Class 10 Maths Chapter 14 Statistics Practice Paper 4

PRACTICE PAPER 04 (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 range of the data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 is:
(a) 10
(b) 15
(c) 18
(d) 26

2. In the formula x̄ = a + h × ( Σ f u /Σf ), u is equal to: i i i i
(a) (x + a)/h i
(b) h(x - a) i
(c) (x - a)/h i
(d) (a - x i )/h

3. The distribution given below shows the number of wickets taken by bowlers in one-day cricket matches: Number of Less than Less than Less than Less than Less than Less than wickets 15 30 45 60 75 90 Number of 2 5 9 17 39 54 bowlers The number of bowlers who have taken 45 or more but less than 60 wickets is:
(a) 22
(b) 8
(c) 17
(d) 39

4. For a given data with 60 observations, the "less than ogive" and "more than ogive" intersect at (25.5, 30). The median of the data is:
(a) 30
(b) 25.5
(c) 55.5
(d) Cannot be determined

5. If the mean of the first n natural numbers is 5n/9, then n is:
(a) 5
(b) 7
(c) 9
(d) 10

6. Consider the data: Class 0-10 10-20 20-30 30-40 40-50 50-60 Frequency 8 10 12 22 30 18 The modal class is:
(a) 30-40
(b) 40-50
(c) 10-20
(d) 50-60

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

8. The median of first 10 prime numbers is:
(a) 11
(b) 12
(c) 13
(d) 14

9. Assertion
(a) : If the median of data is 350 and mean is 320, then the mode is approximately 410. Reason (R): Mode = 3 Median - 2 Mean
(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) : The sum of deviations of the observations from their mean is always zero. Reason (R): Mean is a measure of central tendency.
(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. The mean of six numbers is 23. If one of them is excluded, the mean of remaining numbers is 20. Find the excluded number.

12. Find the mode of the following data: Size of shoes 3 4 5 6 7 8 9 Number of pairs sold 4 18 25 12 5 3 2

13. If the median of the observations 10, 11, 13, 17, x+5, 20, 22, 24, 25 arranged in ascending order is 18, find the value of x.

14. The following table shows marks obtained by students: Marks More than 0 More than 10 More than 20 More than 30 More than 40 No. of students 40 35 28 20 10 Write the frequency distribution table for the above data. SECTION – C Questions 15 to 17 carry 3 marks each.

15. The following table shows the ages of patients treated in a hospital on a particular day: Age (in years) 10-20 20-30 30-40 40-50 50-60 60-70 Number of patients 10 18 22 16 14 8 Find the mode of the data.

16. Find the median of the following data: Class 10-20 20-30 30-40 40-50 50-60 60-70 70-80 Frequency 4 6 10 15 12 8 5

17. If the mean and median of a set of numbers are 8.9 and 9 respectively, find the mode. SECTION – D Question 18 carries 5 marks.

18. The following frequency distribution gives the monthly consumption of electricity of 80 consumers of a locality: Monthly consumption (in 65- 85- 105- 125- 145- 165- 185- units) 85 105 125 145 165 185 205 Number of consumers 4 5 13 20 14 8 4 Find the median, mean and mode of the data. OR If the median of the following frequency distribution is 46, find the missing frequencies: Class 10-20 20-30 30-40 40-50 50-60 60-70 70-80 Total f f Frequency 12 30 1 65 2 25 18 229 Also find the mode of the data.

SECTION – E (Case Study Based Questions) Questions 19 to 20 carry 4 marks each.

19. Mobile Phone Usage Survey A survey was conducted to study the daily screen time (in hours) of students on mobile phones. The data collected from 70 students is given below: Screen Time (hours) Number of Students 0-2 5 2-4 12 4-6 20 6-8 18 8-10 10 10-12 5 Based on the above information, answer the following questions:
(a) Which class interval has the maximum number of students? (1 mark)
(b) Find the mode of the data. (2 marks)
(c) How many students spend less than 6 hours on screen? (1 mark)

20. Mathematics Test Scores In a mathematics test, 100 students scored as follows: Marks Number of Students 0-20 5 20-40 10 40-60 25 60-80 40 80-100 20 Based on the above information, answer the following questions:
(a) Find the median class. (1 mark)
(b) Calculate the median marks. (2 marks)
(c) How many students scored 60 or more marks? (1 mark) DETAILED ANSWER KEY PRACTICE PAPER 04 - STATISTICS Answer 1:
(d) 26 Formula: Range = Maximum value - Minimum value Maximum value = 32 Minimum value = 6 Range = 32 - 6 = 26 ✓ Correct Answer:
(d) 26 Answer 2:
(c) (x i - a)/h Explanation: In step deviation method, u = (x - a)/h, where x is the class mark, a is the assumed mean, i i i and h is the class width.

✓ Correct Answer:
(c) Answer 3:
(b) 8 Explanation: Number of bowlers with 45-60 wickets = (Less than 60) - (Less than 45) = 17 - 9 = 8 ✓ Correct Answer:
(b) 8 Answer 4:
(b) 25.5 Explanation: The x-coordinate (abscissa) of the point of intersection of the less than ogive and more than ogive gives the median of the data. ✓ Correct Answer:
(b) 25.5 Answer 5:
(c) 9 Step 1: Mean of first n natural numbers = (n+1)/2 Step 2: Given: (n+1)/2 = 5n/9 9(n+1) = 10n 9n + 9 = 10n n = 9 ✓ Correct Answer:
(c) 9 Answer 6:
(b) 40-50 Explanation: Modal class is the class with maximum frequency.

Maximum frequency = 30 in class 40-50 ✓ Correct Answer:
(b) 40-50 Answer 7:
(d) centred at the class marks of the classes Explanation: While computing mean of grouped data, we assume that all observations in each class are centered at the class mark (midpoint) of that class. ✓ Correct Answer:
(d) Answer 8:
(b) 12 Step 1: First 10 prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 Step 2: n = 10 (even) Median = (5th term + 6th term)/2 = (11 + 13)/2 = 24/2 = 12 ✓ Correct Answer:
(b) 12 Answer 9:
(a) Both A and R are true and R is the correct explanation of A Verification:

Mode = 3 Median - 2 Mean = 3(350) - 2(320) = 1050 - 640 = 410 ✓ Both assertion and reason are true, and the reason correctly explains the assertion. ✓ Correct Answer:
(a) Answer 10:
(b) Both A and R are true but R is not the correct explanation of A Assertion: Σ(x - x̄) = 0 is TRUE (property of mean) i Reason: Mean is a measure of central tendency is TRUE However, R doesn't explain WHY the sum of deviations is zero. It's just a definition. ✓ Correct Answer:
(b) Answer 11: Step 1: Sum of 6 numbers = 23 × 6 = 138 Step 2: Sum of remaining 5 numbers = 20 × 5 = 100 Step 3: Excluded number = 138 - 100 = 38 ✓ Excluded number = 38 Answer 12:

Finding mode: Number of pairs sold: 4, 18, 25, 12, 5, 3, 2 Maximum frequency = 25 Mode = Size with maximum frequency = 5 ✓ Mode = Size 5 Answer 13: Step 1: Data in ascending order: 10, 11, 13, 17, x+5, 20, 22, 24, 25 n = 9 (odd) Step 2: Median = 5th term = x + 5 Given: Median = 18 Step 3: x + 5 = 18 x = 13 ✓ x = 13 Answer 14: Frequency Distribution Table: Marks Frequency 0-10 40 - 35 = 5 10-20 35 - 28 = 7 20-30 28 - 20 = 8 30-40 20 - 10 = 10 40-50 10 ✓ Table shown above Answer 15: Step 1: Modal class = 30-40 (maximum frequency = 22) Step 2: l = 30, f 1 = 22, f 0 = 18, f 2 = 16, h = 10 Step 3: Mode = l + [(f 1 - f 0 )/(2f 1 - f 0 - f 2 )] × h = 30 + [(22-18)/(44-18-16)] × 10 = 30 + [4/10] × 10 = 30 + 4 = 34 years ✓ Mode = 34 years Answer 16:

Step 1: Calculate cumulative frequency: Class Frequency CF 10-20 4 4 20-30 6 10 30-40 10 20 40-50 15 35 50-60 12 47 60-70 8 55 70-80 5 60 Step 2: N = 60, N/2 = 30 Median class = 40-50 (cf = 35 > 30) Step 3: l = 40, cf = 20, f = 15, h = 10 Median = 40 + [(30-20)/15] × 10 = 40 + 6.67 = 46.67 ✓ Median = 46.67 Answer 17: Given: Mean = 8.9, Median = 9 Using empirical formula: Mode = 3 Median - 2 Mean = 3(9) - 2(8.9) = 27 - 17.8 = 9.2 ✓ Mode = 9.2 Answer 18: Part 1: Finding Median Consumption Frequency CF 65-85 4 4 85-105 5 9 105-125 13 22 125-145 20 42 145-165 14 56 165-185 8 64 185-205 4 68 N/2 = 34, Median class = 125-145 Median = 125 + [(34-22)/20] × 20 = 125 + 12 = 137 units Part 2: Finding Mean Using direct method: Mean = Σf i x i /Σf i = (4×75 + 5×95 + 13×115 + 20×135 + 14×155 + 8×175 + 4×195)/68 = 9300/68 = 136.76 units Part 3: Finding Mode Modal class = 125-145 (f = 20) Mode = 125 + [(20-13)/(40-13-14)] × 20 = 125 + (7/13) × 20 = 125 + 10.77 = 135.77 units ✓ Median = 137, Mean = 136.76, Mode = 135.77 units OR (Alternative Question):

Step 1: f + f = 229 - (12+30+65+25+18) = 79 ... (i) 1 2 Step 2: Median = 46 lies in class 40-50 cf before = 12 + 30 + f = 42 + f 1 1 46 = 40 + [(114.5-(42+f ))/65] × 10 1 6 = [(72.5-f 1 )/65] × 10 39 = 72.5 - f 1 f = 33.5 ≈ 34 (rounding to nearest integer) 1 Step 3: f = 79 - 34 = 45 2 Step 4: Mode Modal class = 40-50 (f = 65) Mode = 40 + [(65-34)/(130-34-45)] × 10 = 40 + (31/51) × 10 = 40 + 6.08 = 46.08 ✓ OR: f₁ = 34, f₂ = 45, Mode = 46.08 Answer 19:
(a) Maximum students = 20 in class 4-6 hours
(b) Finding Mode:

Modal class = 4-6, l = 4, f = 20, f = 12, f = 18, h = 2 1 0 2 Mode = 4 + [(20-12)/(40-12-18)] × 2 = 4 + (8/10) × 2 = 4 + 1.6 = 5.6 hours
(c) Students with < 6 hours = 5 + 12 + 20 = 37 ✓
(a) 4-6 hours
(b) 5.6 hours
(c) 37 students Answer 20:
(a) Finding median class: CF: 5, 15, 40, 80, 100 N/2 = 50, Median class = 60-80 (cf = 80 > 50)
(b) Calculating median: l = 60, cf = 40, f = 40, h = 20 Median = 60 + [(50-40)/40] × 20 = 60 + 5 = 65 marks
(c) Students with ≥ 60 marks = 40 + 20 = 60 ✓
(a) 60-80
(b) 65 marks
(c) 60 students END OF ANSWER KEY