CBSE Class 10 Mathematics Statistics MCQs Set 14

Multiple Choice Questions (MCQs) for Class 10 Mathematics: Chapter 13 Statistics

Review structured MCQ sets for Class 10 Mathematics Chapter 13 Statistics. Built according to official CBSE guidelines, these downloadable questions support daily revision and core concept reinforcement.

Practice Chapter 13 Statistics MCQs for Class 10 Mathematics

Access the complete set of multiple-choice questions for Chapter 13 Statistics below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official CBSE textbooks.

Multiple Choice Questions

Question 1: The class marks of a distribution are 6, 10, 14, 18, 22, 26, 30. The class size and class interval are
(a) class size 6 and class interval 4
(b) class size 4 and class interval 8
(c) class size 2 and class interval 4
(d) class size 4 and class interval 4
Show Answer & Explanation

Answer: (d) class size 4 and class interval 4

Explanation:
1. The class marks differ by 4, so the class width (class size) is 4 and each class interval is 4 units wide, e.g. 4–8, 8–12, ...

Teacher's Note:
1. Prepare the table with class marks and frequencies, apply the standard formula and recheck the arithmetic.
2. Mean uses sum of f times x divided by sum of f; mode and median use the modal class and median class with the class width.

Question 2: The range of the data 70, 65, 71, 36, 55, 61, 62, 41, 40, 39, 35 is
(a) 71
(b) 35
(c) 36
(d) 31
Show Answer & Explanation

Answer: (c) 36

Explanation:
1. Range = largest value − smallest value = 71 − 35 = 36.

Teacher's Note:
1. Prepare the table with class marks and frequencies, apply the standard formula and recheck the arithmetic.
2. Mean uses sum of f times x divided by sum of f; mode and median use the modal class and median class with the class width.

Question 3: The mean of the first 10 even natural numbers is
(a) 10
(b) 11
(c) 12
(d) 10.5
Show Answer & Explanation

Answer: (b) 11

Explanation:
1. The numbers are 2, 4, ..., 20 with sum 110. Mean = 110/10 = 11.

Teacher's Note:
1. Prepare the table with class marks and frequencies, apply the standard formula and recheck the arithmetic.
2. Mean uses sum of f times x divided by sum of f; mode and median use the modal class and median class with the class width.

Question 4: The weights (in kg) of 10 new born babies are 3.4, 3.6, 4.2, 4.5, 3.9, 4.1, 3.8, 4.5, 4.4, 3.6. The mean weight is
(a) 3.9 kg
(b) 4 kg
(c) 4.2 kg
(d) 40 kg
Show Answer & Explanation

Answer: (b) 4 kg

Explanation:
1. The sum of the weights = 40.0 kg and there are 10 babies, so the mean = 40/10 = 4 kg.

Teacher's Note:
1. Prepare the table with class marks and frequencies, apply the standard formula and recheck the arithmetic.
2. Mean uses sum of f times x divided by sum of f; mode and median use the modal class and median class with the class width.

Question 5: Find the mean of the following distribution.
x1012202535
f3101575

(a) 22
(b) 18
(c) 20
(d) 22.5
Show Answer & Explanation

Answer: (c) 20

Explanation:
1. Σf = 40 and Σfx = 800. Mean = 800/40 = 20.

Teacher's Note:
1. Prepare the table with class marks and frequencies, apply the standard formula and recheck the arithmetic.
2. Mean uses sum of f times x divided by sum of f; mode and median use the modal class and median class with the class width.

Question 6: The marks of 15 students are 4, 6, 5, 7, 9, 8, 10, 4, 7, 6, 5, 9, 8, 7, 7. The mode is
(a) 5
(b) 8
(c) 6
(d) 7
Show Answer & Explanation

Answer: (d) 7

Explanation:
1. Counting frequencies: 4 appears 2 times, 5 twice, 6 twice, 7 four times, 8 twice, 9 twice, 10 once. The most frequent value is 7.

Teacher's Note:
1. Prepare the table with class marks and frequencies, apply the standard formula and recheck the arithmetic.
2. Mean uses sum of f times x divided by sum of f; mode and median use the modal class and median class with the class width.

Question 7: The weights (in kg) of 7 students: the mean weight is 55 kg and six of them weigh 52, 54, 55, 53, 56 and 54 kg. The weight of the seventh student is
(a) 55 kg
(b) 58 kg
(c) 61 kg
(d) 64 kg
Show Answer & Explanation

Answer: (c) 61 kg

Explanation:
1. Total weight = 7 × 55 = 385 kg. The six known weights add to 324 kg, so the seventh = 385 − 324 = 61 kg.

Teacher's Note:
1. Prepare the table with class marks and frequencies, apply the standard formula and recheck the arithmetic.
2. Mean uses sum of f times x divided by sum of f; mode and median use the modal class and median class with the class width.

Question 8: The weights (in kg) of 15 students are 31, 35, 27, 29, 32, 43, 37, 41, 34, 28, 36, 44, 45, 42, 30. The median weight is
(a) 35 kg
(b) 36 kg
(c) 34 kg
(d) 37 kg
Show Answer & Explanation

Answer: (a) 35 kg

Explanation:
1. Arranged in order: 27, 28, 29, 30, 31, 32, 34, 35, 36, 37, 41, 42, 43, 44, 45. There are 15 values, so the median is the 8th value = 35 kg.

Teacher's Note:
1. Prepare the table with class marks and frequencies, apply the standard formula and recheck the arithmetic.
2. Mean uses sum of f times x divided by sum of f; mode and median use the modal class and median class with the class width.

Question 9: The mode of the data 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18 is
(a) 18
(b) 14
(c) 25
(d) 28
Show Answer & Explanation

Answer: (b) 14

Explanation:
1. 14 occurs 4 times and 18 occurs 3 times. The mode is 14.

Teacher's Note:
1. Prepare the table with class marks and frequencies, apply the standard formula and recheck the arithmetic.
2. Mean uses sum of f times x divided by sum of f; mode and median use the modal class and median class with the class width.

Question 10: The mode of the data 7, 9, 12, 7, 12, 13, 15, 7, 12, 7, 25, 18, 7 is
(a) 12
(b) 9
(c) 13
(d) 7
Show Answer & Explanation

Answer: (d) 7

Explanation:
1. 7 occurs 5 times, 12 occurs 3 times, and the rest occur once. The mode is 7.

Teacher's Note:
1. Prepare the table with class marks and frequencies, apply the standard formula and recheck the arithmetic.
2. Mean uses sum of f times x divided by sum of f; mode and median use the modal class and median class with the class width.

Question 11: Find the mean of the following frequency distribution.
Class interval0–66–1212–1818–2424–30
Frequency7510126

(a) 17.75
(b) 15.75
(c) 12.75
(d) 20.75
Show Answer & Explanation

Answer: (b) 15.75

Explanation:
1. Class marks: 3, 9, 15, 21, 27. Σf = 40 and Σfx = 630. Mean = Σfx/Σf = 630/40 = 15.75.

Teacher's Note:
1. Prepare the table with class marks and frequencies, apply the standard formula and recheck the arithmetic.
2. Mean uses sum of f times x divided by sum of f; mode and median use the modal class and median class with the class width.

Question 12: Find the mean of the following frequency distribution.
Class interval0–1010–2020–3030–4040–50
Frequency81091211

(a) 28.6
(b) 26.6
(c) 23.6
(d) 31.6
Show Answer & Explanation

Answer: (b) 26.6

Explanation:
1. Class marks: 5, 15, 25, 35, 45. Σf = 50 and Σfx = 1330. Mean = Σfx/Σf = 1330/50 = 26.6.

Teacher's Note:
1. Prepare the table with class marks and frequencies, apply the standard formula and recheck the arithmetic.
2. Mean uses sum of f times x divided by sum of f; mode and median use the modal class and median class with the class width.

Question 13: Find the mean of the following frequency distribution.
Class interval0–1010–2020–3030–4040–50
Frequency912151014

(a) ≈ 26.33
(b) ≈ 28.33
(c) ≈ 23.33
(d) ≈ 31.33
Show Answer & Explanation

Answer: (a) ≈ 26.33

Explanation:
1. Class marks: 5, 15, 25, 35, 45. Σf = 60 and Σfx = 1580. Mean = Σfx/Σf = 1580/60 = ≈ 26.33.

Teacher's Note:
1. Prepare the table with class marks and frequencies, apply the standard formula and recheck the arithmetic.
2. Mean uses sum of f times x divided by sum of f; mode and median use the modal class and median class with the class width.

Question 14: Find the mean of the following frequency distribution.
Class interval0–1010–2020–3030–4040–50
Frequency12118109

(a) 23.6
(b) 25.6
(c) 20.6
(d) 28.6
Show Answer & Explanation

Answer: (a) 23.6

Explanation:
1. Class marks: 5, 15, 25, 35, 45. Σf = 50 and Σfx = 1180. Mean = Σfx/Σf = 1180/50 = 23.6.

Teacher's Note:
1. Prepare the table with class marks and frequencies, apply the standard formula and recheck the arithmetic.
2. Mean uses sum of f times x divided by sum of f; mode and median use the modal class and median class with the class width.

Question 15: Find the mean of the following frequency distribution.
Class interval0–5050–100100–150150–200200–250250–300
Frequency172442453614

(a) ≈ 153.37
(b) ≈ 155.37
(c) ≈ 150.37
(d) ≈ 158.37
Show Answer & Explanation

Answer: (a) ≈ 153.37

Explanation:
1. Class marks: 25, 75, 125, 175, 225, 275. Σf = 178 and Σfx = 27300. Mean = Σfx/Σf = 27300/178 = ≈ 153.37.

Teacher's Note:
1. Prepare the table with class marks and frequencies, apply the standard formula and recheck the arithmetic.
2. Mean uses sum of f times x divided by sum of f; mode and median use the modal class and median class with the class width.

Question 16: Find the mean of the following frequency distribution.
Class interval0–88–1616–2424–3232–40
Frequency8101598

(a) 21.84
(b) 16.84
(c) 24.84
(d) 19.84
Show Answer & Explanation

Answer: (d) 19.84

Explanation:
1. Class marks: 4, 12, 20, 28, 36. Σf = 50 and Σfx = 992. Mean = Σfx/Σf = 992/50 = 19.84.

Teacher's Note:
1. Prepare the table with class marks and frequencies, apply the standard formula and recheck the arithmetic.
2. Mean uses sum of f times x divided by sum of f; mode and median use the modal class and median class with the class width.

Question 17: Find the mode of the following data showing the frequency with which profits (in thousand rupees) are made.
Class interval3–44–55–66–77–88–99–10
Frequency83272550753818

(a) ≈ 4.10
(b) 4
(c) ≈ 3.60
(d) ≈ 3.26
Show Answer & Explanation

Answer: (c) ≈ 3.60

Explanation:
1. The modal class is 3–4 (highest frequency 83). Here l = 3, h = 1, f1 = 83, f0 = 0, f2 = 27. Mode = l + (f1 − f0)/(2f1 − f0 − f2) × h = 3 + 83/139 × 1 = ≈ 3.60.

Teacher's Note:
1. Prepare the table with class marks and frequencies, apply the standard formula and recheck the arithmetic.
2. Mean uses sum of f times x divided by sum of f; mode and median use the modal class and median class with the class width.

Question 18: Find the mode of the following frequency distribution.
Class interval0–55–1010–1515–2020–2525–3030–3535–4040–45
Frequency202432282016171018

(a) ≈ 15.83
(b) ≈ 13.33
(c) 15
(d) ≈ 11.67
Show Answer & Explanation

Answer: (b) ≈ 13.33

Explanation:
1. The modal class is 10–15 (highest frequency 32). Here l = 10, h = 5, f1 = 32, f0 = 24, f2 = 28. Mode = l + (f1 − f0)/(2f1 − f0 − f2) × h = 10 + 8/12 × 5 = ≈ 13.33.

Teacher's Note:
1. Prepare the table with class marks and frequencies, apply the standard formula and recheck the arithmetic.
2. Mean uses sum of f times x divided by sum of f; mode and median use the modal class and median class with the class width.

Question 19: Find the mode of the following frequency distribution.
Class interval0–1010–2020–3030–4040–5050–6060–7070–80
Frequency5871228201010

(a) ≈ 51.67
(b) 50
(c) ≈ 43.33
(d) ≈ 46.67
Show Answer & Explanation

Answer: (d) ≈ 46.67

Explanation:
1. The modal class is 40–50 (highest frequency 28). Here l = 40, h = 10, f1 = 28, f0 = 12, f2 = 20. Mode = l + (f1 − f0)/(2f1 − f0 − f2) × h = 40 + 16/24 × 10 = ≈ 46.67.

Teacher's Note:
1. Prepare the table with class marks and frequencies, apply the standard formula and recheck the arithmetic.
2. Mean uses sum of f times x divided by sum of f; mode and median use the modal class and median class with the class width.

Question 20: Calculate the mode of the following data for students of class X.
Marks20–2930–3940–4950–5960–6970–7980–8990–99
Number of students5121520181064

(a) ≈ 56.64
(b) ≈ 58.64
(c) 55
(d) ≈ 53.64
Show Answer & Explanation

Answer: (a) ≈ 56.64

Explanation:
1. The classes are inclusive, so the true class boundaries are 19.5–29.5, 29.5–39.5, ... The modal class is 49.5–59.5 (frequency 20), with f0 = 15 and f2 = 18 and h = 10. Mode = 49.5 + (20 − 15)/(40 − 15 − 18) × 10 = 49.5 + 50/7 = ≈ 56.64.

Teacher's Note:
1. Prepare the table with class marks and frequencies, apply the standard formula and recheck the arithmetic.
2. Mean uses sum of f times x divided by sum of f; mode and median use the modal class and median class with the class width.

Question 21: Compute the median of the marks obtained by 250 students.
Class interval0–1010–2020–3030–4040–5050–6060–7070–80
Frequency1520252412317152

(a) ≈ 64.35
(b) ≈ 59.35
(c) 50
(d) ≈ 56.85
Show Answer & Explanation

Answer: (b) ≈ 59.35

Explanation:
1. n = 250, so n/2 = 125. The median class is 50–60 (cumulative frequency before it = 96, f = 31). Median = l + (n/2 − cf)/f × h = 50 + (125 − 96)/31 × 10 = ≈ 59.35.

Teacher's Note:
1. Prepare the table with class marks and frequencies, apply the standard formula and recheck the arithmetic.
2. Mean uses sum of f times x divided by sum of f; mode and median use the modal class and median class with the class width.

Question 22: Determine the median age from the following information.
Class interval20–2525–3030–3535–4040–4545–5050–5555–60
Frequency50701001801501207060

(a) 42.5
(b) 35
(c) 38.75
(d) 40
Show Answer & Explanation

Answer: (d) 40

Explanation:
1. n = 800, so n/2 = 400. The median class is 35–40 (cumulative frequency before it = 220, f = 180). Median = l + (n/2 − cf)/f × h = 35 + (400 − 220)/180 × 5 = 40.

Teacher's Note:
1. Prepare the table with class marks and frequencies, apply the standard formula and recheck the arithmetic.
2. Mean uses sum of f times x divided by sum of f; mode and median use the modal class and median class with the class width.

Question 23: The frequency distribution of the ages (in years) of a group of 199 teachers is given below (obtained from the cumulative frequencies 21, 40, 90, 130, 146, 166, 176, 186, 195, 199). Find the median age.
Class interval20–2525–3030–3535–4040–4545–5050–5555–6060–6565–70
Frequency211950401620101094

(a) ≈ 38.69
(b) 35
(c) ≈ 36.19
(d) ≈ 34.94
Show Answer & Explanation

Answer: (c) ≈ 36.19

Explanation:
1. n = 199, so n/2 = 99.5. The median class is 35–40 (cumulative frequency before it = 90, f = 40). Median = l + (n/2 − cf)/f × h = 35 + (99.5 − 90)/40 × 5 = ≈ 36.19.

Teacher's Note:
1. Prepare the table with class marks and frequencies, apply the standard formula and recheck the arithmetic.
2. Mean uses sum of f times x divided by sum of f; mode and median use the modal class and median class with the class width.

Question 24: Calculate the median daily wages (in Rs.) from the following distribution.
Class interval2–44–66–88–1010–1212–1414–1616–1818–2020–2222–2424–26
Frequency31343102175220204139692561

(a) ≈ 14.49
(b) 12
(c) ≈ 13.49
(d) ≈ 12.99
Show Answer & Explanation

Answer: (c) ≈ 13.49

Explanation:
1. n = 1000, so n/2 = 500. The median class is 12–14 (cumulative frequency before it = 336, f = 220). Median = l + (n/2 − cf)/f × h = 12 + (500 − 336)/220 × 2 = ≈ 13.49.

Teacher's Note:
1. Prepare the table with class marks and frequencies, apply the standard formula and recheck the arithmetic.
2. Mean uses sum of f times x divided by sum of f; mode and median use the modal class and median class with the class width.

Question 25: Find the median of the following weekly wage distribution (in Rs.).
Class interval50–5555–6060–6565–7070–7575–8080–100
Frequency6102230161215

(a) ≈ 67.92
(b) ≈ 70.42
(c) 65
(d) ≈ 66.67
Show Answer & Explanation

Answer: (a) ≈ 67.92

Explanation:
1. n = 111, so n/2 = 55.5. The median class is 65–70 (cumulative frequency before it = 38, f = 30). Median = l + (n/2 − cf)/f × h = 65 + (55.5 − 38)/30 × 5 = ≈ 67.92.

Teacher's Note:
1. Prepare the table with class marks and frequencies, apply the standard formula and recheck the arithmetic.
2. Mean uses sum of f times x divided by sum of f; mode and median use the modal class and median class with the class width.

Free study material for Mathematics

Multiple Choice Questions (MCQs) for Class 10 Mathematics Chapter 13 Statistics

Chapter MCQs with Answers for Class 10 Mathematics

Review structured objective questions for Class 10 Mathematics Chapter 13 Statistics. Built according to official CBSE guidelines, these MCQ sets support daily revision and core concept reinforcement.

Expert Practice Material for Class 10 Mathematics

Cross-reference your completed choices with comprehensive NCERT solutions for Class 10 Mathematics to ensure absolute clarity across all sub-topics in this chapter.

Complete Your Chapter Revision

Wrap up your chapter revision by testing your knowledge against standard question formats. Everything on our platform is provided free of charge.

FAQs

Where can I access latest CBSE Class 10 Mathematics Statistics MCQs Set 14?

You can get most exhaustive CBSE Class 10 Mathematics Statistics MCQs Set 14 for free on StudiesToday.com. These MCQs for Class 10 Mathematics are updated for the 2026-27 academic session as per CBSE examination standards.

Are Assertion-Reasoning and Case-Study MCQs included in the Mathematics Class 10 material?

Yes, our CBSE Class 10 Mathematics Statistics MCQs Set 14 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the CBSE paper is now competency-based.

How do practicing Mathematics MCQs help in scoring full marks in Class 10 exams?

By solving our CBSE Class 10 Mathematics Statistics MCQs Set 14, Class 10 students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Mathematics.

Do you provide answers and explanations for CBSE Class 10 Mathematics Statistics MCQs Set 14?

Yes, Mathematics MCQs for Class 10 have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused CBSE exams.

Can I practice these Mathematics Class 10 MCQs online?

Yes, you can also access online interactive tests for CBSE Class 10 Mathematics Statistics MCQs Set 14 on StudiesToday.com as they provide instant answers and score to help you track your progress in Mathematics.