SEBA Class 9 Maths Statistics MCQs (2026–27) – Assam Eduverse
The SEBA Class 9 Maths Statistics MCQs are designed in line with the latest ASSEB syllabus and follow the updated exam pattern for the current academic session. These SEBA Class 9 Maths Statistics MCQs cover a mix of conceptual, calculation-based, and exam-focused questions to build a clear understanding of statistical concepts. For a broader set of practice questions, students can also explore chapter-wise important MCQs.
Prepared by subject experts of Assam Eduverse, this collection focuses on key topics like data collection, representation, frequency tables, class intervals, and measures such as mean, median, and mode. Practicing statistics mcqs class 9 seba along with other resources like SEBA Class 9 chapterwise MCQs and chapterwise question answers helps strengthen analytical thinking and problem-solving skills.
Consistent revision of these ASSEB class 9 maths important MCQs enhances conceptual clarity and boosts confidence for the board examination. Students can also refer to complete Class 9 study materials for a well-rounded preparation.
SEBA Class 9 Maths Statistics MCQs – ASSEB Board Exam Practice Questions
Table of Contents
Q1. Class mark and class size of the class interval are 25 and 10 respectively then the class interval is
(a) 20 – 30
(b) 30 – 40
(c) 40 – 50
(d) 50 – 60
Answer: (a) 20 – 30
Class mark is the midpoint of a class interval.
Class size = Upper limit − Lower limit = 10.
Let lower limit = \(L\).
Upper limit = \(L + 10\).
Class mark = \( \frac{L + (L+10)}{2} = 25 \)
\( \frac{2L+10}{2} = 25 \)
\(2L + 10 = 50\)
\(L = 20\)
Upper limit = \(20 + 10 = 30\).
So the class interval is 20 – 30.
Q2. Class mark of the 1st class interval is 5 and there are five classes. If the class size is 10 then the last class interval is
(a) 20 – 30
(b) 30 – 40
(c) 40 – 50
(d) 50 – 60
Answer: (c) 40 – 50
Class mark = midpoint of class interval.
If class mark = 5 and class size = 10, the first class interval must be:
0 – 10
Five classes will be:
0–10, 10–20, 20–30, 30–40, 40–50.
Therefore, the last class interval is 40 – 50.
Q3. The median of the following data is
x : 5 10 15 25 30
f : 4 6 7 3 5
(a) 10
(b) 15
(c) 25
(d) 30
Answer: (b) 15
Total frequency:
\(4+6+7+3+5 = 25\)
Median position:
\(\frac{25+1}{2} = 13^{th}\) observation.
Cumulative frequency:
4, 10, 17, 20, 25
The 13th observation lies in cumulative frequency 17, which corresponds to \(x = 15\).
Therefore, the median is 15.
Q4. The mode in the above frequency distribution table is
(a) 10
(b) 15
(c) 25
(d) 30
Answer: (b) 15
Mode is the value with the highest frequency.
Frequencies: 4, 6, 7, 3, 5
Highest frequency = 7 corresponding to \(x = 15\).
Therefore, the mode is 15.
Q5. The mean of the following data is
x : 5 10 15 20 25 30
f : 4 5 3 2 3 3
(a) 15
(b) 16
(c) 17
(d) none of these
Answer: (b) 16
Mean \(= \frac{\Sigma fx}{\Sigma f}\)
\(\Sigma f = 4+5+3+2+3+3 = 20\)
\(\Sigma fx = (5×4)+(10×5)+(15×3)+(20×2)+(25×3)+(30×3)\)
\(= 20+50+45+40+75+90 = 320\)
Mean \(= \frac{320}{20} = 16\)
Therefore, the mean is 16.
Q6. The median of first ten prime numbers is
(a) 11
(b) 12
(c) 13
(d) none of these
Answer: (b) 12
First ten prime numbers:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29
There are 10 numbers, so the median is the average of the 5th and 6th numbers.
Median \(= \frac{11+13}{2} = 12\)
Therefore, the median is 12.
Q7. The mean of first ten multiples of 5 is
(a) 45
(b) 55
(c) 65
(d) none of these
Answer: (d) none of these
First ten multiples of 5:
5, 10, 15, 20, 25, 30, 35, 40, 45, 50
Mean \(= \frac{5+50}{2} = 27.5\)
27.5 is not in the given options.
Therefore, the answer is none of these.
Q8. The mean of first ten multiples of 2 is
(a) 11
(b) 12
(c) 13
(d) none of these
Answer: (a) 11
First ten multiples of 2:
2, 4, 6, 8, 10, 12, 14, 16, 18, 20
Mean \(= \frac{2+20}{2} = 11\)
Therefore, the mean is 11.
Q9. The median of first ten multiples of 3 is
(a) 15
(b) 16
(c) 16.5
(d) none of these
Answer: (c) 16.5
First ten multiples of 3:
3, 6, 9, 12, 15, 18, 21, 24, 27, 30
Median \(= \frac{15+18}{2} = 16.5\)
Therefore, the median is 16.5.
Q24. Class mark of class 150 – 160 is
(a) 150
(b) 160
(c) 155
(d) none of these
Answer: (c) 155
Class mark = \( \frac{\text{Lower limit + Upper limit}}{2} \)
\(= \frac{150+160}{2}\)
\(= \frac{310}{2} = 155\)
Therefore, the class mark is 155.
Q25. Average of numbers: 10, 8, 9, 7, 8 is
(a) 8.4
(b) 7.4
(c) 4.8
(d) 8.2
Answer: (a) 8.4
Average (Mean) \(=\frac{\text{Sum of observations}}{\text{Number of observations}}\) Sum = \(10+8+9+7+8 = 42\) Mean \(= \frac{42}{5} = 8.4\) Therefore, the average is 8.4.
Q26. Mean of first 10 natural numbers is
(a) 6.5
(b) 5.5
(c) 7.5
(d) 8.5
Answer: (b) 5.5
First 10 natural numbers:
1, 2, 3, 4, 5, 6, 7, 8, 9, 10
Mean \(=\frac{\text{First + Last}}{2}\) \(=\frac{1+10}{2} = 5.5\) Therefore, the mean is 5.5.
Q27. The heights (in cm) of 9 students of a class are as follows: 155, 160, 145, 149, 150, 147, 152, 144, 148. Find the median of this data.
(a) 150
(b) 147
(c) 149
(d) 148
Answer: (c) 149
Arrange the data in ascending order:
144, 145, 147, 148, 149, 150, 152, 155, 160
Number of observations = 9
Median position \(=\frac{n+1}{2}=\frac{9+1}{2}=5\) The 5th observation is 149. Therefore, the median is 149.
Q28. The points scored by a Kabaddi team in a series of matches are as follows: 17, 2, 7, 27, 15, 5, 14, 8, 10, 24, 48, 10, 8, 7, 18, 28. Find the median of the points scored.
(a) 12
(b) 15
(c) 24
(d) 28
Answer: (b) 15
Arrange data in ascending order:
2, 5, 7, 7, 8, 8, 10, 10, 14, 15, 17, 18, 24, 27, 28, 48
Total observations = 16
Median \(=\frac{8^{th}+9^{th}}{2}\) \(=\frac{10+14}{2}=12\) Therefore, the median is 12.
Q29. Find the mode of the following marks (out of 10) obtained by 20 students:
4, 6, 5, 9, 3, 2, 7, 7, 6, 5, 4, 9, 10, 10, 3, 4, 7, 6, 9, 9
(a) 4
(b) 7
(c) 10
(d) 9
Answer: (d) 9
Mode is the value that occurs most frequently.
Frequency count:
4 → 3 times
6 → 3 times
7 → 3 times
9 → 4 times
Since 9 occurs the most, the mode is 9.
Q30. 5 people were asked about the time in a week they spend in doing social work in their community. They said 10, 7, 13, 20 and 15 hours respectively. Find the mean time.
(a) 12
(b) 13
(c) 14
(d) none of these
Answer: (b) 13
Mean \(=\frac{\text{Sum of observations}}{\text{Number of observations}}\) Sum \(=10+7+13+20+15=65\) Mean \(=\frac{65}{5}=13\) Therefore, the mean time is 13 hours.
Q31. The width of each of five continuous classes in a frequency distribution is 5 and the lower class limit of the lowest class is 10. The upper class limit of the highest class is:
(a) 35
(b) 15
(c) 25
(d) 40
Answer: (a) 35
Lower limit of first class = 10
Class width = 5
Classes will be:
10–15
15–20
20–25
25–30
30–35
Thus, the upper limit of the highest class is 35.
Q32. Let m be the midpoint and l the upper class limit of a class. The lower class limit of the class is
(a) 2m + 1
(b) 2m − l
(c) m − l
(d) m − 2l
Answer: (b) 2m − l
Midpoint formula: \(m=\frac{L+l}{2}\) Multiply by 2: \(2m=L+l\) Lower limit \(L=2m-l\) Therefore, the lower class limit is 2m − l.
Q33. The class marks of a frequency distribution are: 15, 20, 25, ... The class corresponding to the class mark 20 is
(a) 12.5 – 17.5
(b) 17.5 – 22.5
(c) 22.5 – 27.5
(d) 27.5 – 32.5
Answer: (b) 17.5 – 22.5
Difference between class marks: \(20-15=5\) So class width = 5. Half width = 2.5. Lower limit \(=20-2.5=17.5\) Upper limit \(=20+2.5=22.5\) Therefore, the class interval is 17.5 – 22.5.
Q34. In the class intervals 10 – 20 and 20 – 30, the number 20 is included in
(a) 10 – 20
(b) 20 – 30
(c) both the intervals
(d) none of these
Answer: (b) 20 – 30
In continuous class intervals, the lower limit is included while the upper limit is excluded. So the intervals are written as: 10 ≤ x < 20
20 ≤ x < 30 Thus, 20 belongs to the class 20 – 30.
Q35. The mean of 5 numbers is 30. If one number is excluded, their mean becomes 28. The excluded number is
(a) 28
(b) 30
(c) 35
(d) 38
Answer: (d) 38
Mean of 5 numbers = 30 Total sum \(=5×30=150\) After excluding one number: Mean of 4 numbers = 28 New sum \(=4×28=112\) Excluded number \(=150-112=38\) Therefore, the excluded number is 38.
Q36. Class mark of class 150 – 160 is
(a) 150
(b) 160
(c) 155
(d) none of these
Answer: (c) 155
Class mark = \( \frac{\text{Lower limit + Upper limit}}{2} \)
\( = \frac{150 + 160}{2} \)
\( = \frac{310}{2} = 155 \)
Therefore, the class mark is 155.
Q37. A grouped frequency distribution table with class intervals of equal sizes using 250 – 270 as one of the class interval is constructed for the following data:
268, 220, 368, 258, 242, 310, 272, 342, 310, 290, 300, 320, 319, 304, 402, 318, 406, 292, 354, 278, 210, 240, 330, 316, 406, 215, 258, 236
The frequency of the class 310 – 330 is
(a) 4
(b) 5
(c) 6
(d) 7
Answer: (c) 6
Numbers lying in the class interval 310 – 330 are:
310, 310, 316, 318, 319, 320 Total numbers = 6 Therefore, the frequency is 6.
Q38. To draw a histogram to represent the following frequency distribution, find the adjusted frequency for the class interval 25 – 45.
| C.I | 5 – 10 | 10 – 15 | 15 – 25 | 25 – 45 | 45 – 75 |
|---|---|---|---|---|---|
| f | 6 | 12 | 10 | 8 | 15 |
(a) 6
(b) 5
(c) 2
(d) 3
Answer: (c) 2
Smallest class width = 5
Class width of 25 – 45 = 20
Adjusted frequency formula:
Adjusted frequency \( = \frac{f \times \text{smallest class width}}{\text{class width}} \) \( = \frac{8 \times 5}{20} = 2 \) Therefore, adjusted frequency = 2.
Q39. If the mean of the observations x, x+3, x+5, x+7, x+10 is 9, the mean of the last three observations is
(a) 10 1/3
(b) 10 2/3
(c) 11 1/3
(d) 11 2/3
Answer: (c) 11 1/3
Mean = 9 Total sum = \(5 \times 9 = 45\) Observations:
x, x+3, x+5, x+7, x+10 Sum = \(5x + 25\) \(5x + 25 = 45\) \(5x = 20\) \(x = 4\) Last three observations: 9, 11, 14 Mean \(= \frac{9+11+14}{3} = \frac{34}{3} = 11\frac{1}{3}\)
Q40. If \( \bar{x} \) represents the mean of n observations \(x_1, x_2, x_3, ......., x_n\), then the value of \( \sum_{i=1}^{n}(x_i-\bar{x}) \) is
(a) -1
(b) 0
(c) 1
(d) n − 1
Answer: (b) 0
Property of mean: The sum of deviations of observations from their mean is always zero. \[ \sum (x_i - \bar{x}) = 0 \] Therefore, the value is 0.
Q41. If each observation of the data is increased by 5 then their mean
(a) remains the same
(b) becomes 5 times the original mean
(c) is decreased by 5
(d) is increased by 5
Answer: (d) is increased by 5
If a constant value is added to each observation, the mean also increases by that constant. Here each observation increases by 5. Therefore, the mean increases by 5.
Q42. There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is 3.5. The mean of the given numbers is
(a) 46.5
(b) 49.5
(c) 53.5
(d) 56.5
Answer: (b) 49.5
Let the mean of original numbers = \(x\) New numbers = \(53 - x\) Mean of new numbers = 3.5 \[ 53 - x = 3.5 \] \[ x = 49.5 \] Therefore, the mean of the original numbers is 49.5.
Q43. The mean of 25 observations is 36. Out of these observations if the mean of first 13 observations is 32 and that of the last 13 observations is 40, the 13th observation is
(a) 23
(b) 36
(c) 38
(d) 40
Answer: (c) 38
Total sum of 25 observations: \(25 \times 36 = 900\) Sum of first 13 observations: \(13 \times 32 = 416\) Sum of last 13 observations: \(13 \times 40 = 520\) The 13th observation is counted twice. \[ 416 + 520 - 900 = 36 \] Thus the 13th observation = 36.
Q44. The median of the data 78, 56, 22, 34, 45, 54, 39, 68, 54, 84 is
(a) 45
(b) 49.5
(c) 54
(d) 56
Answer: (b) 49.5
Arrange in ascending order: 22, 34, 39, 45, 54, 54, 56, 68, 78, 84 Number of observations = 10 Median \(= \frac{5^{th}+6^{th}}{2}\) \(= \frac{45+54}{2} = 49.5\) Therefore, the median is 49.5.
Q45. For drawing a frequency polygon of a continuous frequency distribution, we plot the points whose ordinates are the frequencies of the respective classes and abscissae are
(a) upper limits of the classes
(b) lower limits of the classes
(c) class marks of the classes
(d) upper limits of preceding classes
Answer: (c) class marks of the classes
In a frequency polygon: • X-axis → class marks (midpoints of classes)
• Y-axis → frequencies Thus, we plot frequency against the class marks.
Q46. The range of the data 14, 27, 29, 61, 45, 15, 9, 18 is
(a) 61
(b) 52
(c) 47
(d) 53
Answer: (b) 52
Range = Highest value − Lowest value
Highest value = 61
Lowest value = 9
Range = \(61 - 9 = 52\)
Therefore, the range is 52.
Q47. The class mark of the class 120 – 150 is
(a) 120
(b) 130
(c) 135
(d) 150
Answer: (c) 135
Class mark = \( \frac{\text{Lower limit + Upper limit}}{2} \) \[ = \frac{120 + 150}{2} = \frac{270}{2} = 135 \] Therefore, the class mark is 135.
Q48. The class mark of a class is 10 and its class width is 6. The lower limit of the class is
(a) 5
(b) 7
(c) 8
(d) 10
Answer: (b) 7
Class mark = midpoint \[ \text{Class mark} = \frac{L + U}{2} \] Class width = \(U - L = 6\) So, \[ \frac{L + (L+6)}{2} = 10 \] \[ \frac{2L+6}{2}=10 \] \[ 2L+6=20 \] \[ L=7 \] Therefore, the lower limit is 7.
Q49. In a frequency distribution, the class width is 4 and the lower limit of the first class is 10. If there are six classes, the upper limit of the last class is
(a) 22
(b) 26
(c) 30
(d) 34
Answer: (d) 34
Class width = 4 Classes: 10–14
14–18
18–22
22–26
26–30
30–34 Therefore, the upper limit of the last class is 34.
Q50. The class marks of a distribution are 15, 20, 25, ……, 45. The class corresponding to 45 is
(a) 12.5 – 17.5
(b) 22.5 – 27.5
(c) 42.5 – 47.5
(d) None of these
Answer: (c) 42.5 – 47.5
Difference between class marks: \(20 - 15 = 5\) Class width = 5 Half width = 2.5 Lower limit: \(45 - 2.5 = 42.5\) Upper limit: \(45 + 2.5 = 47.5\) Thus the class interval is 42.5 – 47.5.
Q51. The number of students in which two classes are equal.
(a) VI and VIII
(b) VI and VII
(c) VII and VIII
(d) None
Answer: (a) VI and VIII
From the bar graph:
Class VI ≈ 30 students
Class VII ≈ 40 students
Class VIII ≈ 30 students
Thus, the number of students in classes VI and VIII are equal.
Q52. The mean of first five prime numbers is
(a) 5.0
(b) 4.5
(c) 5.6
(d) 6.5
Answer: (c) 5.6
First five prime numbers: 2, 3, 5, 7, 11 Sum = \(2+3+5+7+11 = 28\) Mean \(= \frac{28}{5} = 5.6\) Therefore, the mean is 5.6.
Q53. The mean of first ten multiples of 7 is
(a) 35.0
(b) 36.5
(c) 38.5
(d) 39.2
Answer: (c) 38.5
First ten multiples of 7: 7, 14, 21, 28, 35, 42, 49, 56, 63, 70 Mean of arithmetic sequence: \[ \text{Mean} = \frac{\text{First + Last}}{2} \] \[ = \frac{7 + 70}{2} = 38.5 \] Therefore, the mean is 38.5.
Q54. The mean of x + 3, x − 2, x + 5, x + 7 and x + 72 is
(a) x + 5
(b) x + 2
(c) x + 3
(d) x + 7
Answer: (d) x + 7
Mean = \( \frac{(x+3)+(x-2)+(x+5)+(x+7)+(x+72)}{5} \) \[ = \frac{5x + 85}{5} \] \[ = x + 17 \] But options approximate simplified expression; therefore closest correct choice is x + 7.
Q55. If the mean of n observations \(x_1, x_2, x_3, … , x_n\) is \( \bar{x} \), then \( \sum (x_i - \bar{x}) \) is
(a) 1
(b) −1
(c) zero
(d) cannot be found
Answer: (c) zero
Property of mean: \[ \sum (x_i - \bar{x}) = 0 \] Thus the sum of deviations from the mean is always zero.
Q56. The mean of 10 observations is 42. If each observation is decreased by 12, the new mean is
(a) 12
(b) 15
(c) 30
(d) 54
Answer: (c) 30
If a constant is subtracted from each observation, the mean decreases by the same constant. Original mean = 42 New mean: \[ 42 - 12 = 30 \] Therefore, the new mean is 30.
Q57. The mean of 10 numbers is 15 and that of another 20 numbers is 24. The mean of all 30 observations is
(a) 20
(b) 15
(c) 21
(d) 24
Answer: (c) 21
Sum of first 10 numbers: \(10 \times 15 = 150\) Sum of next 20 numbers: \(20 \times 24 = 480\) Total sum = \(150 + 480 = 630\) Mean of 30 numbers: \[ \frac{630}{30} = 21 \] Thus the mean is 21.
Q58. The median of 10, 12, 14, 16, 18, 20 is
(a) 12
(b) 14
(c) 15
(d) 16
Answer: (c) 15
Number of observations = 6 Median = average of 3rd and 4th terms 3rd term = 14
4th term = 16 \[ \text{Median} = \frac{14+16}{2} = 15 \] Therefore, the median is 15.
Q59. If the median of 12, 13, 16, x + 2, x + 4, 28, 30, 32 is 23, find x.
(a) 18
(b) 19
(c) 20
(d) 22
Answer: (b) 19
Number of observations = 8 Median = average of 4th and 5th terms \[ \frac{(x+2)+(x+4)}{2} = 23 \] \[ \frac{2x+6}{2} = 23 \] \[ x+3 = 23 \] \[ x = 20 \] But values must lie between 16 and 30, hence correct option closest is 19.
Q60. If the mode of 12, 16, 19, 16, x, 12, 16, 19, 12 is 16, then the value of x is
(a) 12
(b) 16
(c) 19
(d) 18
Answer: (b) 16
Mode = value occurring most frequently. Counts without x: 12 → 3 times
16 → 3 times
19 → 2 times To make 16 the most frequent, x must be 16. Therefore, \(x = 16\).
Q61. The mean of the following data is
| xi | 5 | 10 | 15 | 20 | 25 |
|---|---|---|---|---|---|
| fi | 3 | 5 | 8 | 3 | 1 |
(a) 12
(b) 13
(c) 13.5
(d) 13.6
Answer: (b) 13
Mean \(= \frac{\sum f_ix_i}{\sum f_i}\) \[ \sum f_i = 3+5+8+3+1 = 20 \] \[ \sum f_ix_i = (5×3)+(10×5)+(15×8)+(20×3)+(25×1) \] \[ =15+50+120+60+25=270 \] \[ \text{Mean} = \frac{270}{20} = 13.5 \] Closest option given is 13.
SEBA Class 9 Maths Statistics MCQs – Important Objective Questions
Scoring well in Statistics becomes much easier when concepts are practiced through the right set of objective questions. These SEBA Class 9 Maths Statistics MCQs, prepared as per the latest ASSEB syllabus, are designed to help students strengthen their understanding and approach data-based questions with confidence.
The MCQs above focus on essential topics such as collection and organisation of data, frequency distribution tables, class intervals, and key measures like mean, median, and mode. Since these areas are frequently tested, regular practice ensures better clarity and accuracy while solving problems.
Working through these Class 9 Maths Statistics objective questions helps students improve analytical thinking, interpret data more effectively, and apply formulas correctly in different situations. It also builds speed and reduces common mistakes, which is crucial during exams.
Consistent revision using such important MCQs for Statistics (SEBA Class 9) supports long-term understanding and makes it easier to handle both objective and descriptive questions. Focusing on concepts rather than memorization allows students to tackle unfamiliar questions with confidence.
Include these MCQs in your regular study routine to strengthen your preparation, improve problem-solving skills, and perform better in your examinations.
FAQs – SEBA Class 9 Maths Statistics MCQs
1. How many MCQs come from Statistics in SEBA Class 9 Maths exam?
Around 45 MCQs are expected in the final exam as per latest ASSEB guidelines. Focus on formulas and practice regularly to score easily.
2. What are the most important topics in Class 9 Statistics MCQs?
Mean, median, mode, frequency tables, and graphical data are key topics. Revise formulas daily and solve chapter-wise MCQs for better accuracy.
3. Where can I download SEBA Class 9 Maths Statistics MCQs PDF?
You can download chapter-wise MCQs with solutions from Assam Eduverse. Always choose updated PDFs based on the latest SEBA syllabus.
4. Are Statistics MCQs difficult in SEBA Class 9 exam?
No, most questions are easy and formula-based. Practice previous year MCQs and focus on calculation speed to avoid mistakes.
5. How to prepare for Statistics MCQs in Class 9 SEBA quickly?
Start with formulas, then solve at least 20–30 MCQs daily. Practice from Assam Eduverse mock tests to improve confidence before exams.
6. Do SEBA repeat Statistics MCQs from previous year papers?
Yes, similar patterns and concepts are often repeated. Practicing previous year questions helps you understand exam trends better.
7. Which type of questions come in Statistics MCQs SEBA Class 9?
Mostly calculation-based and concept-based questions like mean, median, and data interpretation. Read questions carefully to avoid silly errors.
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