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Measures of Central Tendency (Mean, Median, Quartiles and Mode) — ICSE Class 10 Maths Important Questions

14 ICSE Class 10 Maths practice questions on Measures of Central Tendency (Mean, Median, Quartiles and Mode), each with a full model answer, covering 5 topics from the chapter in the question formats used in the exam.

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Reviewed by Classmate AI Team · 30 September 2026

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Measures of Central Tendency (Mean, Median, Quartiles and Mode) — ICSE Class 10 Maths Important Questions

Mean, Median, Mode — Plus Quartiles

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Quick answer

ICSE Measures of Central Tendency questions cover finding the mean by direct, short-cut and step-deviation methods (xˉ=A+∑fd∑f)\left(\bar{x}=A+\dfrac{\sum fd}{\sum f}\right)(x=A+ fd/ f), the median of grouped/ungrouped data, the mode (modal class), and quartiles. Grouped medians and quartiles are read from an ogive, and the mode from a histogram.

Syllabus reference (2026–27): CISCE (Council for the Indian School Certificate Examinations)

About Measures of Central Tendency (Mean, Median, Quartiles and Mode)

In this ICSE Class 10 Maths chapter Measures of Central Tendency you compute the mean of grouped and ungrouped data by the direct, short-cut (assumed-mean) and step-deviation methods, find the median and quartiles from cumulative frequencies, and determine the mode of a distribution. These summary measures describe the centre of a data set.

Mean by direct methodMean by short-cut (assumed mean) methodMean by step-deviation methodMedian and quartilesMode of grouped/ungrouped data

Key concepts & formulas

Mean (direct method)

For grouped data with class marks xix_ix_i and frequencies fif_if_i, xˉ=∑fixi∑fi\bar{x}=\dfrac{\sum f_i x_i}{\sum f_i}x= f_i x_i/ f_i.

Short-cut & step-deviation

With assumed mean AAA and di=xi−Ad_i=x_i-Ad_i=x_i-A: xˉ=A+∑fidi∑fi\bar{x}=A+\dfrac{\sum f_i d_i}{\sum f_i}x=A+ f_i d_i/ f_i. With class size hhh and ui=xi−Ahu_i=\dfrac{x_i-A}{h}u_i=x_i-A/h: xˉ=A+∑fiui∑fi×h\bar{x}=A+\dfrac{\sum f_i u_i}{\sum f_i}\times hx=A+ f_i u_i/ f_i× h.

Median

For nnn ungrouped values in order: if nnn is odd, median =(n+12)=\left(\dfrac{n+1}{2}\right)=(n+1/2)th value; if nnn is even, median === mean of the n2\dfrac{n}{2}n/2th and (n2+1)\left(\dfrac{n}{2}+1\right)(n/2+1)th values. For grouped data, draw the less-than ogive and read the median at cumulative frequency n2\dfrac{n}{2}n/2 (Q1Q_1Q_1 at n4\dfrac{n}{4}n/4, Q3Q_3Q_3 at 3n4\dfrac{3n}{4}3n/4).

Mode

For ungrouped data, the mode is the most frequent value; for grouped data it lies in the modal class (highest frequency) and can be estimated graphically from a histogram.

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Important questions with answers

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Multiple-choice questions (1 mark)

Q1MCQEasy1 mark

The mean of the first five natural numbers 1,2,3,4,51,2,3,4,51,2,3,4,5 is:

  1. (a)

    222

  2. (b)

    333

  3. (c)

    444

  4. (d)

    555

Show model answer

Answer: (b) 333.

xˉ=1+2+3+4+55=155=3.\bar{x}=\dfrac{1+2+3+4+5}{5}=\dfrac{15}{5}=3.x=1+2+3+4+5/5=15/5=3.

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Q2MCQEasy1 mark

The mode of the data 4,5,5,6,6,6,74,5,5,6,6,6,74,5,5,6,6,6,7 is:

  1. (a)

    444

  2. (b)

    555

  3. (c)

    666

  4. (d)

    777

Show model answer

Answer: (c) 666.

The value 666 occurs three times, more often than any other value, so the mode is 666.

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Q3MCQModerate1 mark

The median of 7,10,4,3,11,87,10,4,3,11,87,10,4,3,11,8 (six values) is:

  1. (a)

    7.57.57.5

  2. (b)

    888

  3. (c)

    101010

  4. (d)

    777

Show model answer

Answer: (a) 7.57.57.5.

In order: 3,4,7,8,10,113,4,7,8,10,113,4,7,8,10,11. With n=6n=6n=6, median === mean of the 333rd and 444th values =7+82=7.5.=\dfrac{7+8}{2}=7.5.=7+8/2=7.5.

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Q4MCQHOTS1 mark

The mean of 666 observations is 171717. If one observation 121212 is removed, the mean of the remaining five is:

  1. (a)

    181818

  2. (b)

    171717

  3. (c)

    18.518.518.5

  4. (d)

    161616

Show model answer

Answer: (a) 181818.

Sum of 666 observations =6×17=102=6\times17=102=6×17=102. After removing 121212, sum =102−12=90=102-12=90=102-12=90, so new mean =905=18.=\dfrac{90}{5}=18.=90/5=18.

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Assertion–Reason questions (1 mark)

Q5Assertion–ReasonModerate1 mark

Assertion (A): The step-deviation method gives the same mean as the direct method.

Reason (R): The step-deviation method only changes the arithmetic by using ui=xi−Ahu_i=\dfrac{x_i-A}{h}u_i=x_i-A/h and rescales back with ×h\times h× h, so no information is lost.

  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) All three methods are algebraically equivalent; the step-deviation method merely simplifies computation and rescales by hhh, giving the identical mean. R correctly explains A.

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Very short answer questions (2 marks)

Q6Very ShortEasy2 marks

Find the mean of the following distribution by the direct method.

xxx: 5,10,15,205,10,15,205,10,15,20; fff: 2,4,3,12,4,3,12,4,3,1.

Show model answer

∑f=2+4+3+1=10.\sum f=2+4+3+1=10.f=2+4+3+1=10.

∑fx=(5)(2)+(10)(4)+(15)(3)+(20)(1)=10+40+45+20=115.\sum fx=(5)(2)+(10)(4)+(15)(3)+(20)(1)=10+40+45+20=115.fx=(5)(2)+(10)(4)+(15)(3)+(20)(1)=10+40+45+20=115.

xˉ=∑fx∑f=11510=11.5.\bar{x}=\dfrac{\sum fx}{\sum f}=\dfrac{115}{10}=11.5.x= fx/ f=115/10=11.5.

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Q7Very ShortModerate2 marks

The mean of x,x+2,x+4,x+6,x+8x, x+2, x+4, x+6, x+8x, x+2, x+4, x+6, x+8 is 131313. Find xxx and the median.

Show model answer

Mean =x+(x+2)+(x+4)+(x+6)+(x+8)5=5x+205=x+4.=\dfrac{x+(x+2)+(x+4)+(x+6)+(x+8)}{5}=\dfrac{5x+20}{5}=x+4.=x+(x+2)+(x+4)+(x+6)+(x+8)/5=5x+20/5=x+4.

Given x+4=13⇒x=9.x+4=13\Rightarrow x=9.x+4=13 x=9.

The five values are 9,11,13,15,179,11,13,15,179,11,13,15,17; being in order with n=5n=5n=5, the median is the 333rd value =13.=13.=13.

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Q8Very ShortModerate2 marks

Find the lower quartile, the upper quartile and the interquartile range of the data 3,5,7,8,10,12,143, 5, 7, 8, 10, 12, 143, 5, 7, 8, 10, 12, 14.

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The data is already in ascending order, with n=7n=7n=7.

Q1=(n+14)Q_1=\left(\dfrac{n+1}{4}\right)Q_1=(n+1/4)th term =2=2=2nd term =5=5=5.

Q3=(3(n+1)4)Q_3=\left(\dfrac{3(n+1)}{4}\right)Q_3=(3(n+1)/4)th term =6=6=6th term =12=12=12.

Interquartile range =Q3−Q1=12−5=7=Q_3-Q_1=12-5=7=Q_3-Q_1=12-5=7.

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Short answer questions (3 marks)

Q9Short AnswerModerate3 marks

Using the short-cut (assumed-mean) method, find the mean of the following data. Take A=25A=25A=25.

Class marks xxx: 15,20,25,30,3515,20,25,30,3515,20,25,30,35; frequencies fff: 4,6,10,7,34,6,10,7,34,6,10,7,3.

Show model answer

Let A=25A=25A=25 and d=x−25d=x-25d=x-25.

xxxfffd=x−25d=x-25d=x-25fdfdfd
151515444−10-10-10−40-40-40
202020666−5-5-5−30-30-30
252525101010000000
303030777555353535
353535333101010303030

∑f=30,  ∑fd=−40−30+0+35+30=−5.\sum f=30,\ \ \sum fd=-40-30+0+35+30=-5.f=30, fd=-40-30+0+35+30=-5.

xˉ=A+∑fd∑f=25+−530=25−0.16‾=24.83 (approx).\bar{x}=A+\dfrac{\sum fd}{\sum f}=25+\dfrac{-5}{30}=25-0.1\overline{6}=24.83\ (\text{approx}).x=A+ fd/ f=25+-5/30=25-0.16=24.83 (approx).

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Q10Short AnswerModerate3 marks

Using the step-deviation method, find the mean of the following grouped data. Take A=25A=25A=25 and class size h=10h=10h=10.

Classes: 000–10,1010,1010,10–20,2020,2020,20–30,3030,3030,30–40,4040,4040,40–505050; frequencies: 6,8,12,9,56,8,12,9,56,8,12,9,5.

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Class marks xxx: 5,15,25,35,455,15,25,35,455,15,25,35,45. Let A=25A=25A=25, h=10h=10h=10, u=x−2510u=\dfrac{x-25}{10}u=x-25/10.

xxxfffuuufufufu
555666−2-2-2−12-12-12
151515888−1-1-1−8-8-8
252525121212000000
353535999111999
454545555222101010

∑f=40,  ∑fu=−12−8+0+9+10=−1.\sum f=40,\ \ \sum fu=-12-8+0+9+10=-1.f=40, fu=-12-8+0+9+10=-1.

xˉ=A+∑fu∑f×h=25+−140×10=25−0.25=24.75.\bar{x}=A+\dfrac{\sum fu}{\sum f}\times h=25+\dfrac{-1}{40}\times10=25-0.25=24.75.x=A+ fu/ f× h=25+-1/40×10=25-0.25=24.75.

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Q11Short AnswerHOTS3 marks

The mean of the following distribution is 545454. Find the missing frequency ppp.

Class marks xxx: 10,30,50,70,9010,30,50,70,9010,30,50,70,90; frequencies fff: 4,6,10,p,64,6,10,p,64,6,10,p,6.

Show model answer

∑f=4+6+10+p+6=26+p.\sum f=4+6+10+p+6=26+p.f=4+6+10+p+6=26+p.

∑fx=(10)(4)+(30)(6)+(50)(10)+(70)(p)+(90)(6)\sum fx=(10)(4)+(30)(6)+(50)(10)+(70)(p)+(90)(6)fx=(10)(4)+(30)(6)+(50)(10)+(70)(p)+(90)(6)

=40+180+500+70p+540=1260+70p.=40+180+500+70p+540=1260+70p.=40+180+500+70p+540=1260+70p.

Using xˉ=∑fx∑f=54\bar{x}=\dfrac{\sum fx}{\sum f}=54x= fx/ f=54:

1260+70p26+p=54.\dfrac{1260+70p}{26+p}=54.1260+70p/26+p=54.

Cross-multiplying: 1260+70p=54(26+p)=1404+54p.1260+70p=54(26+p)=1404+54p.1260+70p=54(26+p)=1404+54p.

70p−54p=1404−1260⇒16p=144⇒p=9.70p-54p=1404-1260\Rightarrow 16p=144\Rightarrow p=9.70p-54p=1404-1260 16p=144 p=9.

The missing frequency is p=9.p=9.p=9.

Check: ∑f=35\sum f=35f=35, ∑fx=1260+630=1890\sum fx=1260+630=1890fx=1260+630=1890, xˉ=189035=54.\bar{x}=\dfrac{1890}{35}=54.x=1890/35=54. Correct.

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Long answer questions (5 marks)

Q12Long AnswerModerate5 marks

For the following distribution, (a) find the mean by the step-deviation method (take A=35A=35A=35, h=10h=10h=10), and (b) draw a histogram and use it to estimate the mode.

Classes: 101010–20,2020,2020,20–30,3030,3030,30–40,4040,4040,40–50,5050,5050,50–606060; frequencies: 5,8,15,9,35,8,15,9,35,8,15,9,3.

Show model answer

(a) class marks x=15,25,35,45,55x=15,25,35,45,55x=15,25,35,45,55; A=35A=35A=35, h=10h=10h=10, u=x−3510u=\dfrac{x-35}{10}u=x-35/10.

xxxfffuuufufufu
151515555−2-2-2−10-10-10
252525888−1-1-1−8-8-8
353535151515000000
454545999111999
555555333222666

∑f=40, ∑fu=−10−8+0+9+6=−3.\sum f=40,\ \sum fu=-10-8+0+9+6=-3.f=40, fu=-10-8+0+9+6=-3.

xˉ=35+−340×10=35−0.75=34.25.\bar{x}=35+\dfrac{-3}{40}\times10=35-0.75=34.25.x=35+-3/40×10=35-0.75=34.25.

(b) Mode from the histogram: draw the histogram. The modal class is 303030–404040 (highest frequency 151515). Join the top-left corner of its bar to the top-left corner of the next bar, and the top-right corner of its bar to the top-right corner of the previous bar. From the point where these lines cross, drop a perpendicular to the xxx-axis.

ICSE Class 10 Maths — Measures of Central Tendency (Mean, Median, Quartiles and Mode): For the following distribution, (a) find the mean by the step-deviation method (take A=35, h=

The perpendicular meets the axis at about 35.435.435.4, so the mode ≈35.4\approx35.435.4.

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Q13Long AnswerHOTS5 marks

The following table gives the marks of 505050 students. Draw a less-than ogive and use it to estimate the lower quartile Q1Q_1Q_1, the median and the upper quartile Q3Q_3Q_3. Hence find the interquartile range.

Marks: 000–10,1010,1010,10–20,2020,2020,20–30,3030,3030,30–40,4040,4040,40–505050; students: 6,10,16,12,66,10,16,12,66,10,16,12,6.

Show model answer

Cumulative frequency table:

Marks less than101010202020303030404040505050
Cumulative frequency666161616323232444444505050

Plot the points (10,6),(20,16),(30,32),(40,44),(50,50)(10,6),(20,16),(30,32),(40,44),(50,50)(10,6),(20,16),(30,32),(40,44),(50,50), starting from (0,0)(0,0)(0,0), and join them to get the less-than ogive.

Here n=50n=50n=50. Read across from the cumulative frequencies n4=12.5\dfrac{n}{4}=12.5n/4=12.5, n2=25\dfrac{n}{2}=25n/2=25 and 3n4=37.5\dfrac{3n}{4}=37.53n/4=37.5 to the ogive, then down to the marks axis:

ICSE Class 10 Maths — Measures of Central Tendency (Mean, Median, Quartiles and Mode): The following table gives the marks of 50 students. Draw a less-than ogive and use it to esti

Q1≈16.5Q_1\approx16.5Q_116.5, median ≈25.6\approx25.625.6 and Q3≈34.6Q_3\approx34.6Q_334.6.

Interquartile range =Q3−Q1≈34.6−16.5=18.1=Q_3-Q_1\approx34.6-16.5=18.1=Q_3-Q_134.6-16.5=18.1.

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Case-based questions (4 marks)

Q14Case-basedModerate4 marks

The runs scored by a batsman in 777 innings are: 38,70,48,34,42,55,6338, 70, 48, 34, 42, 55, 6338, 70, 48, 34, 42, 55, 63.

(i) Find the mean number of runs.

(ii) Find the median number of runs.

(iii) If in the next (8th) innings he scores 505050, find the new mean.

(iv) State the range of the original 777 scores.

Show model answer

(i) Sum =38+70+48+34+42+55+63=350=38+70+48+34+42+55+63=350=38+70+48+34+42+55+63=350. Mean =3507=50=\dfrac{350}{7}=50=350/7=50 runs.

(ii) In order: 34,38,42,48,55,63,7034,38,42,48,55,63,7034,38,42,48,55,63,70. With n=7n=7n=7 (odd), median === the (7+12)=4\left(\dfrac{7+1}{2}\right)=4(7+1/2)=4th value =48=48=48 runs.

(iii) New sum =350+50=400=350+50=400=350+50=400 over 888 innings, so new mean =4008=50=\dfrac{400}{8}=50=400/8=50 runs.

(iv) Range === highest −-- lowest =70−34=36=70-34=36=70-34=36 runs.

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All ICSE Class 10 Maths Chapters

GST (Goods and Services Tax)Banking (Recurring Deposit Accounts)Shares and DividendsLinear Inequations (In One Variable)Quadratic EquationsSolving (Simple) Problems Based on Quadratic EquationsRatio and ProportionRemainder and Factor TheoremsMatricesArithmetic ProgressionGeometric ProgressionReflectionSection and Mid-Point FormulaEquation of a LineSimilarity (With Applications to Maps and Models)Loci (Locus and Its Constructions)CirclesTangents and Intersecting ChordsConstructions (Circles)Cylinder, Cone and Sphere (Surface Area and Volume)Trigonometrical IdentitiesHeights and DistancesGraphical Representation (Histograms and Ogives)Measures of Central Tendency (Mean, Median, Quartiles and Mode)Probability

Frequently asked questions

  • Do these Measures of Central Tendency (Mean, Median, Quartiles and Mode) questions follow the latest ICSE syllabus?
    Yes — they are aligned to the CISCE 2026–27 syllabus for ICSE Class 10 Maths, so nothing here is outside the current course.

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