CBSE Class 10 Maths Ch 13 Statistics Question Bank with Solutions

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Class 10 Maths · Chapter 4 · Question Bank + Solutions

CBSE · NCERT

MATHEMATICS – CHAPTER 13 : STATISTICS

Chapter-wise Question Bank (CBSE Board + Competitive · Moderate to HOTS Level)

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Class : XSubject : MathematicsName : ..................

Questions : 89Session : 2026–27Total Marks : 151

General Instructions

1. This question bank is based on NCERT Class X Mathematics, Chapter 13 (Statistics), and is set in the CBSE board pattern with competitive (HOTS) questions mixed in.
2. Section A: Q.1 to Q.50 are MCQs of 1 mark each (50 marks). Section B: Q.51 to Q.60 are Assertion–Reason questions of 1 mark each (10 marks).
3. Section C: Q.61 to Q.70, Very Short Answer, 2 marks each (20 marks). Section D: Q.71 to Q.80, Short Answer, 3 marks each (30 marks).
4. Section E: Q.81 to Q.84, Case-Based, 4 marks each (16 marks). Section F: Q.85 to Q.89, Long Answer, 5 marks each (25 marks).
5. Show all steps for Sections C to F. Complete solutions are given after the questions.

SECTION – A (Multiple Choice Questions · 1 Mark each)

Q.1 – Q.50 Choose the correct option.50 × 1 = 50

1. The class mark of the class 20–30 is –

(a) 20(b) 25(c) 30(d) 10

2. The class mark of the class 100–120 is –

(a) 20(b) 120(c) 110(d) 100

3. The class width (size) of the class interval 15–25 is –

(a) 15(b) 25(c) 10(d) 20

4. The mean of the first five natural numbers is –

(a) 5(b) 2.5(c) 15(d) 3

5. The mean of 5, 7, 9, 11 and 13 is –

(a) 9(b) 8(c) 10(d) 45

6. The mean of the first 10 even natural numbers is –

(a) 11(b) 10(c) 12(d) 20

7. If the mean of x, x + 2, x + 4, x + 6 and x + 8 is 11, then x = –

(a) 7(b) 5(c) 9(d) 11

8. If the mean of x, x + 2, x + 4, x + 6 and x + 8 is 11, the mean of the last three observations is –

(a) 11(b) 13(c) 15(d) 12

9. The sum of the deviations of all observations from their mean is –

(a) 1(b) equal to n(c) the mean(d) 0

10. If each observation of a data set is increased by 5, the mean –

(a) remains the same(b) becomes 5 times(c) decreases by 5(d) increases by 5

11. If each observation of a data set is multiplied by 2, the mean –

(a) remains the same(b) increases by 2(c) becomes half(d) becomes double

12. The mean of 10 observations is 20. If one observation 25 is replaced by 35, the new mean is –

(a) 20(b) 22(c) 21(d) 25

13. The mean of 20 observations is 15 and of 30 observations is 20. The combined mean of all 50 observations is –

(a) 17.5(b) 19(c) 35(d) 18

14. The mean of 25 observations is 36. The mean of the first 13 is 32 and of the last 13 is 40. The 13th observation is –

(a) 32(b) 36(c) 40(d) 38

15. The median of 3, 5, 7, 9 and 11 is –

(a) 5(b) 9(c) 35(d) 7

16. The median of 4, 6, 8, 10, 12 and 14 is –

(a) 8(b) 9(c) 10(d) 12

17. The median of the first 9 natural numbers is –

(a) 5(b) 4.5(c) 9(d) 4

18. The mode of 2, 3, 3, 4, 5, 3, 6 is –

(a) 4(b) 5(c) 2(d) 3

19. For a moderately skewed distribution, the empirical relation among mean, median and mode is –

(a) Mode = 3 Median – 2 Mean(b) Mode = 2 Median – 3 Mean(c) Mode = 3 Mean – 2 Median(d) Mode = Median + Mean

20. If the mean of a distribution is 27 and its median is 33, the mode is –

(a) 39(b) 21(c) 45(d) 30

21. If the mode is 12 and the mean is 15 for a moderately skewed distribution, the median is –

(a) 13.5(b) 14(c) 12(d) 16

22. The mean of grouped data by the direct method is given by –

(a) Σf / Σfx(b) Σfx / Σf(c) Σx / n only(d) Σfx × Σf

23. In the step-deviation method, the mean is x̄ = –

(a) a – h × (Σfu/Σf)(b) a + h × (Σfu/Σf)(c) a + (Σfu/Σf)(d) h + a × (Σfu/Σf)

24. The mode of grouped data is given by mode = –

(a) l + [(f₁ – f₀)/(f₁ + f₀ + f₂)] × h(b) l + [(f₁ – f₀)/(2f₁ – f₀ – f₂)] × h(c) l + [(n/2 – cf)/f] × h(d) l + (f₁ + f₂)/(2f₀) × h

25. The median of grouped data is given by median = –

(a) l + [(n/2 – cf)/f] × h(b) l + [(f₁ – f₀)/(2f₁ – f₀ – f₂)] × h(c) l + (n – cf)/f × h(d) Σfx/Σf

26. The median class is the class whose cumulative frequency is –

(a) equal to n(b) the highest frequency(c) just greater than or equal to n/2(d) just less than n/2

27. The modal class is the class with –

(a) the highest class mark(b) the highest frequency(c) the largest width(d) cumulative frequency n/2

28. For the distribution below, the mean is –

Class interval	0–10	10–20	20–30	30–40	40–50
Frequency	5	8	15	16	6

(a) 25(b) 28(c) 27(d) 30

29. For the distribution below, the median class is –

Class interval	0–10	10–20	20–30	30–40	40–50
Frequency	5	8	15	16	6

(a) 20–30(b) 10–20(c) 30–40(d) 0–10

30. For the distribution below, the median is –

Class interval	0–10	10–20	20–30	30–40	40–50
Frequency	5	8	15	16	6

(a) 25(b) 30(c) 27(d) 28

31. For the distribution below, the modal class is –

Class interval	0–10	10–20	20–30	30–40	40–50
Frequency	5	8	15	16	6

(a) 20–30(b) 30–40(c) 10–20(d) 40–50

32. For the distribution below, the mode is –

Class interval	0–10	10–20	20–30	30–40	40–50
Frequency	5	8	15	16	6

(a) 35(b) 32.5(c) 28(d) 30.91

33. For the distribution below, the cumulative frequency of the class 20–30 is –

Class interval	0–10	10–20	20–30	30–40	40–50
Frequency	5	8	15	16	6

(a) 28(b) 15(c) 13(d) 44

34. For the distribution below, the mean is –

Class interval	10–20	20–30	30–40	40–50	50–60
Frequency	6	12	20	8	4

(a) 35(b) 32(c) 34(d) 33.4

35. For the distribution below, the median is –

Class interval	10–20	20–30	30–40	40–50	50–60
Frequency	6	12	20	8	4

(a) 33.5(b) 35(c) 30(d) 34.5

36. For the distribution below, the mode is –

Class interval	10–20	20–30	30–40	40–50	50–60
Frequency	6	12	20	8	4

(a) 35(b) 33.5(c) 34(d) 36

37. For the distribution below, using assumed mean a = 50 and h = 20, Σfu is –

Class interval	0–20	20–40	40–60	60–80	80–100
Frequency	7	10	15	12	6

(a) –14(b) 12(c) 0(d) 25

38. For the distribution below, the mean is –

Class interval	0–20	20–40	40–60	60–80	80–100
Frequency	7	10	15	12	6

(a) 52(b) 48(c) 50(d) 55

39. For the distribution below, the median is –

Class interval	0–20	20–40	40–60	60–80	80–100
Frequency	7	10	15	12	6

(a) 50(b) 52.5(c) 50.67(d) 40

40. For the distribution below, the mode is –

Class interval	0–20	20–40	40–60	60–80	80–100
Frequency	7	10	15	12	6

(a) 52.5(b) 50(c) 50.67(d) 60

41. For the distribution below, the median is –

Class interval	40–45	45–50	50–55	55–60	60–65	65–70	70–75
Frequency	2	3	8	6	6	3	2

(a) 57.5(b) 55(c) 56.67(d) 60

42. For the distribution below, the modal class is –

Class interval	40–45	45–50	50–55	55–60	60–65	65–70	70–75
Frequency	2	3	8	6	6	3	2

(a) 55–60(b) 60–65(c) 45–50(d) 50–55

43. The mean of the distribution below is 52 if p = –

Class interval	0–20	20–40	40–60	60–80	80–100
Frequency	6	8	p	12	6

(a) 8(b) 6(c) 10(d) 12

44. In a symmetrical distribution –

(a) mean > median > mode(b) mean = median = mode(c) mode > median > mean(d) only mean = median

45. Which measure of central tendency is most affected by extreme values (outliers)?

(a) Median(b) Mode(c) Mean(d) None of these

46. The sum of the frequencies of the distribution is called –

(a) the class width(b) the total frequency (n)(c) the class mark(d) the mean

47. The median of 12, 15, 11, 18 and 14 is –

(a) 15(b) 14(c) 12(d) 14.4

48. The mode of the data 5, 6, 7, 6, 5, 6, 8 is –

(a) 6(b) 5(c) 7(d) 8

49. The mean of the first n natural numbers is –

(a) n/2(b) n(n + 1)(c) (n – 1)/2(d) (n + 1)/2

50. For the distribution below, the mean is –

Class interval	40–45	45–50	50–55	55–60	60–65	65–70	70–75
Frequency	2	3	8	6	6	3	2

(a) 57.17(b) 56.67(c) 53.57(d) 60

SECTION – B (Assertion (A) – Reason (R) · 1 Mark each)

Q.51 – Q.60 Choose: (a) Both A and R true, R explains A; (b) Both true, R does not explain A; (c) A true, R false; (d) A false, R true.10 × 1 = 10

51. Assertion (A): The mean of the first five natural numbers is 3.
Reason (R): Mean = (sum of observations)/(number of observations).

52. Assertion (A): The mode of 2, 3, 3, 4, 5, 3, 6 is 3.
Reason (R): The mode is the observation that occurs most frequently.

53. Assertion (A): To find the mean of grouped data we use the class marks of the classes.
Reason (R): The class mark of a class is the average of its upper and lower limits.

54. Assertion (A): The median of 3, 5, 7, 9 is 7.
Reason (R): For an even number of observations the median is the mean of the two middle observations.

55. Assertion (A): For a moderately skewed distribution, 3 Median = Mode + 2 Mean.
Reason (R): Mode = 2 Median – 3 Mean.

56. Assertion (A): The mean of 5, 7, 9, 11 and 13 is 9.
Reason (R): In an arithmetic progression with an odd number of terms, the mean equals the middle term.

57. Assertion (A): The sum of the deviations of the observations from their mean is zero.
Reason (R): Σ(x – x̄) = Σx – n x̄ and Σx = n x̄.

58. Assertion (A): The median class is the class whose cumulative frequency is just greater than or equal to n/2.
Reason (R): The median divides the distribution into two equal halves.

59. Assertion (A): If every observation is increased by 5, the mean remains the same.
Reason (R): If every observation is multiplied by k, the mean is multiplied by k.

60. Assertion (A): The modal class is the class with the highest frequency.
Reason (R): Mode = l + [(f₁ – f₀)/(2f₁ – f₀ – f₂)] × h.

SECTION – C (Very Short Answer · 2 Marks each)

61. Find the mean of 5, 7, 9, 11 and 13.(2)

62. Find the class marks of the classes 20–30 and 40–60.(2)

63. Find the median of 7, 3, 9, 5 and 11.(2)

64. Find the median of 4, 6, 8, 10, 12 and 14.(2)

65. The mean and median of a moderately skewed distribution are 27 and 33. Find the mode.(2)

66. Find the value of p if the mean of 2, 4, 6, 8 and p is 6.(2)

67. The mean of 20 observations is 15 and the mean of 30 observations is 20. Find the mean of all 50 observations.(2)

68. Find the median class of the distribution:

Class interval	0–10	10–20	20–30	30–40	40–50
Frequency	5	8	15	16	6

(2)

69. Find the mode of the distribution:

Class interval	10–20	20–30	30–40	40–50	50–60
Frequency	6	12	20	8	4

(2)

70. The mean of 25 observations is 36. The mean of the first 13 is 32 and of the last 13 is 40. Find the 13th observation.(2)

SECTION – D (Short Answer · 3 Marks each)

71. Find the mean of the following distribution by the direct method.

Class interval	0–10	10–20	20–30	30–40	40–50
Frequency	5	8	15	16	6

(3)

72. Find the median of the following distribution.

Class interval	0–10	10–20	20–30	30–40	40–50
Frequency	5	8	15	16	6

(3)

73. Find the mode of the following distribution.

Class interval	10–20	20–30	30–40	40–50	50–60
Frequency	6	12	20	8	4

(3)

74. Find the mean of the following distribution by the step-deviation method.

Class interval	0–20	20–40	40–60	60–80	80–100
Frequency	7	10	15	12	6

(3)

75. Find the median of the distribution of the weights (kg) of 30 students.

Class interval	40–45	45–50	50–55	55–60	60–65	65–70	70–75
Frequency	2	3	8	6	6	3	2

(3)

76. Find the mode of the distribution of the weights (kg) of 30 students.

Class interval	40–45	45–50	50–55	55–60	60–65	65–70	70–75
Frequency	2	3	8	6	6	3	2

(3)

77. Find the value of p if the mean of the following distribution is 52.

Class interval	0–20	20–40	40–60	60–80	80–100
Frequency	6	8	p	12	6

(3)

78. The mean of 25 observations is 36. The mean of the first 13 observations is 32 and that of the last 13 observations is 40. Find the 13th observation.(3)

79. Find the mean of the weights (kg) of 30 students.

Class interval	40–45	45–50	50–55	55–60	60–65	65–70	70–75
Frequency	2	3	8	6	6	3	2

(3)

80. The mean and median of a moderately skewed distribution are 27 and 33 respectively. Find the mode, and also find the median if the mode were 12 and the mean 15.(3)

SECTION – E (Case-Based Questions · 4 Marks each)

81. Read the passage and answer the questions that follow.(4)

The marks (out of 50) of 50 students in a unit test are given in the table.

Class interval	0–10	10–20	20–30	30–40	40–50
Frequency	5	8	15	16	6

(i) Write the class mark of the class 30–40 and the modal class.(1)

(ii) Find the median class and the median.(1)

(iii) Find the mean and the mode of the data.(2)

82. Read the passage and answer the questions that follow.(4)

A shopkeeper recorded the ages (in years) of 50 customers on one day.

Class interval	10–20	20–30	30–40	40–50	50–60
Frequency	6	12	20	8	4

(i) Which age group is the modal class?(1)

(ii) Find the mode of the ages.(1)

(iii) Find the mean and the median. Verify that 3 × Median is close to Mode + 2 × Mean.(2)

83. Read the passage and answer the questions that follow.(4)

The time (in minutes) spent daily on mobile phones by 50 students is shown below.

Class interval	0–20	20–40	40–60	60–80	80–100
Frequency	7	10	15	12	6

(i) Write the class mark of the class 40–60.(1)

(ii) If the assumed mean a = 50 and class width h = 20 are used, find u for the class 80–100.(1)

(iii) Find the mean (step-deviation method) and the median.(2)

84. Read the passage and answer the questions that follow.(4)

The weights (in kg) of 30 students of a class are recorded below.

Class interval	40–45	45–50	50–55	55–60	60–65	65–70	70–75
Frequency	2	3	8	6	6	3	2

(i) Find the median class.(1)

(ii) Find the median weight.(1)

(iii) Find the modal class and the mode.(2)

SECTION – F (Long Answer · 5 Marks each)

85. The median of the following distribution is 28.5 and the total frequency is 60. Find the missing frequencies x and y.

Class interval	0–10	10–20	20–30	30–40	40–50	50–60
Frequency	5	x	20	15	y	5

(5)

86. Find the mean of the following distribution by the step-deviation method, and also find its median and mode.

Class interval	0–20	20–40	40–60	60–80	80–100
Frequency	7	10	15	12	6

(5)

87. The weights (in kg) of 30 students are given below. Find the mean, the median and the mode.

Class interval	40–45	45–50	50–55	55–60	60–65	65–70	70–75
Frequency	2	3	8	6	6	3	2

(5)

88. The mean of the following distribution is 52. Find the missing frequency p and then find the median of the distribution.

Class interval	0–20	20–40	40–60	60–80	80–100
Frequency	6	8	p	12	6

(5)

89. Find the mean, median and mode of the following distribution and compare the mode with 3 Median – 2 Mean.

Class interval	0–10	10–20	20–30	30–40	40–50
Frequency	5	8	15	16	6

(5)

********

CBSE · NCERT

MATHEMATICS – CHAPTER 13 : COMPLETE SOLUTIONS

Statistics · Answer Key with Step-by-Step Solutions

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Class : XSubject : MathematicsTotal Marks : 151

SECTION – A (MCQ Solutions)

Quick Answer Key (Q.1 – Q.50)

1. (b)	2. (c)	3. (c)	4. (d)	5. (a)	6. (a)	7. (a)	8. (b)	9. (d)	10. (d)
11. (d)	12. (c)	13. (d)	14. (b)	15. (d)	16. (b)	17. (a)	18. (d)	19. (a)	20. (c)
21. (b)	22. (b)	23. (b)	24. (b)	25. (a)	26. (c)	27. (b)	28. (c)	29. (a)	30. (d)
31. (b)	32. (d)	33. (a)	34. (d)	35. (a)	36. (c)	37. (c)	38. (c)	39. (c)	40. (a)
41. (c)	42. (d)	43. (a)	44. (b)	45. (c)	46. (b)	47. (b)	48. (a)	49. (d)	50. (a)

Solutions with Explanation

1. Ans. (b) 25

Class mark = (20 + 30)/2 = 25.

2. Ans. (c) 110

(100 + 120)/2 = 110.

3. Ans. (c) 10

Upper limit – lower limit = 25 – 15 = 10.

4. Ans. (d) 3

(1 + 2 + 3 + 4 + 5)/5 = 15/5 = 3.

5. Ans. (a) 9

Sum = 45; mean = 45/5 = 9.

6. Ans. (a) 11

2 + 4 + … + 20 = 110; mean = 110/10 = 11.

7. Ans. (a) 7

Mean = x + 4 = 11 ⇒ x = 7.

8. Ans. (b) 13

x = 7, last three: 11, 13, 15 ⇒ mean = 13.

9. Ans. (d) 0

Σ(x – x̄) = Σx – n x̄ = 0.

10. Ans. (d) increases by 5

x̄′ = (Σx + 5n)/n = x̄ + 5.

11. Ans. (d) becomes double

x̄′ = 2Σx/n = 2x̄.

12. Ans. (c) 21

Sum increases by 10 ⇒ mean increases by 10/10 = 1 ⇒ 21.

13. Ans. (d) 18

(20 × 15 + 30 × 20)/50 = 900/50 = 18.

14. Ans. (b) 36

Sum of all = 900; 13 × 32 + 13 × 40 = 936; the 13th is counted twice ⇒ 936 – 900 = 36.

15. Ans. (d) 7

Middle value of the 5 ordered observations is 7.

16. Ans. (b) 9

n = 6 (even): median = (8 + 10)/2 = 9.

17. Ans. (a) 5

The 5th term of 1, 2, …, 9 is 5.

18. Ans. (d) 3

3 occurs most often (three times).

19. Ans. (a) Mode = 3 Median – 2 Mean

3 Median = Mode + 2 Mean.

20. Ans. (c) 45

Mode = 3 × 33 – 2 × 27 = 45.

21. Ans. (b) 14

3 Median = 12 + 30 = 42 ⇒ 14.

22. Ans. (b) Σfx / Σf

x̄ = Σ f_ix_i / Σ f_i.

23. Ans. (b) a + h × (Σfu/Σf)

u = (x – a)/h ⇒ x̄ = a + h ū.

24. Ans. (b) l + [(f₁ – f₀)/(2f₁ – f₀ – f₂)] × h

This is the grouped-data mode formula (modal class l, h, frequencies f₀, f₁, f₂).

25. Ans. (a) l + [(n/2 – cf)/f] × h

cf = cumulative frequency of the class preceding the median class.

26. Ans. (c) just greater than or equal to n/2

Locate n/2 in the cumulative frequency column.

27. Ans. (b) the highest frequency

By definition.

28. Ans. (c) 27

Σf = 50, Σfx = 25 + 120 + 375 + 560 + 270 = 1350 ⇒ mean = 27.

29. Ans. (a) 20–30

cf: 5, 13, 28, 44, 50; n/2 = 25 lies in 20–30.

30. Ans. (d) 28

20 + [(25 – 13)/15] × 10 = 20 + 8 = 28.

31. Ans. (b) 30–40

Highest frequency 16 is in 30–40.

32. Ans. (d) 30.91

30 + [(16 – 15)/(32 – 15 – 6)] × 10 = 30 + 10/11 = 30.91.

33. Ans. (a) 28

5 + 8 + 15 = 28.

34. Ans. (d) 33.4

Σf = 50, Σfx = 90 + 300 + 700 + 360 + 220 = 1670 ⇒ 33.4.

35. Ans. (a) 33.5

cf: 6, 18, 38, 46, 50; class 30–40: 30 + [(25 – 18)/20] × 10 = 33.5.

36. Ans. (c) 34

30 + [(20 – 12)/(40 – 12 – 8)] × 10 = 30 + 4 = 34.

37. Ans. (c) 0

u = –2, –1, 0, 1, 2 ⇒ Σfu = –14 – 10 + 0 + 12 + 12 = 0.

38. Ans. (c) 50

x̄ = 50 + 20 × (0/50) = 50.

39. Ans. (c) 50.67

cf: 7, 17, 32, 44, 50; class 40–60: 40 + [(25 – 17)/15] × 20 = 50.67.

40. Ans. (a) 52.5

40 + [(15 – 10)/(30 – 10 – 12)] × 20 = 40 + 12.5 = 52.5.

41. Ans. (c) 56.67

n = 30, n/2 = 15; cf up to 50–55 is 13; class 55–60: 55 + [(15 – 13)/6] × 5 = 56.67.

42. Ans. (d) 50–55

Highest frequency 8 is in 50–55.

43. Ans. (a) 8

Σf = 32 + p, Σfx = 1680 + 50p. (1680 + 50p) = 52(32 + p) ⇒ 16 = 2p ⇒ p = 8.

44. Ans. (b) mean = median = mode

In a symmetric distribution all three measures coincide.

45. Ans. (c) Mean

The mean uses every observation; the median and mode do not.

46. Ans. (b) the total frequency (n)

n = Σf.

47. Ans. (b) 14

Ordered: 11, 12, 14, 15, 18 ⇒ middle value 14.

48. Ans. (a) 6

6 occurs three times, the maximum.

49. Ans. (d) (n + 1)/2

Sum = n(n + 1)/2 ⇒ mean = (n + 1)/2.

50. Ans. (a) 57.17

Σf = 30, Σfx = 1715 ⇒ mean = 1715/30 = 57.17.

SECTION – B (Assertion – Reason)

51. Ans. (a)

15/5 = 3 using R.

52. Ans. (a)

3 occurs three times, more than any other value.

53. Ans. (b)

Both true; R defines class mark but does not explain why class marks are used (they represent each class).

54. Ans. (d)

A is false: median = (5 + 7)/2 = 6. R is true.

55. Ans. (c)

A is true. R is false: Mode = 3 Median – 2 Mean.

56. Ans. (a)

5, 7, 9, 11, 13 is an AP with middle term 9.

57. Ans. (a)

Substituting Σx = n x̄ gives 0.

58. Ans. (a)

The median lies where half the observations have been counted.

59. Ans. (d)

A is false: the mean increases by 5. R is true.

60. Ans. (b)

Both true; the formula is applied after identifying the modal class and does not explain its definition.

SECTION – C (2 Marks each)

61. Ans.

Sum = 5 + 7 + 9 + 11 + 13 = 45. [1] Mean = 45/5 = 9. [1]

62. Ans.

Class mark = (lower limit + upper limit)/2. [1] (20 + 30)/2 = 25; (40 + 60)/2 = 50. [1]

63. Ans.

Arranged in order: 3, 5, 7, 9, 11. [1] n = 5 (odd) ⇒ the middle (3rd) value = 7. [1]

64. Ans.

The data is already in order; n = 6 (even). [1] Median = (3rd + 4th)/2 = (8 + 10)/2 = 9. [1]

65. Ans.

Empirical relation: Mode = 3 Median – 2 Mean. [1] = 3 × 33 – 2 × 27 = 99 – 54 = 45. [1]

66. Ans.

(2 + 4 + 6 + 8 + p)/5 = 6 ⇒ 20 + p = 30. [1] p = 10. [1]

67. Ans.

Total of first group = 20 × 15 = 300; of second = 30 × 20 = 600. [1] Combined mean = 900/50 = 18. [1]

68. Ans.

n = 50 ⇒ n/2 = 25. [½] Cumulative frequencies: 5, 13, 28, 44, 50. [1] The first cf ≥ 25 is 28, so the median class is 20–30. [½]

69. Ans.

Modal class = 30–40 (f₁ = 20, f₀ = 12, f₂ = 8, l = 30, h = 10). [1] Mode = 30 + [(20 – 12)/(40 – 12 – 8)] × 10 = 30 + 4 = 34. [1]

70. Ans.

Σ(all 25) = 25 × 36 = 900; Σ(first 13) = 416; Σ(last 13) = 520. [1] The 13th observation is counted in both: 416 + 520 – 900 = 36. [1]

SECTION – D (3 Marks each)

71. Ans.

Class	f	x	fx
0–10	5	5	25
10–20	8	15	120
20–30	15	25	375
30–40	16	35	560
40–50	6	45	270
Total	50		1350

Mean = Σfx/Σf = 1350/50 = 27. [3]

72. Ans.

Cumulative frequencies: 5, 13, 28, 44, 50. n = 50 ⇒ n/2 = 25 ⇒ median class 20–30. [1] l = 20, cf = 13, f = 15, h = 10. [½] Median = 20 + [(25 – 13)/15] × 10 [1] = 20 + 8 = 28. [½]

73. Ans.

Modal class = 30–40 (highest frequency 20). [1] l = 30, h = 10, f₁ = 20, f₀ = 12, f₂ = 8. [½] Mode = 30 + [(20 – 12)/(2 × 20 – 12 – 8)] × 10 [1] = 30 + 80/20 = 34. [½]

74. Ans.

Take a = 50, h = 20, u = (x – 50)/20.

Class	f	x	u	fu
0–20	7	10	-2	-14
20–40	10	30	-1	-10
40–60	15	50	0	0
60–80	12	70	1	12
80–100	6	90	2	12
Total	50			0

[2] Mean = a + h(Σfu/Σf) = 50 + 20 × (0/50) = 50. [1]

75. Ans.

cf: 2, 5, 13, 19, 25, 28, 30. n/2 = 15 ⇒ median class 55–60. [1] l = 55, cf = 13, f = 6, h = 5. [½] Median = 55 + [(15 – 13)/6] × 5 [1] = 55 + 1.67 = 56.67 kg. [½]

76. Ans.

Highest frequency 8 ⇒ modal class 50–55. [1] l = 50, h = 5, f₁ = 8, f₀ = 3, f₂ = 6. [½] Mode = 50 + [(8 – 3)/(16 – 3 – 6)] × 5 = 50 + 25/7 [1] = 53.57 kg. [½]

77. Ans.

Class marks: 10, 30, 50, 70, 90. Σf = 32 + p; Σfx = 60 + 240 + 50p + 840 + 540 = 1680 + 50p. [1½] Mean = 52 ⇒ 1680 + 50p = 52(32 + p) = 1664 + 52p. [1] 2p = 16 ⇒ p = 8. [½]

78. Ans.

Sum of 25 observations = 25 × 36 = 900. [1] Sum of the first 13 = 13 × 32 = 416; sum of the last 13 = 13 × 40 = 520. [1] The 13th observation is in both groups: 416 + 520 – 900 = 36. [1]

79. Ans.

Class	f	x	fx
40–45	2	42.5	85
45–50	3	47.5	142.5
50–55	8	52.5	420
55–60	6	57.5	345
60–65	6	62.5	375
65–70	3	67.5	202.5
70–75	2	72.5	145
Total	30		1715

Mean = 1715/30 = 57.17 kg. [3]

80. Ans.

Mode = 3 Median – 2 Mean [½] = 3 × 33 – 2 × 27 = 99 – 54 = 45. [1] Second case: 3 Median = Mode + 2 Mean = 12 + 30 = 42 [1] ⇒ Median = 14. [½]

SECTION – E (Case-Based · 4 Marks each)

81. Ans.

(i) Class mark = (30 + 40)/2 = 35. Modal class (highest frequency 16) = 30–40. [1]

(ii) cf: 5, 13, 28, 44, 50 ⇒ median class 20–30. Median = 20 + [(25 – 13)/15] × 10 = 28. [1]

(iii) Mean = 1350/50 = 27. [1] Mode = 30 + [(16 – 15)/(32 – 15 – 6)] × 10 = 30 + 10/11 = 30.91. [1] [2]

82. Ans.

(i) Highest frequency 20 ⇒ modal class 30–40. [1]

(ii) Mode = 30 + [(20 – 12)/(40 – 12 – 8)] × 10 = 34 years. [1]

(iii) Mean = 1670/50 = 33.4; median = 30 + [(25 – 18)/20] × 10 = 33.5. [1] 3 × 33.5 = 100.5 and 34 + 2 × 33.4 = 100.8 – nearly equal ✓. [1] [2]

83. Ans.

(i) (40 + 60)/2 = 50. [1]

(ii) x = 90 ⇒ u = (90 – 50)/20 = 2. [1]

(iii) Σfu = –14 – 10 + 0 + 12 + 12 = 0 ⇒ mean = 50 + 20 × 0/50 = 50 minutes. [1] cf: 7, 17, 32, 44, 50; median class 40–60: 40 + [(25 – 17)/15] × 20 = 50.67 minutes. [1] [2]

84. Ans.

(i) n = 30, n/2 = 15; cf: 2, 5, 13, 19, 25, 28, 30 ⇒ median class 55–60. [1]

(ii) 55 + [(15 – 13)/6] × 5 = 56.67 kg. [1]

(iii) Highest frequency 8 ⇒ modal class 50–55. [1] Mode = 50 + [(8 – 3)/(16 – 3 – 6)] × 5 = 50 + 25/7 = 53.57 kg. [1] [2]

SECTION – F (Long Answer · 5 Marks each)

85. Ans.

Total: 5 + x + 20 + 15 + y + 5 = 60 ⇒ x + y = 15 …(i) [1] Median = 28.5 lies in the class 20–30, so the median class is 20–30 with l = 20, h = 10, f = 20. [1] cf of the previous class = 5 + x. n/2 = 30. [½] 28.5 = 20 + [(30 – 5 – x)/20] × 10 [1] ⇒ 8.5 = (25 – x)/2 ⇒ 25 – x = 17 ⇒ x = 8. [1] From (i): y = 7. [½]

86. Ans.

Mean: a = 50, h = 20.

Class	f	x	u	fu
0–20	7	10	-2	-14
20–40	10	30	-1	-10
40–60	15	50	0	0
60–80	12	70	1	12
80–100	6	90	2	12
Total	50			0

Mean = 50 + 20 × (0/50) = 50. [2] Median: cf: 7, 17, 32, 44, 50; n/2 = 25 ⇒ class 40–60: 40 + [(25 – 17)/15] × 20 = 50.67. [1½] Mode: modal class 40–60 (f₁ = 15, f₀ = 10, f₂ = 12): 40 + [(15 – 10)/(30 – 10 – 12)] × 20 = 40 + 12.5 = 52.5. [1½]

87. Ans.

Mean: Σfx = 85 + 142.5 + 420 + 345 + 375 + 202.5 + 145 = 1715; mean = 1715/30 = 57.17 kg. [1½] Median: cf: 2, 5, 13, 19, 25, 28, 30; n/2 = 15 ⇒ class 55–60: 55 + [(15 – 13)/6] × 5 = 56.67 kg. [1½] Mode: modal class 50–55 (f₁ = 8, f₀ = 3, f₂ = 6): 50 + [(8 – 3)/(16 – 3 – 6)] × 5 = 50 + 25/7 = 53.57 kg. [2]

88. Ans.

Σf = 32 + p; Σfx = 10 × 6 + 30 × 8 + 50p + 70 × 12 + 90 × 6 = 1680 + 50p. [1] 52 = (1680 + 50p)/(32 + p) ⇒ 1664 + 52p = 1680 + 50p ⇒ p = 8. [1½] Frequencies: 6, 8, 8, 12, 6 (n = 40). cf: 6, 14, 22, 34, 40. [½] n/2 = 20 ⇒ median class 40–60 (l = 40, cf = 14, f = 8, h = 20). [1] Median = 40 + [(20 – 14)/8] × 20 = 40 + 15 = 55. [1]

89. Ans.

Mean: Σfx = 1350, Σf = 50 ⇒ mean = 27. [1] Median: cf: 5, 13, 28, 44, 50; n/2 = 25 ⇒ class 20–30: 20 + [(25 – 13)/15] × 10 = 28. [1½] Mode: modal class 30–40 (f₁ = 16, f₀ = 15, f₂ = 6): 30 + [(16 – 15)/(32 – 15 – 6)] × 10 = 30 + 10/11 = 30.91. [1½] Comparison: 3 Median – 2 Mean = 84 – 54 = 30, which is close to the computed mode 30.91 – the empirical relation holds approximately. [1]

QUICK REVISION · KEY POINTS (CBSE / COMPETITIVE)

1. Mean (direct): x̄ = Σfx/Σf, where x is the class mark = (lower + upper limit)/2.
2. Mean (assumed mean): x̄ = a + Σfd/Σf (d = x – a). Step deviation: x̄ = a + h(Σfu/Σf), u = (x – a)/h.
3. Mode: modal class = highest frequency; Mode = l + [(f₁ – f₀)/(2f₁ – f₀ – f₂)] × h.
4. Median: build the cumulative frequency (cf) column; median class has cf ≥ n/2; Median = l + [(n/2 – cf)/f] × h (cf = cf of the previous class).
5. Empirical relation (moderately skewed data): Mode = 3 Median – 2 Mean.
6. For raw data: median is the middle value (odd n) or the mean of two middle values (even n); mode is the most frequent value.
7. Adding k to every observation adds k to the mean; multiplying by k multiplies the mean by k. Σ(x – x̄) = 0.
8. Combined mean = (n₁x̄₁ + n₂x̄₂)/(n₁ + n₂). Missing-frequency questions: form an equation using the given mean or median.

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