is drawn at random. Find the probability of drawing a red ball and the probability of drawing a non-blue ball. **Solution:** Total balls = 3 + 5 + 2 = 10 P(red) = 3/10 Non-blue balls = red + green = 3 + 2 = 5 **P(non-blue) = 5/10 = 1/2**

Five-mark questions test deep understanding and integration of multiple concepts. Expect questions combining mean/median calculations, graph interpretation, and probability reasoning. **Q1:** The following table shows the marks obtained by 50 students in a Mathematics test: Marks: 0–10, 10–20, 20–30, 30–40, 40–50 No. of Students: 5, 8, 12, 15, 10 Calculate (i) the mean, (ii) identify the median class, and (iii) draw a bar graph representing this data. **Solution:** (i) **Mean Calculation:** Class midpoints: 5, 15, 25, 35, 45 f × x: 5×5=25, 8×15=120, 12×25=300, 15×35=525, 10×45=450 Sum of f × x = 25 + 120 + 300 + 525 + 450 = 1420 Total frequency = 50 Mean = 1420 / 50 = **28.4 marks** (ii) **Median Class:** Cumulative frequencies: 5, 13, 25, 40, 50 Median position = 50/2 = 25th observation The 25th observation lies in the class 20–30 (because cumulative frequency reaches 25 at the end of this class). **Median class = 20–30** (iii) **Bar Graph:** [Bar graph would show marks ranges on x-axis (0–10, 10–20, etc.) and number of students on y-axis (0 to 15). Heights would be 5, 8, 12, 15, and 10 respectively. The tallest bar represents 30–40 with 15 students.] **Q2:** Two fair dice are rolled simultaneously. Draw a sample space table and find the probability that: (i) The sum is 7 (ii) The sum is greater than 9 (iii) Both dice show the same number **Solution:** **Sample Space Table (6×6 = 36 outcomes):** ``` (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6) ``` Total outcomes = 36 (i) **Sum = 7:** (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 outcomes **P(sum = 7) = 6/36 = 1/6** (ii) **Sum > 9:** (4,6), (5,5), (5,6), (6,4), (6,5), (6,6) → 6 outcomes **P(sum > 9) = 6/36 = 1/6** (iii) **Both dice same:** (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) → 6 outcomes **P(same) = 6/36 = 1/6** **Q3:** A school conducted a survey of 300 students about their preferred sport. The results are: Cricket 120, Football 90, Basketball 60, Badminton 30. (i) Calculate the central angle for each sport in a pie chart. (ii) Draw the pie chart. (iii) If a student is selected at random, what is the probability that they prefer either Cricket or Football? **Solution:** (i) **Central Angles:** Cricket: (120/300) × 360° = 0.4 × 360° = **144°** Football: (90/300) × 360° = 0.3 × 360° = **108°** Basketball: (60/300) × 360° = 0.2 × 360° = **72°** Badminton: (30/300) × 360° = 0.1 × 360° = **36°** [Total = 144° + 108° + 72° + 36° = 360° ✓] (ii) **Pie Chart:** [A pie chart divided into four sectors with the angles above, labeled with sport names and percentages: Cricket 40%, Football 30%, Basketball 20%, Badminton 10%.] (iii) **Probability of Cricket or Football:** Students preferring Cricket or Football = 120 + 90 = 210 **P(Cricket or Football) = 210/300 = 7/10 = 0.7**

This question integrates multiple concepts and real-world application — typical of CBSE's emphasis on critical thinking. **Case Study: School Attendance & Event Planning** A school has 150 students in Class 9. Over a month, attendance records showed: Attendance Rate: 70–80%, 80–90%, 90–95%, 95–100% No. of Students: 30, 45, 50, 25 The school wants to (a) reward students with attendance ≥ 90%, (b) identify the typical attendance rate (median), and (c) randomly select a student to serve as Class Captain from those with attendance ≥ 90%. **Question:** (i) Calculate the mean attendance rate. (ii) Identify the median attendance class. (iii) How many students are eligible for the reward? What is the probability that a randomly selected eligible student is chosen as Class Captain? (iv) Draw a bar graph of the attendance distribution. (v) If the school wants 60% of students to have attendance ≥ 85%, is this goal currently met? Justify with calculations. **Solution (Step-by-Step):** **(i) Mean Attendance Rate:** Use class midpoints: 75%, 85%, 92.5%, 97.5% f × x: 30×75=2250, 45×85=3825, 50×92.5=4625, 25×97.5=2437.5 Sum = 2250 + 3825 + 4625 + 2437.5 = 13137.5 Mean = 13137.5 / 150 = **87.58%** (approximately 87.6%) **(ii) Median Attendance Class:** Cumulative frequencies: 30, 75, 125, 150 Median position = 150/2 = 75th student The 75th observation exactly reaches the end of the 80–90% class. **Median class = 80–90%** **(iii) Eligible Students & Probability:** Students with attendance ≥ 90% = 50 + 25 = 75 students Probability a randomly selected eligible student is Class Captain = 1/75 **(iv) Bar Graph:** [A bar chart with x-axis showing attendance ranges (70–80%, 80–90%, 90–95%, 95–100%) and y-axis showing number of students (0–50). Bar heights: 30, 45, 50, 25.] **(v) Goal Achievement Check:** Students with attendance ≥ 85% = (45 × 5/10) + 50 + 25 = 22.5 + 50 + 25 = 97.5 ≈ 98 students (Using midpoint of 80–90% class, 5% of the 10% range accounts for 5% above 85%) Percentage = 98 / 150 ≈ 65.3% **Yes, the goal of 60% is exceeded; currently 65.3% of students meet the criterion.** **Key Skills Tested:** Mean calculation from grouped data, identifying median class using cumulative frequency, probability as a ratio, basic percentage analysis, data interpretation, and graph construction — all integrated into a realistic scenario.

Mastering Data Handling and Probability demands more than reading solutions — it requires repetitive, adaptive practice on these exact question templates. CBSETUTOR.ai's AI-powered tutoring platform is specifically engineered for Class 9 students preparing for CBSE board exams. **Daily Drill Structure:** — **Adaptive Question Generation:** The AI generates 5–7 questions daily based on your performance history, cycling through MCQs, short-answer, and long-answer formats. If you struggle with median identification, the system assigns 3–4 median-focused variants the next day. — **Instant Feedback with Reasoning:** Each answer is evaluated not just for correctness but for method. If you use a wrong formula or skip a step, the AI explains exactly where you went wrong and walks through the correct approach with worked numbers. — **Spaced Repetition:** The system tracks which question types you answered poorly and resurfaces them at optimal intervals (typically 2–3 days later) to strengthen weak areas before they cause board-exam errors. — **Real NCERT & Board Alignment:** Every question is sourced from or mirrors the 2024-25 NCERT Class 9 Maths textbook and past CBSE board papers, ensuring 100% relevance and no surprises on exam day. — **Time Pressure Simulation:** Practice mode allows you to attempt questions under timed conditions (1-mark in 1 min, 3-mark in 4 min, 5-mark in 8 min), building the speed and confidence needed on exam day. — **Concept-Video Integration:** Before drilling questions, you can watch a 5–8-minute concept video (e.g., 'Calculating Mean from Grouped Data') narrated by expert educators, ensuring clarity before problem-solving. Start a 3-day free trial at cbsetutor.ai and experience how personalized AI-driven practice transforms your Data Handling and Probability score.