Class 10 Maths Chapter 13 Statistics Extra Questions (HOTS)

Genuinely harder, HOTS-level practice for Class 10 Maths Chapter 13 (Statistics), going beyond the textbook exercises with fresh numbers, general proofs, and reverse-engineering problems. These Class 10 Mathematics Chapter 13 important questions are handy for last-minute exam practice.

1. Q1 (General proof). Show that the step-deviation mean formula is algebraically equivalent to the direct mean formula.
   Solution: Since uᵢ=(xᵢ-a)/h, xᵢ=a+huᵢ. So Σfᵢxᵢ=aΣfᵢ+hΣfᵢuᵢ. Dividing by Σfᵢ: Σfᵢxᵢ/Σfᵢ = a+h(Σfᵢuᵢ/Σfᵢ) — the direct-method mean equals the step-deviation formula for any a and h. Hence proved — both methods always give the same numeric mean.
2. Q2 (Reverse-engineering). The mean of a distribution with classes 0-20(5), 20-40(f₁), 40-60(10), 60-80(f₂), 80-100(7), 100-120(8), total 50, is 62.8. Find f₁ and f₂.
   Solution: 5+f₁+10+f₂+7+8=50 ⇒ f₁+f₂=20. Step-deviation a=50,h=20: Σfᵢuᵢ=28-f₁+f₂. 62.8=50+[(28-f₁+f₂)/50]×20 ⇒ f₂-f₁=4. Solving: f₁=8, f₂=12.
3. Q3 (Assertion-Reason). Assertion (A): Mode = l+[(f₁-f₀)/(2f₁-f₀-f₂)]×h. Reason (R): The modal class is the class with the highest cumulative frequency.
   (a) Both true, R explains A  (b) Both true, R doesn’t explain A  (c) A true, R false  (d) A false, R true
   Solution: A is the correct mode formula (true). R is false — the modal class has the highest frequency, not highest cumulative frequency (cf is always largest for the last class). Answer: (c).
4. Q4 (Fresh numbers, mean). Marks of 60 students: 0-10(3),10-20(9),20-30(15),30-40(18),40-50(11),50-60(4). Find the mean by step-deviation.
   Solution: a=35,h=10: Σfᵢuᵢ=-23. Mean=35+(-23/60)×10=31.17 marks (approx.).
5. Q5 (Fresh numbers, mode). Daily max temperature (°C) over 50 days: 20-24(6),24-28(10),28-32(16),32-36(12),36-40(6). Find the modal temperature.
   Solution: Modal class 28-32 (f=16): l=28,f₁=16,f₀=10,f₂=12,h=4. Mode=28+[6/10]×4=30.4°C.
6. Q6 (Fresh numbers, median). Weekly wages (Rs hundred) of 45 workers: 0-10(5),10-20(8),20-30(20),30-40(7),40-50(5). Find the median wage.
   Solution: cf:5,13,33,40,45. n=45,n/2=22.5; median class 20-30. l=20,cf=13,f=20,h=10. Median=20+[(22.5-13)/20]×10=24.75, i.e. Rs 2475.
7. Q7 (Verify empirical relationship). For classes 10-20(4),20-30(8),30-40(14),40-50(18),50-60(10),60-70(6), find mean, median, mode and verify 3×Median = Mode+2×Mean.
   Solution: Mean (a=35,step-dev): Σfᵢuᵢ=40, Mean=35+6.67=41.67. Median (n=60,n/2=30,cf 4,12,26,44,54,60,class 40-50): Median=40+2.22=42.22. Mode (modal class 40-50,f=18): Mode=40+3.33=43.33. Check: 3×42.22=126.67; 43.33+2(41.67)=126.67. Verified.

More on This Chapter

• NCERT Solutions – Chapter 13
• Extra Questions (HOTS) – Chapter 13
• Revision Notes – Chapter 13
• Class 10 Maths Book (Catalog Page)

Written by Satish

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