NCERT Solutions for Class 10 Maths Chapter 13 Statistics (2026-27)

Class 10 Maths Chapter 13, Statistics, has three exercises in the current 2023-rationalised, 2026-27 session syllabus — Exercise 13.1 (9 questions, mean of grouped data), Exercise 13.2 (6 questions, mode of grouped data), and Exercise 13.3 (7 questions, median of grouped data) — 22 questions in all. Below are original, independently-derived, step-by-step solutions to all 22. These Class 10 Mathematics Chapter 13 solutions are also useful as quick revision notes before exams.

Why This Chapter Matters for Boards

Statistics is a high-weightage, high-scoring chapter in the CBSE Class 10 Maths board exam, since the mean/median/mode formulas are mechanical once the correct method (direct, assumed-mean, or step-deviation) is chosen. Questions frequently test converting inclusive/discontinuous class intervals to continuous form, finding a missing frequency from a given mean or median, and comparing all three measures for the same data.

Exercise 13.1 Solutions — Mean of Grouped Data

1. Q1. A survey of 20 houses recorded the number of plants: 0-2(1), 2-4(2), 4-6(1), 6-8(5), 8-10(6), 10-12(2), 12-14(3) houses. Find the mean number of plants per house.
   Solution: Direct Method (small values): class marks 1,3,5,7,9,11,13; fᵢxᵢ = 1,6,5,35,54,22,39; Σfᵢ = 20, Σfᵢxᵢ = 162. Mean = 162/20 = 8.1 plants.
2. Q2. Daily wages of 50 workers: 100-120(12), 120-140(14), 140-160(8), 160-180(6), 180-200(10). Find the mean using a suitable method.
   Solution: Step-Deviation, a=150, h=20; uᵢ = -2,-1,0,1,2; fᵢuᵢ = -24,-14,0,6,20; Σfᵢuᵢ = -12. Mean = 150 + (-12/50)×20 = 150 – 4.8 = Rs 145.20.
3. Q3. Pocket allowance data: 11-13(7), 13-15(6), 15-17(9), 17-19(13), 19-21(f), 21-23(5), 23-25(4). Mean is Rs 18. Find f.
   Solution: Assumed mean a=18 (matches given mean). fᵢdᵢ sums to 2f-40. Since mean=a, Σfᵢdᵢ=0 ⇒ 2f-40=0 ⇒ f = 20.
4. Q4. Heart beats/min of 30 women: 65-68(2), 68-71(4), 71-74(3), 74-77(8), 77-80(7), 80-83(4), 83-86(2). Find the mean.
   Solution: Step-Deviation, a=75.5, h=3; Σfᵢuᵢ = 4. Mean = 75.5 + (4/30)×3 = 75.9 beats/min.
5. Q5. Mangoes per box (inclusive classes): 50-52(15), 53-55(110), 56-58(135), 59-61(115), 62-64(25). Find the mean.
   Solution: Convert to continuous classes (49.5-52.5, etc., h=3). Step-Deviation, a=57: Σfᵢuᵢ = 25, Σfᵢ = 400. Mean = 57 + (25/400)×3 = 57.1875 mangoes.
6. Q6. Daily food expenditure of 25 households: 100-150(4), 150-200(5), 200-250(12), 250-300(2), 300-350(2). Find the mean.
   Solution: Step-Deviation, a=225, h=50: Σfᵢuᵢ = -7. Mean = 225 + (-7/25)×50 = 225-14 = Rs 211.
7. Q7. SO₂ concentration (ppm) in 30 localities: 0.00-0.04(4), 0.04-0.08(9), 0.08-0.12(9), 0.12-0.16(2), 0.16-0.20(4), 0.20-0.24(2). Find the mean.
   Solution: Step-Deviation, a=0.14, h=0.04: Σfᵢuᵢ = -31. Mean = 0.14 + (-31/30)×0.04 = 0.099 ppm (approx.).
8. Q8. Days absent for 40 students (unequal class widths): 0-6(11), 6-10(10), 10-14(7), 14-20(4), 20-28(4), 28-38(3), 38-40(1). Find the mean.
   Solution: Assumed-Mean Method (unequal widths, cannot use step-deviation shortcut), a=17: Σfᵢdᵢ = -181. Mean = 17 + (-181/40) = 12.48 days (approx.).
9. Q9. Literacy rate (%) of 35 cities: 45-55(3), 55-65(10), 65-75(11), 75-85(8), 85-95(3). Find the mean.
   Solution: Step-Deviation, a=70, h=10: Σfᵢuᵢ = -2. Mean = 70 + (-2/35)×10 = 69.43% (approx.).

Exercise 13.2 Solutions — Mode of Grouped Data

1. Q1. Ages of patients: 5-15(6), 15-25(11), 25-35(21), 35-45(23), 45-55(14), 55-65(5). Find the mode and mean, and compare.
   Solution: Modal class 35-45 (highest f=23): l=35,f₁=23,f₀=21,f₂=14,h=10. Mode = 35+[2/11]×10 = 36.8 years. Mean (assumed mean a=30): Σfᵢdᵢ=430, Mean=30+430/80=35.38 years. Most patients cluster near 36.8 (mode), average is slightly higher at 35.38.
2. Q2. Lifetimes (h) of 225 components: 0-20(10),20-40(35),40-60(52),60-80(61),80-100(38),100-120(29). Find the modal lifetime.
   Solution: Modal class 60-80 (f=61): l=60,f₁=61,f₀=52,f₂=38,h=20. Mode=60+[9/32]×20=65.625 hours.
3. Q3. Monthly expenditure of 200 families across 8 classes (1000-5000, step 500): 24,40,33,28,30,22,16,7. Find modal and mean expenditure.
   Solution: Modal class 1500-2000 (f=40): l=1500,f₁=40,f₀=24,f₂=33,h=500. Mode=1500+[16/23]×500=Rs 1847.83 (approx.). Mean (step-deviation a=2750): Σfᵢuᵢ=-35, Mean=2750+(-35/200)×500=Rs 2662.50.
4. Q4. Students-per-teacher ratio across states: 15-20(3),20-25(8),25-30(9),30-35(10),35-40(3),40-45(0),45-50(0),50-55(2). Find mode and mean; interpret.
   Solution: Modal class 30-35 (f=10): Mode=30+[1/8]×5=30.6 (approx.). Mean (step-deviation a=32.5): Σfᵢuᵢ=-23, Mean=32.5+(-23/35)×5=29.2 (approx.). Most states cluster near 30.6; nationwide average is slightly lower.
5. Q5. ODI runs of top batsmen (8 classes, 3000-11000, step 1000): 4,18,9,7,6,3,1,1. Find the mode.
   Solution: Modal class 4000-5000 (f=18): l=4000,f₁=18,f₀=4,f₂=9,h=1000. Mode=4000+[14/23]×1000=4608.7 runs (approx.).
6. Q6. Cars passing in 100 periods of 3 min (8 classes, 0-80, step 10): 7,14,13,12,20,11,15,8. Find the mode.
   Solution: Modal class 40-50 (f=20): l=40,f₁=20,f₀=12,f₂=11,h=10. Mode=40+[8/17]×10=44.7 cars (approx.).

Exercise 13.3 Solutions — Median of Grouped Data

1. Q1. Monthly electricity units for 68 consumers: 65-85(4),85-105(5),105-125(13),125-145(20),145-165(14),165-185(8),185-205(4). Find median, mean, mode; compare.
   Solution: n=68,n/2=34; cf: 4,9,22,42,56,64,68 ⇒ median class 125-145. Median=125+[(34-22)/20]×20=137. Mode (modal class 125-145,f=20): Mode=125+[7/13]×20=135.77 (approx.). Mean (step-dev a=135): Mean=135+(7/68)×20=137.06 (approx.). All three are close, showing a fairly symmetric distribution.
2. Q2. Median of the distribution 0-10(5),10-20(x),20-30(20),30-40(15),40-50(y),50-60(5), total 60, is 28.5. Find x and y.
   Solution: 5+x+20+15+y+5=60 ⇒ x+y=15. Median class 20-30 (l=20,cf=5+x,f=20,h=10): 28.5=20+[(30-5-x)/20]×10 ⇒ x=8. So x=8, y=7.
3. Q3. Ages of 100 policy holders given as “below” cumulative data (below 20 to below 60, step 5): 2,6,24,45,78,89,92,98,100. Find the median age.
   Solution: Convert to class frequencies: 2,4,18,21,33,11,3,6,2 (cf same as given). n=100,n/2=50; median class 35-40 (cf just >50 is 78). l=35,cf=45,f=33,h=5. Median=35+[(50-45)/33]×5=35.76 years (approx.).
4. Q4. Leaf lengths (mm) of 40 leaves, inclusive classes 118-126 to 172-180: 3,5,9,12,5,4,2. Find the median length.
   Solution: Convert to continuous classes (117.5-126.5, h=9). n=40,n/2=20; cf:3,8,17,29,34,38,40 ⇒ median class 144.5-153.5. l=144.5,cf=17,f=12,h=9. Median=144.5+[(20-17)/12]×9=146.75 mm.
5. Q5. Lifetime (h) of 400 neon lamps: 1500-2000(14),2000-2500(56),2500-3000(60),3000-3500(86),3500-4000(74),4000-4500(62),4500-5000(48). Find the median lifetime.
   Solution: cf: 14,70,130,216,290,352,400. n=400,n/2=200; median class 3000-3500 (first cf>200 is 216). l=3000,cf=130,f=86,h=500. Median=3000+[(200-130)/86]×500=3406.98 hours (approx.).
6. Q6. Letters in 100 surnames: 1-4(6),4-7(30),7-10(40),10-13(16),13-16(4),16-19(4). Find median, mean, modal number of letters.
   Solution: n=100,n/2=50; cf:6,36,76,92,96,100 ⇒ median class 7-10. Median=7+[(50-36)/40]×3=8.05. Mean (step-dev a=11.5): Σfᵢuᵢ=-106, Mean=11.5+(-106/100)×3=8.32. Mode (modal class 7-10,f=40): Mode=7+[10/34]×3=7.88.
7. Q7. Weights (kg) of 30 students: 40-45(2),45-50(3),50-55(8),55-60(6),60-65(6),65-70(3),70-75(2). Find the median weight.
   Solution: cf:2,5,13,19,25,28,30. n=30,n/2=15; median class 55-60 (cf just>15 is 19). l=55,cf=13,f=6,h=5. Median=55+[(15-13)/6]×5=56.67 kg (approx.).

More on This Chapter

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

FAQ

Q: Is this content updated for the 2026-27 NCERT edition?
A: Yes. The current rationalised syllabus has exactly 3 exercises (13.1, 13.2, 13.3) totalling 22 questions, cross-verified against two independent current sources before writing.

Q: Which formula should I remember for board exams?
A: Mode = l + [(f₁-f₀)/(2f₁-f₀-f₂)] × h, and Median = l + [(n/2-cf)/f] × h — both need the modal/median class identified correctly first.

Written by Satish

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