Class 10 Maths Chapter 14 Statistics Practice Paper 1

PRACTICE PAPER 01 (2025-26) CHAPTER 13: STATISTICS SUBJECT: Mathematics CLASS: X MAX. MARKS: 40 DURATION: 1½ hrs

General Instructions:

1. All questions are compulsory.

2. This question paper contains 20 questions divided into five Sections A, B, C, D and E.

3. Section A comprises of 10 MCQs of 1 mark each. Section B comprises of 4 questions of 2 marks each.

Section C comprises of 3 questions of 3 marks each. Section D comprises of 1 question of 5 marks

and Section E comprises of 2 Case Study Based Questions of 4 marks each.

4. There is no overall choice.

5. Use of Calculators is not permitted. SECTION – A Questions 1 to 10 carry 1 mark each.

1. The relationship between mean, median and mode for a moderately skewed distribution is:
(a) Mode = 3 Median – 2 Mean
(b) Mode = 2 Median – 3 Mean
(c) Mean = 3 Median – 2 Mode
(d) Median = 3 Mode – 2 Mean

2. While computing mean of grouped data, we assume that the frequencies are:
(a) evenly distributed over all the classes
(b) centred at the class marks of the classes
(c) centred at the upper limits of the classes
(d) centred at the lower limits of the classes

3. For the following distribution: Marks 0-10 10-20 20-30 30-40 40-50 Frequency 3 8 12 10 7 The sum of lower limits of the median class and modal class is:
(a) 40
(b) 50
(c) 30
(d) 45

4. If the mean of the first n natural numbers is 15, then the value of n is:
(a) 15
(b) 30
(c) 29
(d) 14

5. The median of the data: 3, 4, 5, 6, 7, 3, 4 is:
(a) 5
(b) 3
(c) 4
(d) 6

6. For a frequency distribution, mean, median and mode are connected by the relation:
(a) Mode = 3 Mean – 2 Median
(b) Mode = 2 Median – 3 Mean
(c) Mode = 3 Median – 2 Mean
(d) Mode = 3 Median + 2 Mean

7. Consider the data: Class 10-20 20-30 30-40 40-50 50-60 Frequency 4 6 8 10 6 The difference between the upper limit of the median class and the lower limit of the modal class is:
(a) 0
(b) 10
(c) 20
(d) 30

8. The class mark of the class 90-120 is:
(a) 90
(b) 105
(c) 115
(d) 120

9. Assertion
(a) : The mean of first 10 odd natural numbers is 10. Reason (R): The sum of first n odd natural numbers is n².
(a) Both assertion
(a) and reason (R) are true and reason (R) is the correct explanation of assertion
(a) .
(b) Both assertion
(a) and reason (R) are true but reason (R) is not the correct explanation of assertion
(a) .
(c) Assertion
(a) is true but reason (R) is false.
(d) Assertion
(a) is false but reason (R) is true.

10. Assertion
(a) : If xi represents class marks and fi represents corresponding frequencies, then mean = (Σfixi)/(Σfi). Reason (R): Class mark is the mid-point of upper and lower class limits.
(a) Both assertion
(a) and reason (R) are true and reason (R) is the correct explanation of assertion
(a) .
(b) Both assertion
(a) and reason (R) are true but reason (R) is not the correct explanation of assertion
(a) .
(c) Assertion
(a) is true but reason (R) is false.
(d) Assertion
(a) is false but reason (R) is true.

SECTION – B Questions 11 to 14 carry 2 marks each.

11. Find the mean of the following distribution using direct method: Class Interval 0-10 10-20 20-30 30-40 40-50 Frequency 5 10 15 8 2

12. The following table shows the ages of patients admitted in a hospital during a week: Age (in years) 5-15 15-25 25-35 35-45 45-55 55-65 No. of patients 6 11 21 23 14 5 Find the modal class.

13. If the median of the following frequency distribution is 32.5, find the values of f and f , if the total 1 2 frequency is 40. Class 0-10 10-20 20-30 30-40 40-50 50-60 Frequency f 5 9 12 f 3 1 2

14. The mean of 25 observations was found to be 78.4. Later it was detected that one observation 96 was wrongly copied as 69. Find the correct mean. SECTION – C Questions 15 to 17 carry 3 marks each.

15. Find the mode of the following frequency distribution: Class 20-30 30-40 40-50 50-60 60-70 70-80 Frequency 8 12 20 15 10 5

16. The following distribution gives the daily income of 50 workers of a factory: Daily Income (in ₹) 100-120 120-140 140-160 160-180 180-200 Number of workers 12 14 8 6 10 Find the median daily income.

17. Using empirical formula, find the mode of a distribution whose mean is 45 and median is 48. SECTION – D Question 18 carries 5 marks.

18. The following table gives the daily wages of workers in a factory: Daily Wages (in ₹) 100-150 150-200 200-250 250-300 300-350 350-400 Number of workers 10 18 25 22 15 10 Find the mean daily wages using step deviation method. Also find the mode and median daily wages. SECTION – E (Case Study Based Questions) Questions 19 to 20 carry 4 marks each.

19. COVID-19 Pandemic Data Analysis During the COVID-19 pandemic, a health organization collected data about the age distribution of infected patients in a city. The data is presented below: Age Group (in years) Number of Patients 0-20 15 20-40 28 40-60 42 60-80 30 80-100 10 Based on the above information, answer the following questions:
(a) Find the modal class. (1 mark)
(b) Find the mode of the given data. (2 marks)
(c) Which age group had the least number of infected patients? (1 mark)

20. Sports Survey in a School A school conducted a survey to find out the time (in hours per week) spent by students on sports activities. The data collected is shown below: Time (hours per week) Number of Students 0-5 8 5-10 16 10-15 x 15-20 12 20-25 6 If the mean time spent is 11.5 hours per week and the total number of students is 60, answer the following:
(a) Find the value of x. (2 marks)
(b) Find the median class. (1 mark)
(c) Find the median time spent by students on sports. (1 mark) DETAILED ANSWER KEY PRACTICE PAPER 01 - STATISTICS Answer 1:
(a) Mode = 3 Median – 2 Mean Explanation: This is the empirical relationship between mean, median and mode for a moderately skewed distribution.

✓ Correct Answer:
(a) Answer 2:
(b) centred at the class marks of the classes Explanation: While computing mean of grouped data, we assume that all observations in a class are centered at the class mark (mid-point) of that class. ✓ Correct Answer:
(b) Answer 3:
(a) 40 Step 1: Calculate cumulative frequency: N = 3 + 8 + 12 + 10 + 7 = 40 N/2 = 20 Step 2: Median class (cf just greater than 20): Cumulative frequencies: 3, 11, 23, 33, 40 Median class = 20-30 (lower limit = 20) Step 3: Modal class (highest frequency = 12):

Modal class = 20-30 (lower limit = 20) Step 4: Sum = 20 + 20 = 40 ✓ Correct Answer:
(a) 40 Answer 4:
(c) 29 Step 1: Mean of first n natural numbers = (n+1)/2 Step 2: Given: (n+1)/2 = 15 n + 1 = 30 n = 29 ✓ Correct Answer:
(c) 29 Answer 5:
(c) 4 Step 1: Arrange data in ascending order: 3, 3, 4, 4, 5, 6, 7 Step 2: Number of observations n = 7 (odd) Median = Middle term = 4th term = 4 ✓ Correct Answer:
(c) 4 Answer 6:
(c) Mode = 3 Median – 2 Mean Explanation: This is the empirical formula connecting mean, median and mode.

✓ Correct Answer:
(c) Answer 7:
(a) 0 Step 1: Modal class (highest frequency = 10): Modal class = 40-50, lower limit = 40 Step 2: Find median class: N = 34, N/2 = 17 Cumulative frequencies: 4, 10, 18, 28, 34 Median class = 30-40, upper limit = 40 Step 3: Difference = 40 - 40 = 0 ✓ Correct Answer:
(a) 0 Answer 8:
(b) 105 Formula: Class mark = (Upper limit + Lower limit)/2 Calculation: Class mark = (90 + 120)/2 = 210/2 = 105 ✓ Correct Answer:
(b) 105 Answer 9:
(a) Both A and R are true and R is the correct explanation of A Verification of Assertion:

First 10 odd natural numbers: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19 Sum = 100 = 10² (using formula n²) Mean = 100/10 = 10 ✓ Verification of Reason: Sum of first n odd natural numbers = n² ✓ Reason correctly explains why the mean is 10. ✓ Correct Answer:
(a) Answer 10:
(b) Both A and R are true but R is not the correct explanation of A Assertion A: Mean = Σfixi/Σfi is TRUE (formula for mean) Reason R: Class mark is mid-point of limits is TRUE However, R doesn't explain A. It just defines what xi represents, not why the formula works.

✓ Correct Answer:
(b) Answer 11: Direct Method: Class Frequency (f ) Class mark (x ) f x i i i i 0-10 5 5 25 10-20 10 15 150 20-30 15 25 375 30-40 8 35 280 40-50 2 45 90 Total Σf = 40 Σf x = 920 i i i Mean = Σf x /Σf = 920/40 = 23 i i i ✓ Mean = 23 Answer 12: Modal class: The class with maximum frequency Frequencies: 6, 11, 21, 23, 14, 5 Maximum frequency = 23 ✓ Modal class = 35-45 years Answer 13: Step 1: f + 5 + 9 + 12 + f + 3 = 40 1 2 f + f + 29 = 40 1 2 f 1 + f 2 = 11 ... (i) Step 2: Median = 32.5 lies in class 30-40 Step 3: Find cumulative frequency just before median class:

Cf = f + 5 + 9 = f + 14 1 1 Step 4: Using median formula: Median = l + [(N/2 - cf)/f] × h 32.5 = 30 + [(20 - (f + 14))/12] × 10 1 2.5 = [(6 - f 1 )/12] × 10 3 = 6 - f 1 f = 3 1 Step 5: From equation (i): f = 11 - 3 = 8 2 ✓ f = 3, f = 8 1 2 Answer 14: Step 1: Incorrect sum of observations = 78.4 × 25 = 1960 Step 2: Correct sum = 1960 - 69 + 96 = 1987 Step 3: Correct mean = 1987/25 = 79.48 ✓ Correct mean = 79.48 Answer 15: Step 1: Modal class (maximum frequency = 20) = 40-50 l = 40, f = 20, f = 12, f = 15, h = 10 1 0 2 Step 2: Mode = l + [(f - f )/(2f - f - f )] × h 1 0 1 0 2 Step 3: Mode = 40 + [(20 - 12)/(40 - 12 - 15)] × 10 = 40 + [8/13] × 10 = 40 + 80/13 = 40 + 6.15 = 46.15 ✓ Mode = 46.15 Answer 16:

Step 1: Calculate cumulative frequency: Daily Income Frequency Cumulative Frequency 100-120 12 12 120-140 14 26 140-160 8 34 160-180 6 40 180-200 10 50 Step 2: N = 50, N/2 = 25 Median class = 120-140 (cf = 26, just greater than 25) Step 3: l = 120, cf = 12, f = 14, h = 20 Median = l + [(N/2 - cf)/f] × h = 120 + [(25 - 12)/14] × 20 = 120 + (13/14) × 20 = 120 + 18.57 = 138.57 ✓ Median daily income = ₹138.57 Answer 17: Given: Mean = 45, Median = 48 Empirical Formula: Mode = 3 Median - 2 Mean Calculation:

Mode = 3(48) - 2(45) = 144 - 90 = 54 ✓ Mode = 54 Answer 18: Part 1: Mean using Step Deviation Method Let assumed mean A = 225, h = 50 Wages f x u = (x -225)/50 f u i i i i i i 10 125 -2 -20 100-150 150-200 18 175 -1 -18 200-250 25 225 0 0 250-300 22 275 1 22 300-350 15 325 2 30 350-400 10 375 3 30 Total 100 44 Mean = A + (Σf u /Σf ) × h i i i = 225 + (44/100) × 50 = 225 + 22 = 247 Part 2: Mode Modal class = 200-250 (highest frequency = 25) l = 200, f = 25, f = 18, f = 22, h = 50 1 0 2 Mode = 200 + [(25-18)/(50-18-22)] × 50 = 200 + [7/10] × 50 = 200 + 35 = 235 Part 3: Median N/2 = 50, Median class = 200-250 (cf = 28+25 = 53 > 50) l = 200, cf = 28, f = 25, h = 50 Median = 200 + [(50-28)/25] × 50 = 200 + 44 = 244 ✓ Mean = ₹247, Mode = ₹235, Median = ₹244 Answer 19:

(a) Modal class: Class with maximum frequency = 42 Modal class = 40-60 years
(b) Mode calculation: l = 40, f = 42, f = 28, f = 30, h = 20 1 0 2 Mode = l + [(f - f )/(2f - f - f )] × h 1 0 1 0 2 = 40 + [(42-28)/(84-28-30)] × 20 = 40 + [14/26] × 20 = 40 + 10.77 = 50.77 years
(c) Least number of patients = 10 in age group 80-100 years ✓
(a) 40-60 years
(b) 50.77 years
(c) 80-100 years Answer 20:
(a) Finding x: Total students: 8 + 16 + x + 12 + 6 = 60 x + 42 = 60 x = 18 Verification using mean: Class f i x i f i x i 0-5 8 2.5 20 5-10 16 7.5 120 10-15 18 12.5 225 15-20 12 17.5 210 20-25 6 22.5 135 Total 60 690 Mean = 690/60 = 11.5 ✓
(b) Median class:

N/2 = 30, cf: 8, 24, 42... Median class = 10-15 hours
(c) Median: l = 10, cf = 24, f = 18, h = 5 Median = 10 + [(30-24)/18] × 5 = 10 + 1.67 = 11.67 hours ✓
(a) x = 18
(b) 10-15 hours
(c) 11.67 hours END OF ANSWER KEY