Class 10 Mathematics Practice Sheet: CBSE Class 10 Mathematics Statistics (Mean, Median and Mode) Worksheet Set 11
Review targeted academic worksheets with the CBSE Class 10 Mathematics Statistics (Mean, Median and Mode) Worksheet Set 11. Built according to official educational standards for the 2026-27 term, these downloadable Class 10 Mathematics resources support effective daily practice and detailed self-evaluation for Chapter 13 Statistics.
Download Chapter 13 Statistics Worksheet PDF with Answers
View or download the dedicated CBSE Class 10 Mathematics Statistics (Mean, Median and Mode) Worksheet Set 11 resource below. Engaging with these practice papers under focused study conditions ensures continuous academic progress and mastery of the 2026-27 curriculum for Chapter 13 Statistics.
Questions
Part A: Concept and Short Questions
Q1. Find the mean of the first ten prime numbers.
View Answer & Explanation
Answer: Sum = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29 = 129. Mean = 129/10 = 12.9.
Teacher's Note:
1. List the ten primes and add them.
2. Divide the sum by 10.
Q2. Which measure of central tendency is given by the abscissa of the point of intersection of the less than ogive and the more than ogive?
View Answer & Explanation
Answer: It is the median of the data.
Teacher's Note:
1. The two ogives meet at the cumulative frequency N/2.
2. The corresponding class value is the median.
Q3. If the mode of a data set is 45 and the mean is 27, find the median.
View Answer & Explanation
Answer: Mode = 3 Median − 2 Mean, so 45 = 3 Median − 54 and Median = 33.
Teacher's Note:
1. Use the empirical relation among the three measures.
2. Rearrange for the median.
Q4. Find the mode of the following data: value 35, 38, 40, 42, 44 with frequencies 5, 9, 10, 7, 2.
View Answer & Explanation
Answer: The highest frequency is 10, which belongs to the value 40. Mode = 40.
Teacher's Note:
1. For ungrouped data the mode is the value with the largest frequency.
2. Compare the frequencies.
Q5. Write the median class of the following distribution. Classes 0–10, 10–20, 20–30, 30–40, 40–50, 50–60, 60–70 with frequencies 4, 4, 8, 10, 12, 8, 4.
View Answer & Explanation
Answer: Total = 50, so N/2 = 25. Cumulative frequencies are 4, 8, 16, 26, … The class where it first reaches 25 or more is 30–40.
Teacher's Note:
1. Compute the cumulative frequency column.
2. Find the first class with cumulative frequency at least N/2.
Q6. The wickets taken by a bowler in 10 cricket matches are 2, 6, 4, 5, 0, 2, 1, 3, 2, 3. Find the mode.
View Answer & Explanation
Answer: The value 2 occurs three times, more than any other value. Mode = 2.
Teacher's Note:
1. Count how often each value appears.
2. The most frequent value is the mode.
Q7. How can the median of a frequency distribution be found graphically?
View Answer & Explanation
Answer: Draw the less than type and more than type ogives on the same graph. The x-coordinate (abscissa) of their point of intersection is the median.
Teacher's Note:
1. Both ogives need the same axes and scale.
2. Read the value on the x-axis.
Part B: Standard Questions
Q8. Find the median of the following heights. Height (cm): 160–162, 163–165, 166–168, 169–171, 172–174 with frequencies 15, 117, 136, 118, 14.
View Answer & Explanation
Answer: Convert to continuous classes: 159.5–162.5, 162.5–165.5, 165.5–168.5, … N = 400, N/2 = 200. Cumulative frequency before the median class 165.5–168.5 is 132. Median = 165.5 + (200 − 132)/136 × 3 = 167 cm.
Teacher's Note:
1. Make the classes continuous by adjusting the limits by 0.5.
2. Then apply the median formula.
Q9. Find the median IQ. IQ: 75–84, 85–94, 95–104, 105–114, 115–124, 125–134, 135–144 with frequencies 8, 11, 26, 31, 18, 4, 2.
View Answer & Explanation
Answer: Make classes continuous (74.5–84.5, …). N = 100, N/2 = 50; median class 104.5–114.5, cf = 45, f = 31. Median = 104.5 + (50 − 45)/31 × 10 = 106.1.
Teacher's Note:
1. Adjust class limits first.
2. Locate the median class using cumulative frequency.
Q10. Find the median of the following. Class: 0–10, 10–20, 20–30, 30–40, 40–50, 50–60 with frequencies 5, 8, 20, 15, 7, 5.
View Answer & Explanation
Answer: N = 60, N/2 = 30. Cumulative frequencies: 5, 13, 33, … so the median class is 20–30 with cf = 13 and f = 20. Median = 20 + (30 − 13)/20 × 10 = 28.5.
Teacher's Note:
1. Build the cumulative frequency column.
2. Substitute in the median formula.
Q11. A teacher records the absence of 40 students for a term. Days: 0–6, 6–10, 10–14, 14–20, 20–28, 28–38, 38–40 with students 11, 10, 7, 4, 4, 3, 1. Write the data as a less than type cumulative frequency distribution.
View Answer & Explanation
Answer: Less than 6: 11; less than 10: 21; less than 14: 28; less than 20: 32; less than 28: 36; less than 38: 39; less than 40: 40.
Teacher's Note:
1. Add the frequencies progressively.
2. Label each total with the upper class limit.
Q12. Using the assumed mean method, find the mean. Class: 0–10, 10–20, 20–30, 30–40, 40–50 with frequencies 7, 8, 12, 13, 10.
View Answer & Explanation
Answer: Take a = 25. Deviations d = −20, −10, 0, 10, 20 with f = 7, 8, 12, 13, 10. Σfd = −140 − 80 + 0 + 130 + 200 = 110 and Σf = 50. Mean = 25 + 110/50 = 27.2.
Teacher's Note:
1. Class marks are 5, 15, 25, 35, 45.
2. Add the average deviation to the assumed mean.
Q13. The mean of the distribution below is 25. Find p. Class: 0–10, 10–20, 20–30, 30–40, 40–50 with frequencies 5, 18, 15, p, 6.
View Answer & Explanation
Answer: Σf = 44 + p and Σfx = 25 + 270 + 375 + 35p + 270 = 940 + 35p. So 940 + 35p = 25(44 + p), giving 10p = 160 and p = 16.
Teacher's Note:
1. Use class marks 5, 15, 25, 35, 45.
2. Form an equation using mean = Σfx/Σf.
Q14. Find the mean by the step deviation method. Class: 0–10, 10–20, 20–30, 30–40, 40–50 with frequencies 7, 12, 13, 10, 8.
View Answer & Explanation
Answer: Take a = 25, h = 10, so u = −2, −1, 0, 1, 2. Σfu = −14 − 12 + 0 + 10 + 16 = 0 and Σf = 50. Mean = 25 + (0/50) × 10 = 25.
Teacher's Note:
1. Divide each deviation by the class size to get u.
2. Σfu is zero, so the mean equals the assumed mean.
Q15. The median of the following distribution is 50. Find p. Class: 20–30, 30–40, 40–50, 50–60, 60–70, 70–80, 80–90 with frequencies 25, 15, p, 6, 24, 12, 8.
View Answer & Explanation
Answer: N = 90 + p. If the median class is 40–50, then 50 = 40 + ((90 + p)/2 − 40)/p × 10, which gives 45 + p/2 − 40 = p and p = 10. Check: N = 100, N/2 = 50 and cumulative frequency up to 40–50 is 50, so the class is correct.
Teacher's Note:
1. Assume 40–50 is the median class and verify afterwards.
2. Simplify the equation to get p.
Q16. Find the median of the following data. Marks less than: 10, 30, 50, 70, 90, 110, 130, 150 with cumulative frequencies 0, 10, 25, 43, 65, 87, 96, 100.
View Answer & Explanation
Answer: Class frequencies: 10–30: 10; 30–50: 15; 50–70: 18; 70–90: 22; 90–110: 22; 110–130: 9; 130–150: 4. N/2 = 50, median class 70–90 with cf = 43. Median = 70 + (50 − 43)/22 × 20 = 76.36.
Teacher's Note:
1. Convert the cumulative data into class frequencies.
2. Apply the median formula.
Q17. If the mean of a data set is 60 and the median is 50, find the mode using the empirical relationship.
View Answer & Explanation
Answer: Mode = 3 Median − 2 Mean = 150 − 120 = 30.
Teacher's Note:
1. Use Mode = 3 Median − 2 Mean.
2. Substitute the values.
Part C: Long Answer Questions
Q18. If the mean of the distribution is 18, find p. Variate x: 13, 15, 17, 19, 20 + p, 23 with frequencies 8, 2, 3, 4, 5p, 6.
View Answer & Explanation
Answer: Σf = 23 + 5p and Σfx = 104 + 30 + 51 + 76 + 5p(20 + p) + 138 = 399 + 100p + 5p². Setting equal to 18(23 + 5p) = 414 + 90p gives 5p² + 10p − 15 = 0, i.e. p² + 2p − 3 = 0. So p = 1 (p = −3 is rejected).
Teacher's Note:
1. Form the equation Σfx = 18 Σf.
2. Reject the negative root because frequency cannot be negative.
Q19. Find the mean, median and mode. Class: 0–10, 10–20, 20–30, 30–40, 40–50, 50–60, 60–70 with frequencies 5, 8, 15, 20, 14, 8, 5.
View Answer & Explanation
Answer: N = 75. Mean = 2615/75 ≈ 34.87. Median class is 30–40 (N/2 = 37.5, cf = 28): median = 30 + (37.5 − 28)/20 × 10 = 34.75. Modal class 30–40: mode = 30 + (20 − 15)/(40 − 15 − 14) × 10 ≈ 34.55.
Teacher's Note:
1. Find the class marks for the mean.
2. Use the median and mode formulas with the correct classes.
Q20. The median of the following data is 52.5 and the total frequency is 100. Find x and y. Class: 0–10, 10–20, 20–30, 30–40, 40–50, 50–60, 60–70, 70–80, 80–90, 90–100 with frequencies 2, 5, x, 12, 17, 20, y, 9, 7, 4.
View Answer & Explanation
Answer: Sum of the known frequencies is 76, so x + y = 24. Median class 50–60: cf = 36 + x, so 52.5 = 50 + (50 − 36 − x)/20 × 10, giving 14 − x = 5, x = 9 and y = 15.
Teacher's Note:
1. Use the total to get one equation.
2. Use the median formula for the second.
Q21. Find the median of the data. Classes: 20–30, 30–40, 40–50, 50–60, 60–70, 70–80, 80–90 with frequencies 10, 8, 12, 24, 6, 25, 15.
View Answer & Explanation
Answer: N = 100, N/2 = 50; cumulative frequencies are 10, 18, 30, 54, … so the median class is 50–60 with cf = 30. Median = 50 + (50 − 30)/24 × 10 = 58.33.
Teacher's Note:
1. The median class is where cf first exceeds 50.
2. Use the median formula.
Q22. Find the mean marks. Marks: below 10, 20, 30, 40, 50, 60, 70, 80, 90, 100 with number of students 5, 9, 17, 29, 45, 60, 70, 78, 83, 85.
View Answer & Explanation
Answer: Class frequencies: 5, 4, 8, 12, 16, 15, 10, 8, 5, 2 for classes 0–10, …, 90–100. Σfx = 4115 and Σf = 85, so mean = 4115/85 ≈ 48.41.
Teacher's Note:
1. Find individual frequencies by differences.
2. Use class marks 5, 15, …, 95.
Q23. Find the median age. Age below: 10, 20, 30, 40, 50, 60, 70, 80 with number of persons 200, 500, 900, 1200, 1400, 1500, 1550, 1560.
View Answer & Explanation
Answer: Class frequencies: 200, 300, 400, 300, 200, 100, 50, 10. N = 1560, N/2 = 780; median class 20–30 with cf = 500. Median = 20 + (780 − 500)/400 × 10 = 27 years.
Teacher's Note:
1. Convert the cumulative data to class frequencies.
2. Apply the median formula.
Q24. The median of the data is 32.5 and the total frequency is 40. Find x and y. Class: 0–10, 10–20, 20–30, 30–40, 40–50, 50–60, 60–70 with frequencies x, 5, 9, 12, y, 3, 2.
View Answer & Explanation
Answer: x + y = 9. Median class 30–40 has cf = x + 14: 32.5 = 30 + (20 − x − 14)/12 × 10, so 6 − x = 3, x = 3 and y = 6.
Teacher's Note:
1. Use the total frequency to get x + y.
2. Use the median formula for the other equation.
Q25. The monthly electricity consumption of 68 consumers is shown. Units: 65–85, 85–105, 105–125, 125–145, 145–165, 165–185, 185–205 with consumers 4, 5, 13, 20, 14, 8, 4. Find the median and the number of families consuming 125 or more units.
View Answer & Explanation
Answer: N = 68, N/2 = 34; cf before 125–145 is 22. Median = 125 + (34 − 22)/20 × 20 = 137 units. Families using 125 or more units = 20 + 14 + 8 + 4 = 46.
Teacher's Note:
1. Locate the median class from cumulative frequencies.
2. Add the frequencies of the classes from 125 onwards.
Q26. The mileage of 50 cars is given. Mileage (km/l): 10–12, 12–14, 14–16, 16–18 with 7, 12, 18, 13 cars. Find the mean mileage. Does it support a claim of 16 km/l?
View Answer & Explanation
Answer: Mean = (11×7 + 13×12 + 15×18 + 17×13)/50 = 724/50 = 14.48 km/l. Since this is below 16 km/l, the data does not support the claim.
Teacher's Note:
1. Use the class marks 11, 13, 15, 17.
2. Compare the mean with the claimed value.
Free study material for Mathematics
Free CBSE Practice Worksheets: Class 10 Mathematics Chapter 13 Statistics
Download Chapter Worksheets: Class 10 Mathematics
Review targeted practice exercises for Class 10 Mathematics Chapter 13 Statistics. Curated to match official CBSE guidelines, these printable problem sets support daily revision and improve overall test readiness.
Concept Clarification for Chapter 13 Statistics
Designed around the official curriculum for Class 10 Mathematics, these practice sheets guarantee standard compliance. Reviewing step-by-step solutions after completion sharpens your accuracy and clarifies complex sub-topics within Chapter 13 Statistics.
Effective Revision Strategies for School Exams
Consistent engagement with these exercises builds familiarity with recurring exam themes. If specific areas within Chapter 13 Statistics cause trouble, utilize our dedicated NCERT solutions for Class 10 Mathematics to clear up doubts immediately.
FAQs
You can download the latest chapter-wise printable worksheets for Class 10 Mathematics Chapter 13 Statistics for free from StudiesToday.com. These have been made as per the latest CBSE curriculum for this academic year.
Yes, Class 10 Mathematics worksheets for Chapter 13 Statistics focus on activity-based learning and also competency-style questions. This helps students to apply theoretical knowledge to practical scenarios.
Yes, we have provided solved worksheets for Class 10 Mathematics Chapter 13 Statistics to help students verify their answers instantly.
Yes, our Class 10 Mathematics test sheets are mobile-friendly PDFs and can be printed by teachers for classroom.
For Chapter 13 Statistics, regular practice with our worksheets will improve question-handling speed and help students understand all technical terms and diagrams.