Statistics CBQs — CBSE Class 10 (2027) with step-wise solutions

Sample Statistics CBQs

1. Question 1MCQ1 mark
   A transport survey recorded how long 25 commuters waited at a bus stop one morning: Waiting time (in minutes): 0-4, 4-8, 8-12, 12-16, 16-20 Number of commuters: 3, 7, 8, 5, 2 What is the mean waiting time?
   • 9.36 minutes
   • 10 minutes
   • 9.6 minutes
   • 8.36 minutes
   Show step-wise solution

   1. [1 mark] Class marks 2, 6, 10, 14, 18; Σfx = 6 + 42 + 80 + 70 + 36 = 234 and Σf = 25, so mean = 234 ÷ 25 = 9.36 minutes.

   Answer: 9.36 minutes

2. Question 2Assertion-Reason1 mark
   The times taken by 30 students to reach school are: Time (in minutes): 0-10, 10-20, 20-30, 30-40 Number of students: 5, 12, 8, 5 Assertion (A): The modal class of the data is 10-20. Reason (R): The modal class is the class with the highest frequency.
   • (A) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
   • (B) Both Assertion and Reason are true but Reason is NOT the correct explanation of Assertion.
   • (C) Assertion is true but Reason is false.
   • (D) Assertion is false but Reason is true.
   Show step-wise solution

   1. [1 mark] (A): The highest frequency, 12, belongs to 10-20, so A is true; R is the definition of the modal class and is exactly why A holds.

   Answer: (A)

3. Question 3Short answer2 marks
   A bookstore manager records the time (in minutes) that 60 customers spend in the store during one Saturday, to decide how long a typical browsing visit lasts. Time (in minutes) : Number of customers 0–10 : 8 10–20 : 14 20–30 : 22 30–40 : 10 40–50 : 6 Find the modal time spent in the store.
   Show step-wise solution

   1. [1 mark] Highest frequency = 22, so the modal class is 20–30: l = 20, h = 10, f₁ = 22, f₀ = 14, f₂ = 10.
   2. [1 mark] Mode = l + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h = 20 + [(22 − 14)/(44 − 14 − 10)] × 10 = 20 + (8/20) × 10 = 24 minutes.

   Answer: 24 minutes

4. Question 4Case study4 marks
   Under the Jal Jeevan Mission, the resident welfare association of a housing society in Pune fitted water meters in 40 flats and recorded the average water used per flat per day for one month. The data are given below. Water used (litres per day): 100-150, 150-200, 200-250, 250-300, 300-350, 350-400 Number of flats: 4, 7, 12, 9, 5, 3 Based on the above information, answer the following questions: (i) Write the modal class of the data. (ii) The society calls a flat a 'heavy user' if it uses 300 litres or more per day. How many flats are heavy users? (iii) Using the assumed mean method with assumed mean 225, find the mean water used per flat per day.
   Show step-wise solution

   1. [1 mark] (i) The highest frequency is 12, for the class 200-250, so the modal class is 200-250.
   2. [1 mark] (ii) The classes 300-350, 350-400 have frequencies 5 + 3 = 8, so the answer is 8 flats.
   3. [1 mark] (iii) Class marks x = 125, 175, 225, 275, 325, 375; with a = 225, d = x − a = −100, −50, 0, 50, 100, 150; Σf = 40 and Σfd = 650.
   4. [1 mark] (iii) Mean = a + Σfd/Σf = 225 + (650)/40 = 241.25 litres.

   Answer: (i) 200-250 (ii) 8 flats (iii) 241.25 litres

5. Question 5Long answer5 marks
   A car assembly unit assembles a limited number of cars daily, depending on the prevailing demand. The table presents an analysis: cars assembled per day (0-4: 33 days, 4-8: 18 days, 8-12: 21 days, 12-16: 11 days, 16-20: 7 days). (i) If the demand of the cars is doubled, estimate how many cars on an average should be assembled per day to meet the increased demand? (ii) At least on how many days, less than average number of cars were assembled?
   Show step-wise solution

   1. [1 mark] Class marks xᵢ: 2, 6, 10, 14, 18; frequencies fᵢ: 33, 18, 21, 11, 7; Σfᵢ = 90.
   2. [1 mark] Σfᵢ·xᵢ = 33·2 + 18·6 + 21·10 + 11·14 + 7·18 = 66 + 108 + 210 + 154 + 126 = 664.
   3. [1 mark] Mean = Σfᵢxᵢ/Σfᵢ = 664/90 ≈ 7.38 cars per day.
   4. [1 mark] (i) For doubled demand, average required = 2 × 7.38 ≈ 14.76 ≈ 15 cars per day.
   5. [1 mark] (ii) The mean 7.38 lies in class 4-8; all 33 days in class 0-4 had fewer than 7.38 cars assembled, so on at least 33 days production was below average.

   Answer: (i) ~15 cars/day; (ii) at least 33 days.

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