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Statistics — CBSE Class 10 Maths Important Questions

13 hand-picked CBSE Class 10 Maths important questions for Statistics, each with a full model answer — the formats and topics most likely to appear in your board exam.

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Statistics — CBSE Class 10 Maths Important Questions

Mean, Median, Mode — No Mix-Ups

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Quick answer

For grouped data: Mean =a+∑fidi∑fi=a+\dfrac{\sum f_id_i}{\sum f_i}=a+ f_id_i/ f_i (assumed-mean method), Median =l+N2−cff×h=l+\dfrac{\frac{N}{2}-cf}{f}\times h=l+N/2-cff× h, Mode =l+f1−f02f1−f0−f2×h=l+\dfrac{f_1-f_0}{2f_1-f_0-f_2}\times h=l+f_1-f_0/2f_1-f_0-f_2× h. The empirical relation is Mode=3 Median−2 Mean\text{Mode}=3\,\text{Median}-2\,\text{Mean}Mode=3\,Median-2\,Mean.

Syllabus reference (2026–27): NCERT (National Council of Educational Research and Training) · CBSE (Central Board of Secondary Education)

About Statistics

This chapter is about the three measures of central tendency for grouped (continuous) data: mean, median and mode. It is a scoring, formula-driven chapter worth roughly 6-8 marks, usually including a case-study and a 5-mark question, often with a missing-frequency twist. Identify the correct class first (median class = the class where cumulative frequency first reaches N/2N/2N/2; modal class = the class with the highest frequency) before substituting into the formula.

Mean of grouped data (direct, assumed-mean, step-deviation)Median of grouped dataMode of grouped dataEmpirical relationship: Mode = 3 Median minus 2 MeanFinding missing frequencies; identifying median and modal class

Key concepts & formulas

Mean (assumed-mean method)

Take an assumed mean aaa (a convenient class mark), compute di=xi−ad_i=x_i-ad_i=x_i-a, then xˉ=a+∑fidi∑fi\bar{x}=a+\dfrac{\sum f_id_i}{\sum f_i}x=a+ f_id_i/ f_i. The direct method xˉ=∑fixi∑fi\bar{x}=\dfrac{\sum f_ix_i}{\sum f_i}x= f_ix_i/ f_i gives the same answer, where xix_ix_i is the class mark.

Median of grouped data

Median=l+N2−cff×h\text{Median}=l+\dfrac{\frac{N}{2}-cf}{f}\times hMedian=l+N/2-cff× h, where l=l=l= lower limit of the median class, N=∑fiN=\sum f_iN= f_i, cf=cf=cf= cumulative frequency of the class before the median class, f=f=f= frequency of the median class, h=h=h= class width.

Mode of grouped data

Mode=l+f1−f02f1−f0−f2×h\text{Mode}=l+\dfrac{f_1-f_0}{2f_1-f_0-f_2}\times hMode=l+f_1-f_0/2f_1-f_0-f_2× h, where l=l=l= lower limit of the modal class (highest-frequency class), f1=f_1=f_1= its frequency, f0=f_0=f_0= frequency of the preceding class, f2=f_2=f_2= frequency of the following class, h=h=h= class width.

Empirical relationship

For a moderately skewed distribution, Mode=3 Median−2 Mean\text{Mode}=3\,\text{Median}-2\,\text{Mean}Mode=3\,Median-2\,Mean. If any two are known, the third can be estimated. The median can also be read as the xxx-coordinate where the two ogives intersect.

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Important questions with answers

Try each on paper first, then reveal the model answer to check your method.

Question typeCountMarks
MCQ41
Assertion–Reason11
Very Short22
Short Answer33
Long Answer25
Case-based14

Multiple-choice questions (1 mark)

Q1MCQEasy1 mark

The class mark of the class interval 151515–353535 is:

  1. (a)

    252525

  2. (b)

    202020

  3. (c)

    505050

  4. (d)

    101010

Show model answer

Answer: (a) 252525

Class mark =lower limit+upper limit2=15+352=25.=\dfrac{\text{lower limit}+\text{upper limit}}{2}=\dfrac{15+35}{2}=25.=lower limit+upper limit/2=15+35/2=25.

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Q2MCQEasy1 mark

For a distribution, the median is 150150150 and the mean is 148148148. Using the empirical relationship, the mode is:

  1. (a)

    154154154

  2. (b)

    152152152

  3. (c)

    150150150

  4. (d)

    146146146

Show model answer

Answer: (a) 154154154

Mode=3 Median−2 Mean=3(150)−2(148)=450−296=154.\text{Mode}=3\,\text{Median}-2\,\text{Mean}=3(150)-2(148)=450-296=154.Mode=3\,Median-2\,Mean=3(150)-2(148)=450-296=154.

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Q3MCQModerate1 mark

Which measure of central tendency for grouped data can be obtained graphically as the xxx-coordinate of the point of intersection of the two ogives?

  1. (a)

    Median

  2. (b)

    Mean

  3. (c)

    Mode

  4. (d)

    Range

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Answer: (a) Median

The xxx-coordinate of the point where the 'less than' and 'more than' ogives intersect gives the median of the data.

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Q4MCQHOTS1 mark

For the distribution below, the sum of the lower limits of the median class and the modal class is:

Class0–55–1010–1515–2020–25
Frequency101512209
  1. (a)

    252525

  2. (b)

    151515

  3. (c)

    303030

  4. (d)

    353535

Show model answer

Answer: (a) 252525

N=66, N2=33.N=66,\ \dfrac{N}{2}=33.N=66, N/2=33. Cumulative frequencies: 10,25,37,57,66.10,25,37,57,66.10,25,37,57,66. The c.f. first reaches 333333 in class 101010–151515, so the median class is 101010–151515 (lower limit 101010). The highest frequency 202020 is in 151515–202020, so the modal class is 151515–202020 (lower limit 151515). Sum =10+15=25.=10+15=25.=10+15=25.

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Assertion–Reason questions (1 mark)

Q5Assertion–ReasonEasy1 mark

Assertion (A): The median of a grouped frequency distribution can be found graphically.

Reason (R): The xxx-coordinate of the point of intersection of the 'less than' and 'more than' ogives gives the median.

  1. (a)

    Both A and R are true and R is the correct explanation of A

  2. (b)

    Both A and R are true but R is not the correct explanation of A

  3. (c)

    A is true but R is false

  4. (d)

    A is false but R is true

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Answer: (a) Both A and R are true and R is the correct explanation of A.

The median can indeed be found graphically, precisely by the ogive-intersection method described in R, which is why R correctly explains A.

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Very short answer questions (2 marks)

Q6Very ShortEasy2 marks

The mean of 555 observations is 181818. If one observation is removed, the mean of the remaining observations becomes 161616. Find the value of the removed observation.

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Sum of 555 observations =5×18=90.=5\times 18=90.=5× 18=90.

Sum of remaining 444 observations =4×16=64.=4\times 16=64.=4× 16=64.

Removed observation =90−64=26.=90-64=26.=90-64=26.

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Q7Very ShortModerate2 marks

For the following distribution, write the median class and the modal class.

Class0–1010–2020–3030–4040–50
Frequency581575
Show model answer

N=40, N2=20.N=40,\ \dfrac{N}{2}=20.N=40, N/2=20. Cumulative frequencies: 5,13,28,35,40.5,13,28,35,40.5,13,28,35,40. The c.f. first reaches 202020 in class 202020–303030, so the median class is 202020–303030.

The highest frequency is 151515 (class 202020–303030), so the modal class is also 202020–303030.

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Short answer questions (3 marks)

Q8Short AnswerModerate3 marks

Find the mean daily wage from the following data.

Daily wage (Rs)100–120120–140140–160160–180180–200
No. of workers12148610
Show model answer

Class marks xix_ix_i: 110,130,150,170,190.110,130,150,170,190.110,130,150,170,190.

∑fixi=110(12)+130(14)+150(8)+170(6)+190(10)\sum f_ix_i=110(12)+130(14)+150(8)+170(6)+190(10)f_ix_i=110(12)+130(14)+150(8)+170(6)+190(10)
=1320+1820+1200+1020+1900=7260.=1320+1820+1200+1020+1900=7260.=1320+1820+1200+1020+1900=7260.

N=∑fi=50.N=\sum f_i=50.N= f_i=50. Mean =726050=Rs 145.20.=\dfrac{7260}{50}=\text{Rs }145.20.=7260/50=Rs 145.20.

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Q9Short AnswerModerate3 marks

Find the mode of the following distribution of the ages of patients admitted to a hospital.

Age (years)0–1010–2020–3030–4040–50
No. of patients81636346
Show model answer

The highest frequency is 363636, so the modal class is 202020–303030.

Here l=20, f1=36, f0=16, f2=34, h=10.l=20,\ f_1=36,\ f_0=16,\ f_2=34,\ h=10.l=20, f_1=36, f_0=16, f_2=34, h=10.

Mode=l+f1−f02f1−f0−f2×h=20+36−1672−16−34×10=20+2022×10=20+9.09=29.09 years.\text{Mode}=l+\dfrac{f_1-f_0}{2f_1-f_0-f_2}\times h=20+\dfrac{36-16}{72-16-34}\times 10=20+\dfrac{20}{22}\times 10=20+9.09=29.09\ \text{years}.Mode=l+f_1-f_0/2f_1-f_0-f_2× h=20+36-16/72-16-34× 10=20+20/22× 10=20+9.09=29.09 years.

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Q10Short AnswerHOTS3 marks

The median of the following distribution is 28.528.528.5 and the total frequency is 606060. Find the missing frequencies xxx and yyy.

Class0–1010–2020–3030–4040–5050–60
Frequency5xxx2015yyy5
Show model answer

Total: 5+x+20+15+y+5=60⇒x+y=15.5+x+20+15+y+5=60\Rightarrow x+y=15.5+x+20+15+y+5=60 x+y=15.

N2=30\dfrac{N}{2}=30N/2=30; since the median 28.528.528.5 lies in 202020–303030, that is the median class (l=20, f=20, h=10l=20,\ f=20,\ h=10l=20, f=20, h=10, cf=5+xcf=5+xcf=5+x).

Median=l+N2−cff×h⇒28.5=20+30−(5+x)20×10.\text{Median}=l+\dfrac{\frac{N}{2}-cf}{f}\times h\Rightarrow 28.5=20+\dfrac{30-(5+x)}{20}\times 10.Median=l+N/2-cff× h 28.5=20+30-(5+x)/20× 10.

8.5=25−x2⇒25−x=17⇒x=8.8.5=\dfrac{25-x}{2}\Rightarrow 25-x=17\Rightarrow x=8.8.5=25-x/2 25-x=17 x=8. Then y=15−8=7.y=15-8=7.y=15-8=7.

x=8, y=7.x=8,\ y=7.x=8, y=7.

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Long answer questions (5 marks)

Q11Long AnswerModerate5 marks

For the following frequency distribution, find the mean and the median. Hence, using the empirical relationship, estimate the mode.

Class0–1010–2020–3030–4040–50
Frequency5815166
Show model answer

Mean: class marks 5,15,25,35,45.5,15,25,35,45.5,15,25,35,45. ∑fixi=25+120+375+560+270=1350, N=50.\sum f_ix_i=25+120+375+560+270=1350,\ N=50.f_ix_i=25+120+375+560+270=1350, N=50. Mean =135050=27.=\dfrac{1350}{50}=27.=1350/50=27.

Median: N2=25.\dfrac{N}{2}=25.N/2=25. Cumulative frequencies: 5,13,28,44,50.5,13,28,44,50.5,13,28,44,50. Median class =20=20=20–303030 (l=20, cf=13, f=15, h=10l=20,\ cf=13,\ f=15,\ h=10l=20, cf=13, f=15, h=10).

Median=20+25−1315×10=20+1215×10=20+8=28.\text{Median}=20+\dfrac{25-13}{15}\times 10=20+\dfrac{12}{15}\times 10=20+8=28.Median=20+25-13/15× 10=20+12/15× 10=20+8=28.

Mode (empirical): Mode=3 Median−2 Mean=3(28)−2(27)=84−54=30.\text{Mode}=3\,\text{Median}-2\,\text{Mean}=3(28)-2(27)=84-54=30.Mode=3\,Median-2\,Mean=3(28)-2(27)=84-54=30.

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Q12Long AnswerHOTS5 marks

The mean of the following distribution is 62.862.862.8 and the sum of all frequencies is 505050. Find the missing frequencies f1f_1f_1 and f2f_2f_2.

Class0–2020–4040–6060–8080–100100–120
Frequency5f1f_1f_110f2f_2f_278
Show model answer

Total: 5+f1+10+f2+7+8=50⇒f1+f2=20.5+f_1+10+f_2+7+8=50\Rightarrow f_1+f_2=20.5+f_1+10+f_2+7+8=50 f_1+f_2=20. …(1)

Class marks: 10,30,50,70,90,110.10,30,50,70,90,110.10,30,50,70,90,110.

∑fixi=5(10)+f1(30)+10(50)+f2(70)+7(90)+8(110)=2060+30f1+70f2.\sum f_ix_i=5(10)+f_1(30)+10(50)+f_2(70)+7(90)+8(110)=2060+30f_1+70f_2.f_ix_i=5(10)+f_1(30)+10(50)+f_2(70)+7(90)+8(110)=2060+30f_1+70f_2.

Mean =∑fixi50=62.8⇒2060+30f1+70f2=3140⇒3f1+7f2=108.=\dfrac{\sum f_ix_i}{50}=62.8\Rightarrow 2060+30f_1+70f_2=3140\Rightarrow 3f_1+7f_2=108.= f_ix_i/50=62.8 2060+30f_1+70f_2=3140 3f_1+7f_2=108. …(2)

From (1), f1=20−f2.f_1=20-f_2.f_1=20-f_2. Substituting: 3(20−f2)+7f2=108⇒60+4f2=108⇒f2=12.3(20-f_2)+7f_2=108\Rightarrow 60+4f_2=108\Rightarrow f_2=12.3(20-f_2)+7f_2=108 60+4f_2=108 f_2=12. Then f1=8.f_1=8.f_1=8.

f1=8, f2=12.f_1=8,\ f_2=12.f_1=8, f_2=12.

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Case-based questions (4 marks)

Q13Case-basedModerate4 marks

A company tested the life (in hours) of 200200200 electric bulbs. The results are shown below.

Life (hours)300–400400–500500–600600–700700–800
No. of bulbs1456604030

(i) Write the modal class.

(ii) Write the median class.

(iii) Find the median life of the bulbs.

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(i) Highest frequency is 606060 (class 500500500–600600600), so the modal class is 500500500–600600600.

(ii) N=200, N2=100.N=200,\ \dfrac{N}{2}=100.N=200, N/2=100. Cumulative frequencies: 14,70,130,170,200.14,70,130,170,200.14,70,130,170,200. The c.f. first reaches 100100100 in 500500500–600600600, so the median class is 500500500–600600600.

(iii) l=500, cf=70, f=60, h=100.l=500,\ cf=70,\ f=60,\ h=100.l=500, cf=70, f=60, h=100.
Median=500+100−7060×100=500+3060×100=500+50=550 hours.\text{Median}=500+\dfrac{100-70}{60}\times 100=500+\dfrac{30}{60}\times 100=500+50=550\ \text{hours}.Median=500+100-70/60× 100=500+30/60× 100=500+50=550 hours.

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All CBSE Class 10 Maths Chapters

Real NumbersPolynomialsPair of Linear Equations in Two VariablesQuadratic EquationsArithmetic ProgressionsTrianglesCoordinate GeometryIntroduction to TrigonometrySome Applications of TrigonometryCirclesAreas Related to CirclesSurface Areas and VolumesStatisticsProbability

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Frequently asked questions

  • Are these Statistics important questions free?
    Yes. All 13 CBSE Class 10 Maths important questions for Statistics are free, with full model answers and no login required.
  • Do these Statistics questions follow the latest CBSE syllabus?
    Yes — they are aligned to the NCERT 2026–27 syllabus for CBSE Class 10 Maths, so nothing here is outside the current course.
  • How should I practise the Statistics important questions?
    Attempt each question on paper first, then reveal the model answer to check your method — not just the final result. Re-do anything you got wrong the same day.
  • What types of questions are covered for Statistics?
    A full mix — multiple-choice questions, assertion–reason questions, very short answer questions, short answer questions, long answer questions, case-based questions — so every format in the CBSE paper is covered.

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