Statistics: CBSE Class 10 Previous Year Questions

198 different questions from Statistics (NCERT Chapter 13) asked in CBSE Class 10 Maths board exams 2022–2026. Pick a mark group to practise, each with answers and step-by-step solutions.

1 mark questions69 questions 2 marks questions15 questions 3 marks questions27 questions 4 marks questions14 questions 5 marks questions73 questions

Most asked Statistics questions

Q81 markMCQStatistics CBSE 2026 · Basic 430/4/1

While calculating mean of a grouped frequency distribution using step deviation method $(u = \frac{x - a}{h})$ it was found that $x$ = 62, a = 47.5, h = 5. The value of $u$ is :

(A)3
(B)14.5
(C)2.9
(D)3.1

Show answer & solution

Answer: (C) 2.9

$x = a + h u$ .
$62 = 47.5 + 5 u$ .
$u = \frac{14.5}{5} = 2.9$ .

Also asked in: 2026 Basic 430/4/2, 2026 Basic 430/4/3

Q201 markAssertion–ReasonStatistics CBSE 2026 · Basic 430/5/1

Assertion (A) : Median of a data is the value of $\frac{N}{2}$ , where N represents sum of all frequencies.
Reason (R) : Median divides the whole distribution in two equal parts.

(A)Both, Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B)Both, Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C)Assertion (A) is true, but Reason (R) is false.
(D)Assertion (A) is false, but Reason (R) is true.

Show answer & solution

Answer: (D) Assertion (A) is false, but Reason (R) is true.

$\frac{N}{2}$ only tells us the position used to locate the median (median class); the median is the value of the observation there, not $\frac{N}{2}$ itself. A is false.
The median is the middle value, which divides the distribution into two equal parts. R is true.

Also asked in: 2026 Basic 430/5/2, 2026 Basic 430/5/3

Q171 markMCQStatistics CBSE 2026 · Standard 30/1/1

If the mean and mode of a data are 12 and 21 respectively, then its median is :

(A)6
(B)13.5
(C)15
(D)14

Show answer & solution

Answer: (C) 15

Empirical relation: $3 Median = Mode + 2 Mean$ .
$3 Median = 21 + 24 = 45$ .
Median $= 15$ .

Also asked in: 2026 Standard 30/1/2, 2026 Standard 30/1/3

Q171 markMCQStatistics CBSE 2026 · Standard 30/2/1

The mean and median of a frequency distribution are 43 and 43.4 respectively. The mode of the distribution is :

(A)43.4
(B)42.4
(C)44.2
(D)49.3

Show answer & solution

Answer: (C) 44.2

Mode = 3 Median - 2 Mean
$= 3 \times 43.4 - 2 \times 43 = 130.2 - 86 = 44.2$

Also asked in: 2026 Standard 30/2/2, 2026 Standard 30/2/3

Q161 markMCQStatistics CBSE 2026 · Standard 30/3/1

The median and mode of a distribution are 25.2 and 26.1 respectively. The mean of the distribution is :

(A)24.75
(B)24.25
(C)24.3
(D)25.5

Show answer & solution

Answer: (A) 24.75

Empirical relation: Mode $= 3$ Median $- 2$ Mean
$26.1 = 3 \times 25.2 - 2 \times$ Mean $= 75.6 - 2 \times$ Mean
Mean $= \frac{75.6 - 26.1}{2} = \frac{49.5}{2} = 24.75$

Also asked in: 2026 Standard 30/3/2, 2026 Standard 30/3/3

Q181 markMCQStatistics CBSE 2026 · Standard 30/4/1

Mean and Median of a frequency distribution are 43 and 40 respectively. The value of mode is

(A)34
(B)43
(C)38.5
(D)41.5

Show answer & solution

Answer: (A) 34

Mode = 3 Median $-$ 2 Mean
$= 3 \times 40 - 2 \times 43 = 120 - 86 = 34$

Also asked in: 2026 Standard 30/4/2, 2026 Standard 30/4/3

Q61 markMCQStatistics CBSE 2026 · Standard 30/5/1

While calculating mean of a grouped frequency distribution, step deviation method was used $(\frac{x - a}{h} = u)$ . It was found that $x ˉ = 64$ , h = 5 and a = 62.5. The value of $u ˉ$ is

(A)0.5
(B)1.5
(C)0.3
(D)7.5

Show answer & solution

Answer: (C) 0.3

$x ˉ = a + h u ˉ$
$64 = 62.5 + 5 u ˉ$
$5 u ˉ = 1.5$ , so $u ˉ = 0.3$

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

Q374 marksCase StudyStatistics CBSE 2026 · Basic 430/4/1

CENTRAL POLLUTION CONTROL BOARD'S AIR QUALITY STANDARDS
AIR QUALITY INDEX (AQI) | CATEGORY
0-50 | Good
51-100 | Satisfactory
101-200 | Moderate
201-300 | Poor
301-400 | Very Poor
401-500 | Severe
The Air Quality Index (AQI) is a scale from 0 to 500 that indicates air quality, with higher numbers signifying more pollution and greater health concerns.
Mansi collected the daily data of AQI of her city for a month and presented it as given below :
AQI Range : 1 – 100 | 101 – 200 | 201 – 300 | 301 – 400 | 401 – 500
Number of Days : 3 | 9 | 12 | 4 | 2
(i) Convert the data to continuous frequency distribution. (1)
(ii) What is the quality of air in most of the days of the month ? (1)
(iii) (a) Using table formed in part (i), find mode of the data. (2)
OR
(b) Using table formed in part (i), find median of the data. (2)

Show answer & solution

Answer: (i) 0.5 – 100.5, 100.5 – 200.5, 200.5 – 300.5, 300.5 – 400.5, 400.5 – 500.5 with frequencies 3, 9, 12, 4, 2 (ii) Poor (iii) (a) Mode =

200.5 + \frac{300}{11} \approx 227.77

(b) Median = 225.5

(i) Subtract 0.5 from each lower limit and add 0.5 to each upper limit: 0.5 – 100.5 (3), 100.5 – 200.5 (9), 200.5 – 300.5 (12), 300.5 – 400.5 (4), 400.5 – 500.5 (2).
(ii) The highest frequency (12 days) is in the class 201 – 300, which is the 'Poor' category.
(iii) (a) Modal class 200.5 – 300.5: l = 200.5, $f_{1} = 12$ , $f_{0} = 9$ , $f_{2} = 4$ , h = 100.
Mode = $l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 200.5 + \frac{3}{11} \times 100 \approx 200.5 + 27.27 = 227.77$ .
(iii) (b) n = 30, $\frac{n}{2} = 15$ ; cumulative frequencies 3, 12, 24, 28, 30, so median class is 200.5 – 300.5 with cf = 12, f = 12.
Median = $l + \frac{\frac{n}{2} - c f}{f} \times h = 200.5 + \frac{15 - 12}{12} \times 100 = 225.5$ .

Also asked in: 2026 Basic 430/4/2, 2026 Basic 430/4/3

Q161 markMCQStatistics CBSE 2025 · Basic 430/1/1

The class mark of the median class of the following data is :
Class Interval: 10 – 25, 25 – 40, 40 – 55, 55 – 70, 70 – 85, 85 – 100
Frequency: 2, 3, 7, 6, 6, 6

(A)40
(B)55
(C)47.5
(D)62.5

Show answer & solution

Answer: (D) 62.5

$n = 30$ , so $\frac{n}{2} = 15$ .
Cumulative frequencies: 2, 5, 12, 18, 24, 30.
15 lies in the class 55 – 70, which is the median class.
Class mark $= \frac{55 + 70}{2} = 62.5$ .

Also asked in: 2025 Basic 430/1/2, 2025 Basic 430/1/3

Q171 markMCQStatistics CBSE 2025 · Basic 430/1/1

The following distribution shows the number of runs scored by some batsmen in test matches :
Runs Scored: 3000 – 4000, 4000 – 5000, 5000 – 6000, 6000 – 7000
Number of Batsmen: 5, 10, 9, 8
The lower limit of the modal class is :

(A)3000
(B)4000
(C)5000
(D)6000

Show answer & solution

Answer: (B) 4000

The highest frequency is 10, for the class 4000 – 5000.
So the modal class is 4000 – 5000 and its lower limit is 4000.

Also asked in: 2025 Basic 430/1/2, 2025 Basic 430/1/3

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