Chapter 13: Statistics

Important questions — the ones the board asked most (20)

1. Q1Long answer5 marksAsked: 2026

   The mean of the following frequency distribution is 35. If the sum of frequencies is 25, find the values of x and y : Class: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, 60-70 Frequency: 1, x, 5, 7, y, 3, 1

   

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   Model answer · Mean, missing frequencies

   1. 1.Given: the mean of the distribution is 35 and the sum of frequencies is 25. We find the mean by the direct method, x̄ = Σf_i x_i / Σf_i, where x_i is the class mark of each class. The table of class marks (x_i) and products f_i x_i is: Class | Frequency (f_i) | Class mark (x_i) | f_i x_i 0-10 | 1 | 5 | 5 10-20 | x | 15 | 15x 20-30 | 5 | 25 | 125 30-40 | 7 | 35 | 245 40-50 | y | 45 | 45y 50-60 | 3 | 55 | 165 60-70 | 1 | 65 | 65 Total | Σf_i = 17 + x + y | | Σf_i x_i = 605 + 15x + 45y2 marks
   2. 2.The sum of frequencies is given as 25: Σf_i = 17 + x + y = 25 ⇒ x + y = 8 ... (i)0.5 marks
   3. 3.Using the formula for the mean: Mean (x̄) = Σf_i x_i / Σf_i = (605 + 15x + 45y)/25 = 350.5 marks
   4. 4.Multiplying both sides by 25: 605 + 15x + 45y = 875 ⇒ 15x + 45y = 270 Dividing both sides by 15: ⇒ x + 3y = 18 ... (ii)1 mark
   5. 5.Solving (i) and (ii): subtract (i) from (ii): (x + 3y) − (x + y) = 18 − 8 2y = 10 ⇒ y = 5 Putting y = 5 in (i): x = 8 − 5 = 3 Therefore, x = 3 and y = 5.1 mark

   • Draw the full table with the class mark (x_i) and f_i x_i columns, and write the totals Σf_i = 17 + x + y and Σf_i x_i = 605 + 15x + 45y.
   • Label the two equations "x + y = 8 ... (i)" and "x + 3y = 18 ... (ii)" before you solve them.
   • End with the final statement "x = 3 and y = 5".

   Marks split as in the board’s published marking scheme

2. Q2Long answer5 marksAsked: 2026

   The monthly expenditure on fruits in 200 families of a Housing Society is given below. Find the value of x and also find the mode and mean expenditure on fruits.

   

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   Model answer · mode and mean, missing frequency

   1. 1.For the 200 families, x = 30, the modal monthly expenditure on fruits is Rs 1847.82 and the mean monthly expenditure on fruits is Rs 2662.50. The working follows. The unknown frequency x belongs to the class 3000 - 3500. The class mark x_i of each class is its mid-point. Expenditure (in Rs) | Number of families (f_i) | Class mark (x_i) | f_i x_i 1000 - 1500 | 24 | 1250 | 30000 1500 - 2000 | 40 | 1750 | 70000 2000 - 2500 | 33 | 2250 | 74250 2500 - 3000 | 28 | 2750 | 77000 3000 - 3500 | x | 3250 | 3250x 3500 - 4000 | 22 | 3750 | 82500 4000 - 4500 | 16 | 4250 | 68000 4500 - 5000 | 7 | 4750 | 33250 Total | Σf_i = 170 + x | | Σf_i x_i = 435000 + 3250x1.5 marks
   2. 2.There are 200 families in total, so Σf_i = 170 + x = 200 x = 200 - 170 = 300.5 marks
   3. 3.The maximum class frequency is 40, so 1500 - 2000 is the modal class. For this class, the lower limit l = 1500 and the class size h = 500. The frequency of the modal class is f_1 = 40. The frequency of the class preceding the modal class is f_0 = 24, and the frequency of the class succeeding the modal class is f_2 = 33.0.5 marks
   4. 4.Mode = l + ((f_1 - f_0)/(2f_1 - f_0 - f_2)) × h => Mode = 1500 + ((40 - 24)/(2 × 40 - 24 - 33)) × 500 = 1500 + (16/23) × 5001 mark
   5. 5.= 1500 + 347.82 = 1847.82 Hence the modal monthly expenditure is Rs 1847.82.0.5 marks
   6. 6.Putting x = 30 into the total of the f_i x_i column: Σf_i x_i = 435000 + 3250 × 30 = 435000 + 97500 = 5325000.5 marks
   7. 7.By the direct method, Mean = Σf_i x_i / Σf_i Mean = 532500 / 200 = 2662.50 Hence the mean monthly expenditure is Rs 2662.50.0.5 marks

   • In the table, show the class mark column and the f_i x_i column, and write the totals as "170 + x" and "435000 + 3250x".
   • Write the values of l, h, f_1, f_0 and f_2 before you put them into the mode formula.
   • Finish each result with its unit: "the modal monthly expenditure is Rs 1847.82" and "the mean monthly expenditure is Rs 2662.50".

   Marks split as in the board’s published marking scheme

3. Q3Long answer5 marksAsked: 2026

   The marks obtained by 80 students of class X in a mock test of Mathematics are given below in the table. Find median and the mode of the data :

   

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   Model answer · Median and mode, grouped data

   1. 1.The median of the data is 52 marks and the mode is 55 marks. Given: the marks of 80 students in a Mathematics mock test, in classes of size h = 10. To find the median, we add a cumulative frequency column to the frequency table: Marks | Number of students (f) | Cumulative frequency (cf) 0 - 10 | 5 | 5 10 - 20 | 8 | 13 20 - 30 | 6 | 19 30 - 40 | 6 | 25 40 - 50 | 12 | 37 50 - 60 | 15 | 52 60 - 70 | 12 | 64 70 - 80 | 8 | 72 80 - 90 | 5 | 77 90 - 100 | 3 | 801 mark
   2. 2.n = 80 n/2 = 400.5 marks
   3. 3.The cumulative frequency just greater than n/2 = 40 is 52, and it belongs to the class 50 - 60. So 50 - 60 is the median class. For this class: lower limit (l) = 50, cumulative frequency of the class before it (cf) = 37, frequency (f) = 15, class size (h) = 10.1 mark
   4. 4.Median = l + ((n/2 − cf)/f) × h Median = 50 + ((40 − 37)/15) × 100.5 marks
   5. 5.= 50 + 2 Median = 52 marks0.5 marks
   6. 6.The maximum class frequency is 15, and the class with this frequency is 50 - 60. So 50 - 60 is the modal class. Lower limit (l) of the modal class = 50, class size (h) = 10, frequency (f1) of the modal class = 15, frequency (f0) of the class preceding the modal class = 12, frequency (f2) of the class succeeding the modal class = 12.1 mark
   7. 7.Mode = l + ((f1 − f0)/(2f1 − f0 − f2)) × h Mode = 50 + ((15 − 12)/(2 × 15 − 12 − 12)) × 10 = 50 + (3/6) × 10 = 50 + 5 Mode = 55 marks0.5 marks

   • Add a cumulative frequency column to the table before you pick the median class.
   • Write the values of l, cf, f, f1, f0, f2 and h before you put them into each formula.
   • Write the unit "marks" after the final median and the final mode.

   Marks split as in the board’s published marking scheme

4. Q4Long answer5 marksAsked: 2026

   An SBI Health Insurance agent found the following data for distribution of ages of 100 policy holders. Health insurance policies are given to persons whose age is 15 years or more but less than 60 years. | Age (in years) | Number of policy holders | | --- | --- | | 15-20 | 2 | | 20-25 | 4 | | 25-30 | 18 | | 30-35 | 21 | | 35-40 | 33 | | 40-45 | 11 | | 45-50 | 3 | | 50-55 | 6 | | 55-60 | 2 | Find the mode age and the median age of the policy holders.

   How the model answer begins

   1. The mode age of the policy holders is about 36.7 years and the median age is about 35.7 years. They are worked out below. Given: the ages of 100 policy holders. The frequency distribution table with the cumulative…

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5. Q5Long answer5 marksAsked: 2026

   If the median of the following distribution is 32. | Class | Frequency | | --- | --- | | 0-10 | x | | 10-20 | 5 | | 20-30 | 9 | | 30-40 | 12 | | 40-50 | y | | 50-60 | 3 | | 60-70 | 2 | | Total | 40 | 5, then find the values of x and y.

6. Q6Long answer5 marksAsked: 2026

   An SBI health insurance agent found the following data for distribution of ages of 100 policy holders. The health insurance policies are given to persons of age 15 years and onwards, but less than 60 years. Age (in yrs) | Number of policy holders 15 - 20 | 2 20 - 25 | 4 25 - 30 | 18 30 - 35 | 21 35 - 40 | 33 40 - 45 | 11 45 - 50 | 3 50 - 55 | 6 55 - 60 | 2 Find the modal age and median age of the policy holders.

   How the model answer begins

   1. Given: the ages of 100 policy holders, with class size h = 5. The table below gives each class, its frequency (f) and its cumulative frequency (cf): Age (in years) | Number of policy holders (f) | cf 15 - 20 | 2 | 2 20…

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7. Q7Long answer5 marksAsked: 2026

   (a) The median of the following data is 137. Find the values of x and y, given that total of frequencies is 68. [Table: Class 65-85, 85-105, 105-125, 125-145, 145-165, 165-185, 185-205; Frequency 4, 5, x, 20, 14, y, 4] OR (b) Find mean and mode of the following distribution : [Table: Class 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, 60-70; Frequency 3, 6, 11, 10, 13, 3, 4]

8. Q8Long answer5 marksAsked: 2026

   (a) The median of the following data is 137. Find the values of x and y, if the sum of all the frequencies is 68. | Class | Frequency | | --- | --- | | 65-85 | 4 | | 85-105 | 5 | | 105-125 | x | | 125-145 | 20 | | 145-165 | 14 | | 165-185 | y | | 185-205 | 4 |

   How the model answer begins

   1. Given: the sum of all the frequencies is n = 68. The cumulative frequency table is: Class | Frequency | Cumulative frequency 65 - 85 | 4 | 4 85 - 105 | 5 | 9 105 - 125 | x | 9 + x 125 - 145 | 20 | 29 + x 145 - 165 | 14…

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9. Q9Long answer5 marksAsked: 2026

   Find the mean and mode of the following frequency distribution : | Class | Frequency | | --- | --- | | 0-15 | 9 | | 15-30 | 15 | | 30-45 | 35 | | 45-60 | 20 | | 60-75 | 11 | | 75-90 | 13 | | 90-105 | 17 |

   How the model answer begins

   1. The mean of the given frequency distribution is 52 and its mode is about 38.5. The working is shown below. Given: a frequency distribution of 7 classes, each of class size h = 15. We use the step deviation method. The…

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10. Q10Long answer5 marksAsked: 2025

    (a) The lengths of 40 leaves of a plant are measured, correct to the nearest millimetre and data obtained is represented in the following table : Length in (mm) | Number of leaves 100 - 120 | 8 120 - 140 | 9 140 - 160 | 12 160 - 180 | 5 180 - 200 | 6 Find the median length (in mm) of the leaves. OR (b) A class teacher has the following absentees record of 30 students of a class. Number of days | 0 - 4 | 4 - 8 | 8 - 12 | 12 - 16 | 16 - 20 | 20 - 24 Number of Absent students | 1 | 8 | x | 6 | 5 | y If the mean number of days a student was absent is 12, find the values of x and y.

    How the model answer begins

    1. Given: the lengths of 40 leaves in classes of width 20 mm. We make the cumulative frequency table: Length (in mm) | Number of leaves (f) | Cumulative frequency (cf) 100 - 120 | 8 | 8 120 - 140 | 9 | 17 140 - 160 | 12 |…

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11. Q11Long answer5 marksAsked: 2025

    The weights of 30 students of a class are given in the following distribution table : Weight (in kg) | Number of students 40 - 45 | 2 45 - 50 | 5 50 - 55 | 8 55 - 60 | 6 60 - 65 | 6 65 - 70 | 3 Find the median weight of the students.

    How the model answer begins

    1. The median weight of the students is 55 kg. It is found as follows. Given: the weights of 30 students in class intervals of 5 kg, with the number of students (frequency) in each class. To find the median, we first find…

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12. Q12Long answer5 marksAsked: 2025

    The following table shows the daily expenditure of 25 households of a locality. Daily Expenditure (in ₹) | Number of Households 500 - 750 | 4 750 - 1000 | 2x + 1 1000 - 1250 | 12 1250 - 1500 | x 1500 - 1750 | 2 Find the value of x. Hence find the mean daily expenditure.

    How the model answer begins

    1. Given: the total number of households is 25. So, 4 + (2x + 1) + 12 + x + 2 = 25 3x + 19 = 25 3x = 6 x = 2 So the frequencies are 4, 2x + 1 = 5, 12, x = 2 and 2.

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13. Q13Long answer5 marksAsked: 2025

    Find 'median' and 'mode' of the following data : Class 100-105, 105-110, 110-115, 115-120, 120-125, 125-130. Frequency 6, 8, 10, 4, 9, 3.

    How the model answer begins

    1. Cumulative frequencies: 6, 14, 24, 28, 37, 40; N = 40.

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14. Q14Long answer5 marksAsked: 2025

    (a) Find 'mean' and 'mode' of the following data : Class: 10-25, 25-40, 40-55, 55-70, 70-85, 85-100 Number of Students: 12, 10, 15, 13, 8, 12 OR (b) The following table shows the ages of patients admitted in a hospital during a year : Age (in years): 5-15, 15-25, 25-35, 35-45, 45-55, 55-65 Number of Patients: 7, 10, 21, 22, 15, 5 Find 'mode' and 'median' of the above data.

    How the model answer begins

    1. (a) Given: the class intervals and the number of students (frequency fᵢ). We find the mean by the step deviation method, taking the assumed mean a = 47.5 and the class size h = 15, so that uᵢ = (xᵢ − a)/h, where xᵢ is…

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15. Q15Long answer5 marksAsked: 2025

    Find 'mean' and 'mode' of the following data : Class: 15-20, 20-25, 25-30, 30-35, 35-40, 40-45 Frequency: 12, 10, 15, 11, 7, 5

    How the model answer begins

    1. Given data: classes 15-20, 20-25, 25-30, 30-35, 35-40, 40-45 with frequencies 12, 10, 15, 11, 7, 5. We find the mean by the step deviation method, taking the assumed mean a = 27.5 and class size h = 5, where u_i = (x_i…

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16. Q16Long answer5 marksAsked: 2025

    Find the 'mean' and 'mode' of the following data: Marks: 0-5, 5-10, 10-15, 15-20, 20-25, 25-30 Number of students: 2, 3, 8, 15, 14, 8

    How the model answer begins

    1. Class marks xᵢ: 2.5, 7.5, 12.5, 17.5, 22.5, 27.5; fᵢxᵢ: 5, 22.5, 100, 262.5, 315, 220.

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17. Q17Long answer5 marksAsked: 2024

    The daily wages of 50 workers are shown in the following distribution: [Table: Daily wages (in Rs) 100-120, 120-140, 140-160, 160-180, 180-200; Number of workers 12, 14, 8, 6, 10] Find the mean and median of the above data.

    How the model answer begins

    1. The mean daily wage of the 50 workers is Rs 145.20 and the median daily wage is Rs 138.57 (approx.). Given: n = 50 workers, class size h = 20. Take the assumed mean a = 150 and ui = (xi − 150)/20, where xi is the class…

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18. Q18Long answer5 marksAsked: 2024

    Find the mean and median of the following distribution : Class: 0-20, 20-40, 40-60, 60-80, 80-100, 100-120 Frequency: 5, 8, 10, 12, 7, 8

    How the model answer begins

    1. Given: a grouped frequency distribution with six classes of class size h = 20. We take the assumed mean a = 50, the class mark of 40 - 60, and use u_i = (x_i − a)/h. The class mark x_i is the mid-point of each class.…

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19. Q19Long answer5 marksAsked: 2024

    The marks obtained by 45 students of a class in a test are given below : Marks: 40-45, 45-50, 50-55, 55-60, 60-65, 65-70. No. of Students: 8, 9, 10, 9, 5, 4. Find the mean and median marks.

    How the model answer begins

    1. The mean marks are 53.2 (approx.) and the median marks are 52.75. Given: the marks of n = 45 students in class intervals of class size h = 5. Take the assumed mean a = 52.5 (the mid-value of 50 - 55) and u = (x − a)/h,…

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20. Q20Long answer5 marksAsked: 2024

    The following frequency distribution table gives the monthly consumption of electricity of 70 consumers of a locality. | Monthly consumption (in units) | Number of consumers | | --- | --- | | 65-85 | 7 | | 85-105 | 8 | | 105-125 | 7 | | 125-145 | 20 | | 145-165 | 14 | | 165-185 | 9 | | 185-205 | 5 | Find the median of the data.

    How the model answer begins

    1. The median monthly consumption of electricity of the 70 consumers is 138 units. It is worked out below. Given: the number of consumers in each class of monthly consumption, with n = N = 70. First, find the cumulative…

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61 more questions of this chapter, each with its model answer, in the Ask Medha app.

Objective questions with the board’s key (10)

1. 1. CBSE 2026

   If the mean and mode of a data are 12 and 21 respectively, then its median is :

   • (A) 6
   • (B) 13.5
   • (C) 15 ✓
   • (D) 14

   Answer: (C)

2. 2. CBSE 2026

   The mean and median of a frequency distribution are 43 and 43.4 respectively. The mode of the distribution is :

   • (A) 43.4
   • (B) 42.4
   • (C) 44.2 ✓
   • (D) 49.3

   Answer: (C)

3. 3. CBSE 2026

   Assertion (A) : The mean of first 'n' natural numbers is (n - 1)/2. Reason (R): The sum of first 'n' natural numbers is n(n + 1)/2.

   • (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
   • (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
   • (C) Assertion (A) is true, but Reason (R) is false.
   • (D) Assertion (A) is false, but Reason (R) is true. ✓

   Answer: (D)

4. 4. CBSE 2026

   The median and mode of a distribution are 25.2 and 26.1 respectively. The mean of this distribution is :

   • (A) 24.75 ✓
   • (B) 24.25
   • (C) 24.3
   • (D) 25.5

   Answer: (A)

5. 5. CBSE 2026

   While calculating mean of a grouped frequency distribution using step deviation method (u = (x - a)/h) it was found that x_bar = 62, a = 47.5, h = 5. The value of u_bar is :

   • (A) 3
   • (B) 14.5
   • (C) 2.9 ✓
   • (D) 3.1

   Answer: (C)

6. 6. CBSE 2026

   While calculating mean of a grouped frequency distribution using step deviation method (u = (x - a)/h) it was found that x̄ = 62, a = 47.5, h = 5. The value of ū is :

   • (A) 3
   • (B) 14.5
   • (C) 2.9 ✓
   • (D) 3.1

   Answer: (C)

7. 7. CBSE 2026

   While calculating mean of a grouped frequency distribution, step deviation method was used ((x-a)/h = u). It was found that x̄ = 64, h = 5 and a = 62.5. The value of ū is

   • (A) 0.5
   • (B) 1.5
   • (C) 0.3 ✓
   • (D) 7.5

   Answer: (C)

8. 8. CBSE 2026

   While calculating mean of a grouped frequency distribution, step deviation method was used ((x-a)/h = u). It was found that x_bar = 64, h = 5 and a = 62.5. The value of u is

   • (A) 0.5
   • (B) 1.5
   • (C) 0.3 ✓
   • (D) 7.5

   Answer: (C)

9. 9. CBSE 2026

   While calculating mean of a grouped frequency distribution, step deviation method was used ((x-a)/h = u). It was found that x_bar = 64, h = 5 and a = 62.5. The value of u_bar is

   • (A) 0.5
   • (B) 1.5
   • (C) 0.3 ✓
   • (D) 7.5

   Answer: (C)

10. 10. CBSE 2026

    Assertion (A) : Median of a data is the value of N/2, where N represents sum of all frequencies. Reason (R) : Median divides the whole distribution in two equal parts.

    • (A) Both, Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
    • (B) Both, Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
    • (C) Assertion (A) is true, but Reason (R) is false.
    • (D) Assertion (A) is false, but Reason (R) is true. ✓

    Answer: (D)

Questions parents and students ask

How many marks does "Statistics" carry in CBSE Class 10 Maths?

In the CBSE papers of 2023–2026, this chapter carried about 6 marks and about 2 questions a year; by marks it ranks 6 in Maths.

Which are the most important (VVI) questions of "Statistics"?

The 20 questions above are from the board’s real papers, each with the years it was asked; the most-asked come first with their full model answers. All 81 written questions of this chapter, with answers, are in the Ask Medha app.

Do the answers follow the board’s marking scheme?

Yes. CBSE publishes its marking schemes; every model answer here is written point by point to the marks the board’s own scheme gives, and checked by a second, independent model.