School Assignments for Class 10 Mathematics: Chapter 13 Statistics
Review targeted academic assignments with the CBSE Class 10 Mathematics Statistics Assignment Set 20. Built according to official CBSE standards for the 2026-27 term, these downloadable Class 10 Mathematics worksheets support effective daily practice for Chapter 13 Statistics.
Practice Class 10 Mathematics Assignments: Chapter 13 Statistics
View or download the dedicated CBSE Class 10 Mathematics Statistics Assignment Set 20 resource below. Engaging with these assignments under focused study conditions ensures continuous academic progress and mastery of the 2026-27 curriculum.
Exercise-1 Subjective Questions: Introduction and Arithmetic Mean
Question A-1. Find the mean of the first five multiples of 6.
Answer:
The numbers are 6, 12, 18, 24, 30.
Mean = \( \dfrac{90}{5} \) = 18.
Teacher's Note:
1. Write the numbers out, then add and divide by 5.
Question A-2. If the average of 11, 12, 13, 14 and x is 13, then find the value of x.
Answer:
Sum = 13 × 5 = 65, and 11 + 12 + 13 + 14 = 50.
x = 65 − 50 = 15.
Teacher's Note:
1. Sum of observations = mean × number of observations.
Question A-3. The mean of six numbers is 15. If 2 is taken away from every number, then find the new mean.
Answer:
If a constant is subtracted from every observation, the mean is reduced by the same constant.
New mean = 15 − 2 = 13.
Teacher's Note:
1. Subtracting a constant shifts the mean by that constant.
| x | 10 | 15 | p | 25 | 30 |
| f | 5 | 10 | 7 | 8 | 2 |
Answer:
Σfᵢ = 32, so Σfᵢxᵢ = 18.75 × 32 = 600.
50 + 150 + 7p + 200 + 60 = 600, so 7p = 140 and p = 20.
Teacher's Note:
1. Mean × Σf = Σfx.
| Class interval | 50-70 | 70-90 | 90-110 | 110-130 | 130-150 | 150-170 |
| No. of workers | 18 | 12 | 13 | 27 | 8 | 22 |
Answer:
Class marks 60, 80, 100, 120, 140, 160. By the direct method, Σfᵢ = 100 and Σfᵢxᵢ = 1080 + 960 + 1300 + 3240 + 1120 + 3520 = 11220.
Mean = 11220 ÷ 100 = 112.20.
Check by step deviation (a = 100, h = 20): uᵢ = −2, −1, 0, 1, 2, 3; Σfᵢuᵢ = −36 − 12 + 0 + 27 + 16 + 66 = 61; mean = 100 + (61/100) × 20 = 112.2.
Teacher's Note:
1. Both methods must give the same mean.
| C.I. | 0-80 | 80-160 | 160-240 | 240-320 | 320-400 | Total |
| Freq. | 20 | 25 | f₁ | f₂ | 10 | 100 |
Answer:
20 + 25 + f₁ + f₂ + 10 = 100, so f₁ + f₂ = 45 …(i)
Class marks 40, 120, 200, 280, 360: 800 + 3000 + 200f₁ + 280f₂ + 3600 = 188 × 100 = 18800, so 5f₁ + 7f₂ = 285 …(ii)
From (i) and (ii): 5(45 − f₂) + 7f₂ = 285, so f₂ = 30. f₁ = 15 and f₂ = 30.
Teacher's Note:
1. Use the total and the mean to get two equations.
Median
Question B-1. If the mean and median of a set of numbers are 11 and 10 respectively, then find the mode.
Answer:
Mode = 3 Median − 2 Mean = 30 − 22 = 8.
Teacher's Note:
1. Use the empirical relation.
| Rent (in Rs.) | 15-25 | 25-35 | 35-45 | 45-55 | 55-65 | 65-75 | 75-85 | 85-95 |
| No. of houses | 8 | 10 | 15 | 25 | 40 | 20 | 15 | 7 |
Answer:
N = 140, N/2 = 70. Cumulative frequencies: 8, 18, 33, 58, 98, … so the median class is 55 - 65: ℓ = 55, C = 58, f = 40, h = 10.
Median = \( 55 + \dfrac{70 - 58}{40} \times 10 = \) 58.
Teacher's Note:
1. The median class is 55 - 65 because 98 is the first cumulative frequency above 70.
| Marks obtained | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
| Frequency | 8 | 10 | 12 | 22 | 30 | 18 |
Answer:
N = 100, N/2 = 50. Cumulative frequencies: 8, 18, 30, 52, … The first value above 50 is 52.
Median class = 30 - 40.
Teacher's Note:
1. Median class = class containing the (N/2)th observation.
| No. of accidents | 0 | 1 | 2 | 3 | 4 | 5 |
| No. of days | 46 | f₁ | f₂ | 25 | 10 | 5 |
Answer:
46 + f₁ + f₂ + 25 + 10 + 5 = 200, so f₁ + f₂ = 114 …(i)
Σfx = f₁ + 2f₂ + 75 + 40 + 25 = 1.46 × 200 = 292, so f₁ + 2f₂ = 152 …(ii)
(ii) − (i): f₂ = 38 and f₁ = 76. f₁ = 76, f₂ = 38.
Cumulative frequencies: 46, 122, 160, … N/2 = 100 lies in the value 1, so the median = 1.
Teacher's Note:
1. For a discrete distribution, the median is the value whose cumulative frequency first exceeds N/2.
Mode
| C.I. | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | 80-90 | 90-100 |
| Freq. | 3 | 5 | 16 | 12 | 13 | 20 | 5 | 4 | 1 | 1 |
Answer:
The modal class is 50 - 60: ℓ = 50, f₁ = 20, f₀ = 13, f₂ = 5, h = 10.
Mode = \( 50 + \dfrac{20 - 13}{40 - 13 - 5} \times 10 = 50 + 3.18 = \) 53.18.
Teacher's Note:
1. Take f₀ = 13 (class 40 - 50) and f₂ = 5 (class 60 - 70).
| C.I. | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
| Freq. | 5 | 8 | 7 | 12 | 28 | 20 | 10 | 10 |
Answer:
The modal class is 40 - 50: ℓ = 40, f₁ = 28, f₀ = 12, f₂ = 20, h = 10.
Mode = \( 40 + \dfrac{28 - 12}{56 - 12 - 20} \times 10 = 40 + 6.67 = \) 46.67.
Teacher's Note:
1. Apply the mode formula to the class with frequency 28.
| Classes | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
| Frequency | 5 | 10 | 18 | 30 | 20 | 12 | 5 |
Answer:
N = 100. Class marks 5, 15, …, 65: Σfᵢxᵢ = 25 + 150 + 450 + 1050 + 900 + 660 + 325 = 3560, so mean = 35.6.
Median class: N/2 = 50, cumulative frequencies 5, 15, 33, 63, … so the median class is 30 - 40: median = 30 + (50 − 33)/30 × 10 = 35.67.
Modal class 30 - 40: mode = 30 + (30 − 18)/(60 − 18 − 20) × 10 = 35.45.
Teacher's Note:
1. Check the relation: 3 × median − 2 × mean = 35.8, close to the mode 35.45.
Question C-4. If the ratio of mean and mode is 5 : 2, then find the ratio of median : mode.
Answer:
Let mean = 5k and mode = 2k. Mode = 3 Median − 2 Mean ⇒ 2k = 3 Median − 10k ⇒ Median = 4k.
Median : Mode = 4k : 2k = 2 : 1.
Teacher's Note:
1. Express all three in terms of k.
| Class interval | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
| Frequency | 12 | 20 | x | 26 | 6 | 7 |
Answer:
The mode 42 lies in 40 - 50, so this is the modal class: ℓ = 40, f₁ = 26, f₀ = x, f₂ = 6, h = 10.
\( 42 = 40 + \dfrac{26 - x}{52 - x - 6} \times 10 \Rightarrow \dfrac{26 - x}{46 - x} = 0.2 \Rightarrow 26 - x = 9.2 - 0.2x \Rightarrow \) x = 21.
Teacher's Note:
1. The class containing the mode is the modal class.
Free study material for Mathematics
Download Practice Assignments: Class 10 Mathematics Chapter 13 Statistics
Chapter Practice Questions for Class 10 Mathematics
Review targeted chapter assignments for Class 10 Mathematics Chapter 13 Statistics. Built according to official CBSE guidelines, these downloadable problem sets help students build accuracy and prepare effectively for school tests.
Maximize Exam Scores with Chapter Practice Sets
- Exam Alignment: Questions strictly follow contemporary CBSE sample papers and grading schemes.
- Comprehensive Coverage: Features objective drills, case studies, and structured descriptive problems for Chapter 13 Statistics.
- Pacing & Precision: Regular problem-solving builds critical calculation speed and test-taking accuracy.
Effective Strategy for Class 10 Mathematics Assignments
- Textbook Review: Always study the core NCERT book for Class 10 Mathematics prior to beginning the assignment.
- Independent Attempt: Solve Chapter 13 Statistics questions on your own initially before cross-checking with expert solutions.
- Error Tracking: Record challenging concepts in a dedicated notebook and practice online MCQ tests for revision.
FAQs
You can download free PDF assignments for Class 10 Mathematics Chapter 13 Statistics from StudiesToday.com. These practice sheets have been updated for the 2026-27 session covering all concepts from latest NCERT textbook.
Yes, our teachers have given solutions for all questions in the Class 10 Mathematics Chapter 13 Statistics assignments. This will help you to understand step-by-step methodology to get full marks in school tests and exams.
Yes. These assignments are designed as per the latest CBSE syllabus for 2026. We have included huge variety of question formats such as MCQs, Case-study based questions and important diagram-based problems found in Chapter 13 Statistics.
Practicing topicw wise assignments will help Class 10 students understand every sub-topic of Chapter 13 Statistics. Daily practice will improve speed, accuracy and answering competency-based questions.
Yes, all printable assignments for Class 10 Mathematics Chapter 13 Statistics are available for free download in mobile-friendly PDF format.