HOTS Questions on Class 10 Maths Chapter 13 ‘Statistics’ with Answers

Higher‑Order Thinking Skills (HOTS) question set for Class 10 Maths Chapter 13: ‘Statistics’ is designed to move students beyond routine calculations and cultivate problem‑solving abilities using statistical reasoning. Aligned with CBSE and NCERT learning outcomes, the tasks emphasise conceptual understanding of measures of central tendency, empirical relation and interpretation and representation of data. Each question includes a clear answer and brief reasoning to support classroom discussion and self-study. Download the PDF to access the same HOTS questions in a printable format, making offline learning and exam revision easier.

Solved HOTS Questions on Statistics

Question 1: The mean of the following distribution is 18. Find the value of f.

Solution: 

Class midpoints (xᵢ): 12, 14, 16, 18, 20, 22, 24

Σf = 3 + 6 + 9 + 13 + f + 5 + 4 = 40 + f  

Σfᵢxᵢ = 3(12) + 6(14) + 9(16) + 13(18) + f(20) + 5(22) + 4(24) 

= 36 + 84 + 144 + 234 + 20f + 110 + 96 

= 704 + 20f

Mean = Σfᵢxᵢ/Σf 

⇒ 18 = (704 + 20f)/(40 + f)

18(40 + f) = 704 + 20f 

⇒ 720 + 18f = 704 + 20f 

⇒ 16 = 2f

∴ f = 8

Question 2: A student computes the average of 10 two-digit positive integers. By mistake, one number's digits are swapped. This makes the mean 2.7 less than the original. Find the difference between the digits of that number.

Solution: 

Let the digits of the wrongly-written number be 'a' (tens) and 'b' (units), so the number is 10a+b. When swapped, it becomes 10b+a.

Difference in the number = (10a+b) – (10b+a) = 9(a–b)

This reduces the total sum by 9(a–b), so mean reduces by 9(a–b)/10 = 2.7

9(a–b) = 27 ⇒ a–b = 3

∴ Difference between digits = 3

Question 3: The mean of 11 numbers is 35. The mean of the first 6 numbers is 32 and the mean of the last 6 numbers is 37. Find the 6th number.

Solution: 

Sum of all 11 = 35 × 11 = 385

Sum of first 6 = 32 × 6 = 192 

Sum of last 6 = 37 × 6 = 222

Sum of first 6 + Sum of last 6 = 192 + 222 = 414, but this counts the 6th number twice.

414 = 385 + 6th number 

⇒ 6th number = 414 – 385

∴ 6th number = 29

Question 4: The median of the following frequency distribution is 35. Find the value of x and also identify the modal class.

Solution: 

Σf = 2 + 3 + x + 6 + 5 + 3 + 2 = 21 + x. 

Since median = 35 lies in 30–40, median class = 30–40.

cf before 30 - 40 = 2 + 3 + x = 5 + x

f = 6; l = 30; h = 10; n/2 = (21 + x)/2

35 = 30 + [((21 + x)/2 – (5 + x)) / 6] × 10

5 = [(21 + x – 10 – 2x)/2 / 6] × 10 

⇒ 5 = [(11 – x)/12] × 10 

⇒ 6 = (11 – x) × 10/12 

⇒ 6×12 = 10(11 – x) 

⇒ 72 = 110 – 10x 

⇒ 10x = 38 

⇒ x = 3.8 ≈ 4 (round to nearest whole number or exact value x = 4)

With x = 4: frequencies are 2,3,4,6,5,3,2 

⇒ Maximum frequency = 6 in class 30 - 40.

∴ x = 4, Modal Class = 30 - 40

Question 5: The mean of the following data is 62.8 and the total number of observations is 50. Find the missing frequencies f₁ and f₂.

Solution: 

Σf = 5 + f₁ + 10 + f₂ + 7 + 8 

= 30 + f₁ + f₂ = 50 

⇒ f₁ + f₂ = 20  …(i)

xᵢ: 10, 30, 50, 70, 90, 110 

⇒ Σfᵢxᵢ = 5(10) + f₁(30) + 10(50) + f₂(70) + 7(90) + 8(110) 

= 50 + 30f₁ + 500 + 70f₂ + 630 + 880 

= 2060 + 30f₁ + 70f₂

Mean = 62.8 

⇒ 62.8 = (2060 + 30f₁ + 70f₂)/50 

⇒ 3140 = 2060 + 30f₁ + 70f₂ 

⇒ 30f₁ + 70f₂ = 1080 

⇒ 3f₁ + 7f₂ = 108 …(ii)

From (i): f₁ = 20 – f₂. 

Substituting in (ii): 3(20 – f₂) + 7f₂ = 108 

⇒ 60 – 3f₂ + 7f₂ = 108 

⇒ 4f₂ = 48 

⇒ f₂ = 12

∴ f₁ = 8, f₂ = 12 

Question 6: The weights of 35 students of a class were recorded as a cumulative frequency distribution as shown below. Draw a 'less than' type ogive and find the median weight from the graph. Verify using the formula.

 

Solution: 

From graph: n/2 = 35/2 = 17.5. 

Draw a horizontal line at y = 17.5. Where it meets the ogive, drop a perpendicular to the x-axis. The x-value gives the median ≈ 46.5 kg.

Verification using formula: 

Convert to frequency table: 

 

n/2 = 17.5 

⇒ cf just ≥ 17.5 is 28 

⇒ Median class = 46–48.

l = 46, cf = 14, f = 14, h = 2 

⇒ Median = 46 + [(17.5–14)/14] × 2 

= 46 + (3.5/14)×2 

= 46 + 0.5 

= 46.5 kg 

∴ Median Weight = 46.5 kg

Question 7: he following table gives the distribution of income of 100 families in a town. Draw 'less than' and 'more than' ogives for this data. Use the graph to find the median income. Also verify using the formula.

Solution: 

From the graph: The two ogives intersect at approx. x = 243. 

Draw perpendicular to x-axis 

⇒ Median ≈ ₹243 thousand.

Formula verification: n = 100, n/2 = 50 

⇒ cf just ≥ 50 is 53 

⇒ Median class = 200 - 250.

l = 200, cf = 28, f = 25, h = 50 

⇒ Median = 200 + [(50–28)/25]×50 

= 200 + (22/25)×50 

= 200 + 44 = ₹244 thousand

∴ Median Income ≈ ₹243 - 244 thousand

Questions PDF with worked-out examples for Class 10 Chapter 13: Statistics, perfect for last-minute CBSE exam revision.

Class 10 Chapter 13: Statistics HOTS PDF

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