Important Questions for Class 10 Maths Chapter 13: Statistics

Statistics is the branch of mathematics that deals with collecting, organising, and interpreting numerical data. Class 10 Maths Chapter 13 extends the study of mean, median, and mode from ungrouped to grouped data, introduces cumulative frequency, and covers less than and more than type distributions and ogives.

Important questions class 10 maths chapter 13 on this page are structured in the exact NCERT order. You get very short answer questions, short answer questions, long answer questions, extra questions, formulas, and notes in one place.

Most marks in this chapter are lost on median questions (wrong median class) and mode questions (wrong modal class identification). A student who can build a cumulative frequency table in under two minutes and apply the median formula without errors will handle the highest-weightage questions confidently. All questions and solutions are available section by section below. 

Key Takeaways

Topic What to Know
Class Mark (Upper Limit + Lower Limit) / 2
Mean: Direct Method x = Sfixi / Sfi
Mean: Assumed Mean x = a + (Sfidi / Sfi), where di = xi - a
Mean: Step Deviation x = a + h(Sfiui / Sfi), where ui = (xi - a)/h
Mode Formula l + [(f1 - f0)/(2f1 - f0 - f2)] x h
Median Formula l + [(n/2 - cf)/f] x h
Empirical Relation 3 Median = Mode + 2 Mean
Ogive Cumulative frequency curve (less than or more than type)

Introduction to Statistics Class 10

Statistics at Class 10 level deals with grouped frequency distributions. When raw data is large, it is organised into class intervals. The midpoint of each class interval is called the class mark, and it represents the entire class in mean calculations.

Class mark = (Upper Limit + Lower Limit) / 2. Class size = Upper Limit - Lower Limit.

Three key concepts carry maximum weightage in CBSE 2026 board exams: mean (especially step-deviation method), median (from cumulative frequency), and mode (using the modal class formula). The empirical relationship 3 Median = Mode + 2 Mean connects all three and appears in short-answer questions.

Important Questions Class 10 Maths All Chapters

Chapter No Chapter Name
Chapter 1 Real Numbers
Chapter 2 Polynomials
Chapter 3 Pair of Linear Equations in Two Variables
Chapter 4 Quadratic Equations
Chapter 5 Arithmetic Progressions
Chapter 6 Triangles
Chapter 7 Coordinate Geometry
Chapter 8 Introduction to Trigonometry
Chapter 9 Some Applications of Trigonometry
Chapter 10 Circles
Chapter 11 Areas Related to Circles
Chapter 12 Surface Areas and Volumes
Chapter 13 Statistics
Chapter 14 Probability

Very Short Answer Questions on Statistics

One-mark questions test classification, formula identification, and quick calculation. These cbse class 10 statistics important questions test the definitions and formula components that examiners most frequently check.

  1. In the formula x = a + h(Sfiui / Sfi), uáµ¢ represents: (a) Lower limits of the classes (b) Upper limits (c) Mid-points of the classes (d) Frequencies (c) Mid-points of the classes. ui = (xi - a)/h, where xi is the class mark.
  2. Construction of a cumulative frequency table is useful in determining: (a) Mean (b) Median (c) Mode (d) All of these (b) Median. Cumulative frequency is used to locate the median class.
  3. While computing mean of grouped data, frequencies are assumed to be: (b) Centred at the class marks of the class.
  4. If mode = 80 and mean = 68, find the median using the empirical formula. 3 Median = Mode + 2 Mean = 80 + 136 = 216. Median = 72.
  5. If mode = 65 and mean = 66, find median. 3 Median = 65 + 132 = 197. Median = 65.67.
  6. For what value of k is the mode of the data 7: 2, 4, 7, k, 7, 4, 2, 7? k = 7. Mode is the most frequent value. 7 already appears 3 times; k = 7 confirms it.
  7. If Sfi = 20 and Sfixi = 300, find the mean. Mean = 300/20 = 15.
  8. In the median formula, what does cf represent? cf is the cumulative frequency of the class preceding the median class.

Important formulas for Class 10 Maths Chapter 13 Statistics including mean, median, mode, range, cumulative frequency, histogram, and grouped data formulas.

Short Answer Questions on Statistics

Two-mark and three-mark questions test direct application of formulas from the NCERT textbook. These class 10 statistics practice questions cover the most tested short-answer formats in CBSE 2026 papers.

  1. Find the mode of the following data.
Class 10-20 20-30 30-40 40-50 50-60
Frequency 5 8 15 10 6

Modal class = 30-40 (highest frequency = 15). l = 30, f1 = 15, f0 = 8, f2 = 10, h = 10. Mode = 30 + [(15 - 8)/(30 - 8 - 10)] x 10 = 30 + [7/12] x 10 = 35.83.

  1. The mean and mode of a distribution are 26 and 32 respectively. Find the median. 3 Median = 32 + 52 = 84. Median = 28.
  2. The values of mean and mode are 54 and 63 respectively. Find the median. 3 Median = 63 + 108 = 171. Median = 57.
  3. Construct the cumulative frequency distribution (less than type).
Class 0-10 10-20 20-30 30-40 40-50
Frequency 5 8 12 7 3

 

Marks Cumulative Frequency
Less than 10 5
Less than 20 13
Less than 30 25
Less than 40 32
Less than 50 35
  1. Find the median from the data above. n = 35, n/2 = 17.5. Cumulative frequencies: 5, 13, 25, 32, 35. Median class = 30-40. l = 30, cf = 13, f = 12, h = 10. Median = 30 + [(17.5 - 13)/12] x 10 = 33.75.
  2. If the mean of the following data is 18, find the value of p.
xi 10 15 p 25 30
fi 3 5 8 3 1

Sfi = 20. Sfixi = 30 + 75 + 8p + 75 + 30 = 210 + 8p. Mean = (210 + 8p)/20 = 18 → 210 + 8p = 360 → 8p = 150 → p = 18.75.

  1. The following frequency distribution has mean 28.5. Find the value of p.
Class 0-10 10-20 20-30 30-40 40-50 50-60 Total
Frequency 5 p 20 15 y 5 60

5 + p + 20 + 15 + y + 5 = 60 → p + y = 15. Median class = 20-30. l = 20, f = 20, cf = 5 + p, h = 10. 28.5 = 20 + [(30 - 5 - p)/20] x 10 → 8.5 = (25 - p)/2 → p = 8, y = 7.

Long Answer Questions on Statistics

Four-mark questions combine mean, median, and mode for the same data, or involve missing frequency and ogive construction. These class 10 statistics hard questions are the highest-weightage items in CBSE 2026 board papers.

  1. The following distribution gives the monthly electricity consumption of 68 consumers. Find the median, mean, and mode.
Monthly consumption (units) Frequency
65-85 4
85-105 5
105-125 13
125-145 20
145-165 14
165-185 8
185-205 4

n = 68, n/2 = 34. Cumulative frequencies: 4, 9, 22, 42, 56, 64, 68. Median class = 125-145. l = 125, cf = 22, f = 20, h = 20. Median = 125 + [(34 - 22)/20] x 20 = 125 + 12 = 137.

Modal class = 125-145. f1 = 20, f0 = 13, f2 = 14, l = 125, h = 20. Mode = 125 + [(20 - 13)/(40 - 13 - 14)] x 20 = 125 + [7/13] x 20 = 135.76.

Mean: a = 135, h = 20. u values: -3, -2, -1, 0, 1, 2, 3. Sfiui = -12 - 10 - 13 + 0 + 14 + 16 + 12 = 7. Mean = 135 + (7/68) x 20 = 137.06.

  1. The median of the following data is 525. Find x and y. Total frequency = 100.
Class 0-100 100-200 200-300 300-400 400-500 500-600 600-700 700-800 800-900 900-1000
Frequency 2 5 x 12 17 20 y 9 7 4

x + y = 24 ... (1). Median class = 500-600. l = 500, f = 20, cf = 36 + x, h = 100. 525 = 500 + [(50 - 36 - x)/20] x 100 → 25 = 5(14 - x) → x = 9, y = 15.

  1. A doctor examined 30 ladies. Their heartbeats per minute were recorded. Find the mean using the most appropriate method.
Heartbeats/min 65-68 68-71 71-74 74-77 77-80 80-83 83-86
Frequency 2 4 3 8 7 4 2

a = 75.5, h = 3. ui: -3, -2, -1, 0, 1, 2, 3. fiui: -6, -8, -3, 0, 7, 8, 6. Sfiui = 4, Sfi = 30. Mean = 75.5 + (4/30) x 3 = 75.9 beats/minute.

  1. The daily pocket allowance mean is Rs 18. Find the missing frequency f.
Daily allowance (Rs) 11-13 13-15 15-17 17-19 19-21 21-23 23-25
Frequency 7 6 9 13 f 5 4

a = 18, h = 2. Sfi = 44 + f. Sfiui = f - 20. 18 + 2(f - 20)/(44 + f) = 18 → f - 20 = 0 → f = 20.

Important Questions on Mean of Grouped Data

Mean of grouped data uses the class mark as the representative value for each class interval. All three methods give the same answer. The class 10 statistics most important questions on mean test which method is most appropriate based on the data size.

  1. Find the mean using the direct method.
Class 0-10 10-20 20-30 30-40 40-50
Frequency 5 10 15 8 2

Class marks: 5, 15, 25, 35, 45. Sfi = 40. Sfixi = 25 + 150 + 375 + 280 + 90 = 920. Mean = 920/40 = 23.

  1. Find the mean for the number of plants in 20 houses.
Plants 0-2 2-4 4-6 6-8 8-10 10-12 12-14
Houses 1 2 1 5 6 2 3

Sfi = 20. Sfixi = 1 + 6 + 5 + 35 + 54 + 22 + 39 = 162. Mean = 162/20 = 8.1 plants per house.

  1. Find the mean using the step-deviation method.
Daily wages (Rs) 500-520 520-540 540-560 560-580 580-600
Workers 12 14 8 6 10

a = 550, h = 20. ui: -2, -1, 0, 1, 2. Sfi = 50. Sfiui = -24 - 14 + 0 + 6 + 20 = -12. Mean = 550 + (-12/50) x 20 = Rs 545.20.

Important Questions on Mode of Grouped Data

Mode of grouped data requires identifying the modal class first: the class with the highest frequency. The mode formula then pinpoints the exact value inside that class. These cbse class 10 statistics important questions on mode are consistently tested in board exams.

  1. Find the mode of ages of patients admitted in a hospital.
Age (years) 5-15 15-25 25-35 35-45 45-55 55-65
Patients 6 11 21 23 14 5

Modal class = 35-45. l = 35, f1 = 23, f0 = 21, f2 = 14, h = 10. Mode = 35 + [(23 - 21)/(46 - 21 - 14)] x 10 = 35 + [2/11] x 10 = 36.82 years.

  1. Find the mode of runs scored by top batsmen.
Runs 3000-4000 4000-5000 5000-6000 6000-7000 7000-8000 8000-9000 9000-10000 10000-11000
Batsmen 4 18 9 7 6 3 1 1

Modal class = 4000-5000. l = 4000, f1 = 18, f0 = 4, f2 = 9, h = 1000. Mode = 4000 + [(18 - 4)/(36 - 4 - 9)] x 1000 = 4000 + [14/23] x 1000 = 4608.7 runs.

  1. Find the mode for cars passing a road.
Cars 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80
Frequency 7 14 13 12 20 11 15 8

Modal class = 40-50. l = 40, f1 = 20, f0 = 12, f2 = 11, h = 10. Mode = 40 + [(20 - 12)/(40 - 12 - 11)] x 10 = 40 + [8/17] x 10 = 44.7 cars.

Important Questions on Median of Grouped Data

Median of grouped data requires building a cumulative frequency table, finding n/2, and identifying the median class. These class 10 statistics extra questions with answers are among the most scoring items for CBSE 2026 board exams.

  1. Find the median height of 51 girls of Class X.
Height (cm) Less than 140 Less than 145 Less than 150 Less than 155 Less than 160 Less than 165
Girls 4 11 29 40 46 51

Class intervals: Below 140 (4), 140-145 (7), 145-150 (18), 150-155 (11), 155-160 (6), 160-165 (5). n = 51, n/2 = 25.5. Cumulative frequencies: 4, 11, 29, 40, 46, 51. Median class = 145-150. l = 145, cf = 11, f = 18, h = 5. Median = 145 + [(25.5 - 11)/18] x 5 = 149.03 cm.

  1. The median of the distribution is 28.5. Find x and y. Total = 60.
Class 0-10 10-20 20-30 30-40 40-50 50-60
Frequency 5 x 20 15 y 5

x + y = 15. Median class = 20-30. l = 20, f = 20, cf = 5 + x, h = 10. 28.5 = 20 + [(30 - 5 - x)/20] x 10 → x = 8, y = 7.

  1. Find the median weight of 30 students.
Weight (kg) 40-45 45-50 50-55 55-60 60-65 65-70 70-75
Students 2 3 8 6 6 3 2

n = 30, n/2 = 15. Cumulative frequencies: 2, 5, 13, 19, 25, 28, 30. Median class = 55-60. l = 55, cf = 13, f = 6, h = 5. Median = 55 + [(15 - 13)/6] x 5 = 56.67 kg.

Important Questions on Cumulative Frequency Distribution

Cumulative frequency is the running total of frequencies from the first class up to a given class. The less than type uses upper limits; the more than type uses lower limits. These are essential for drawing ogives and for answering more than type questions class 10 statistics problems.

  1. Construct the less than type cumulative frequency distribution.
Class 0-20 20-40 40-60 60-80 80-100
Frequency 10 15 25 18 12

 

Marks Cumulative Frequency (Less than)
Less than 20 10
Less than 40 25
Less than 60 50
Less than 80 68
Less than 100 80
  1. Construct the more than type distribution from the same data.
Marks Cumulative Frequency (More than)
More than or equal to 0 80
More than or equal to 20 70
More than or equal to 40 55
More than or equal to 60 30
More than or equal to 80 12
  1. For what is a cumulative frequency table useful? Cumulative frequency table is used to find the median of grouped data and to draw ogives.

More Than Type Questions Class 10 Statistics: Less Than and More Than Distributions

Less than type distribution uses upper class limits; more than type uses lower class limits. These form the two types of ogives, and the x-coordinate of their intersection gives the median.

  1. The following data gives marks obtained by 53 students. Construct both distributions.
Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100
Students 5 3 4 3 3 4 7 9 7 8

Less than type:

Marks Cumulative Frequency
Less than 10 5
Less than 20 8
Less than 30 12
Less than 40 15
Less than 50 18
Less than 60 22
Less than 70 29
Less than 80 38
Less than 90 45
Less than 100 53

More than type:

Marks Cumulative Frequency
More than or equal to 0 53
More than or equal to 10 48
More than or equal to 20 45
More than or equal to 30 41
More than or equal to 40 38
More than or equal to 50 35
More than or equal to 60 31
More than or equal to 70 24
More than or equal to 80 15
More than or equal to 90 8
  1. The median of the above distribution found using the formula is 66.4. How can you verify this using ogives? Draw both ogives on the same graph. The x-coordinate of the point where the two ogives intersect gives the median. This should come out to approximately 66.4.

Class 10 Statistics Extra Questions with Answers

These maths class 10 statistics extra questions go beyond the standard NCERT exercises. They appear in pre-board and annual examinations for CBSE 2026.

  1. Find the mean age from the following frequency distribution.
Age (years) 5-15 15-25 25-35 35-45 45-55 55-65
Frequency 6 11 21 23 14 5

a = 40, h = 10. ui: -3, -2, -1, 0, 1, 2. Sfi = 80. Sfiui = -18 - 22 - 21 + 0 + 14 + 10 = -37. Mean = 40 + (-37/80) x 10 = 35.38 years.

  1. Find mean, median, and mode for the following data.
Class 10-20 20-30 30-40 40-50 50-60
Frequency 6 12 20 10 2

n = 50. a = 35, h = 10. Sfiui = -12 - 12 + 0 + 10 + 4 = -10. Mean = 35 + (-10/50) x 10 = 33.

Median: n/2 = 25. Cumulative frequencies: 6, 18, 38, 48, 50. Median class = 30-40. l = 30, cf = 18, f = 20, h = 10. Median = 30 + [(25 - 18)/20] x 10 = 33.5.

Modal class = 30-40. f1 = 20, f0 = 12, f2 = 10, l = 30, h = 10. Mode = 30 + [(20 - 12)/(40 - 12 - 10)] x 10 = 34.44.

  1. Calculate the median height of 51 students.
Height (cm) 140-145 145-150 150-155 155-160 160-165
Frequency 5 15 17 8 6

n = 51, n/2 = 25.5. Cumulative frequencies: 5, 20, 37, 45, 51. Median class = 150-155. l = 150, cf = 20, f = 17, h = 5. Median = 150 + [(25.5 - 20)/17] x 5 = 151.62 cm.

Statistics Class 10 Formulas: Mean Median Mode

Students searching for class 10 statistics formulas mean median mode and class 10 statistics mean median mode formula before board exams need these exact identities from the NCERT 2026 chapter.

Mean of Grouped Data:

Method Formula
Direct Method x = Sfixi / Sfi
Assumed Mean Method x = a + (Sfidi / Sfi), di = xi - a
Step-Deviation Method x = a + h x (Sfiui / Sfi), ui = (xi - a)/h

Mode of Grouped Data: Mode = l + [(f1 - f0) / (2f1 - f0 - f2)] x h where l = lower limit of modal class, h = class size, f1 = frequency of modal class, f0 = frequency of preceding class, f2 = frequency of succeeding class.

Median of Grouped Data: Median = l + [(n/2 - cf) / f] x h where l = lower limit of median class, n = total observations, cf = cumulative frequency of preceding class, f = frequency of median class, h = class size.

Empirical Relationship: 3 Median = Mode + 2 Mean. This connects all three central tendencies and is used to find one when the other two are given.

Q.1 Water in a canal, 30 dm. wide and 12 dm. deep is flowing with a velocity of 10 km/hr. How much area will it irrigate in 30 minutes, if 8 cm of standing water is required for irrigation?

Marks:3
Ans

The width of the canal = 30 dm
= 3 m [10 dm = 1m]
Depth of the canal = 12 dm
= 1.2 m
Length of flowing water 1/2 hour = 5 km
= 5000 m
Volume of the water flowing in 1/2 hour = 5000 — 3 — 1.2
Let x m2 area is irrigated in 1/2 hr. Then,
x — (8/100) = 5000 — 3 — 1.2
x = 1800000/8
= 225000 m2
Hence, the canal irrigates 225000 m2 area in 1/2 hour.

Q.2 A toy is in the form of a cone mounted on a hemisphere of diameter 7 cm. The total height of the toy is 14.5 cm. Find the volume and total surface area of the toy.

Marks:4
Ans

Totalheightofthetoy=14.5cmDiameterofthehemisphere=7cmRadiusofthehemisphere=72=3.5cmHeightoftheconicalportion=14.5−3.5=11cmVolumeofsolid=23r3+13r2h=13r22r+h=13—227—7222—72+11=231cm3Totalsurfaceareaofthetoy=Surfaceareaofaconicalpart+Surfaceareaofahemisphericalpart=rl+2r2=rl+2r=227—72—h2+r2+2rcm2=227—72—112+722+272cm2=203.94cm2

Q.3 The slant height of the frustum of a cone is 4 cm and the circumferences of its circular ends are 18 cm and 6cm.Find the curved surface area of the frustum.

Marks:4
Ans

Given,Slantheightofthefrustuml=4cmLetradiiofthetopandbottomofthefrustumoftheconeber1andr2.Bythequestioncircumferenceofthetop2r1andbottom2r2ofthefrustumand6cmand18cmrespectively.⇒r1=62cmandr2=182cmSince,Curvedsurfaceareaofthefrustum=r1+r2—l=62+182—4cm2=3+9—4cm2=48cm2

Q.4 A circus tent is cylindrical upto a height of 3 m and conical above it. If the diameter of the base is 105 m and the vertical height of the conical part is 7.26 m, find the total canvas used in making the tent.

Marks:5
Ans

Totalcanvasused=TotalcurvedsurfaceareaoftentTotalcurvedsurfaceareaofthetent=Curvedsurfaceareaofthecylindricalportion+CurvedsurfaceareaoftheconicalportionRadiusofthecylindricalportion=D2=1052=52.5mHeightofthecylindricalportion=3mCurvedsurface​​areaofthecylindricalportion=2rH=2—227—52.5—3m2Fortheconicalpart:l=slantheight=h2+r2=7.262+52.52=52.9953mCurvedsurfaceareaoftheconicalpart=rl=227—52.5—53Totalcurvedsurfacearea=2—22—7.5—3+22—7.5—53=22—7.5—6+53=22—7.5—59=9735m2

Q.5 Determine the ratio of the volume of a cube to that of a sphere which will exactly fit inside the cube.

Marks:2
Ans

LettheedgeofthecubebexcmRadiusoftheSphereinsidethecube=x2Volumeofthecube=x3Volumeofthesphere=43r3=43—227—x38Ratioofthevolume=x31121x3=2111=21:11

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FAQs (Frequently Asked Questions)

In the less than type, cumulative frequencies are calculated by adding from the first class onwards and upper class limits are used. In the more than type, cumulative frequencies are found by subtracting from the total and lower class limits are used. When both ogives are drawn on the same graph, their intersection gives the median.

The empirical relationship is 3 Median = Mode + 2 Mean. This formula is used to find the third measure when any two are given. For example, if mean = 26 and mode = 32, then 3 Median = 32 + 52 = 84, so Median = 28.

Standard NCERT exercises test direct application of formulas. Hard questions combine two or three concepts in one problem, such as finding missing frequencies using both mean and total frequency conditions simultaneously, or comparing mean, median, and mode for the same dataset. Practising these builds the speed and accuracy needed for CBSE 2026 board exams.

The median formula with missing frequency carries the highest marks at 4 marks. The step-deviation method for mean and the mode formula each carry 3 marks. Constructing less than and more than type distributions and applying the empirical relationship appear as 2-mark and 3-mark questions. These four areas cover the bulk of marks from this chapter.

The most common mistake is identifying the wrong median class or modal class. For median, students sometimes pick the class where cf equals n/2 exactly rather than the class whose cf first exceeds n/2. For mode, students sometimes pick the class adjacent to the highest frequency instead of the class with the highest frequency itself. Both errors lead to completely wrong answers even when the formula application is correct.