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Mean (Direct/Assumed) PYQs

Class 10 Statistics Board Questions (2014 – 2026)

Topic Overview

Master Mean calculation for grouped data. Direct Method is fast for small numbers, while Assumed Mean Method ($A + \frac{\sum fd}{\sum f}$) reduces calculation errors in larger datasets.

Q1 2024
00:00
In a frequency distribution, the mid value of a class is 10 and the width of the class is 6. The lower limit of the class is:
(a)(A) 6
(b)(B) 7
(c)(C) 8
(d)(D) 12
Lower limit $= \text{Mid value} - \text{Width}/2 = 10 - 3 = 7$.
Ans: (B) 7
Q2 2026
00:00
The mean of the first 10 prime numbers is:
(a)(A) 12.9
(b)(B) 15.5
(c)(C) 12.0
(d)(D) 13.4
First 10 primes: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29.
Sum $= 129$. Mean $= 129/10 = 12.9$.
Ans: (A) 12.9
Q3 2021
00:00
The mean of first $n$ natural numbers is:
(a)(A) $n/2$
(b)(B) $(n+1)/2$
(c)(C) $n(n+1)/2$
(d)(D) $n+1$
Mean $= [n(n+1)/2] / n = (n+1)/2$.
Ans: (B)
Q4 2023
00:00
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate. | Literacy Rate (%) | 45-55 | 55-65 | 65-75 | 75-85 | 85-95 | |-------------------|-------|-------|-------|-------|-------| | Number of cities | 3 | 10 | 11 | 8 | 3 |
$\sum f = 35, \sum fx = 2430$. Mean $= 69.43$.
Ans: 69.43%
Q5 2025
00:00
Find the missing frequency $f$ for the following distribution, if the mean of the data is 18. | Class | 11-13 | 13-15 | 15-17 | 17-19 | 19-21 | 21-23 | 23-25 | |-------|-------|-------|-------|-------|-------|-------|-------| | Freq | 7 | 6 | 9 | 13 | $f$ | 5 | 4 |
$18(44+f) = 752 + 20f \Rightarrow f = 20$.
Ans: 20
Q6 2026
00:00
The daily income of 50 workers of a factory is given below. Calculate the mean income using the Assumed Mean Method. | Daily Income (₹) | 500-520 | 520-540 | 540-560 | 560-580 | 580-600 | |------------------|---------|---------|---------|---------|---------| | No. of workers | 12 | 14 | 8 | 6 | 10 |
Let $A=550$. $\sum fd = -240, \sum f = 50$. Mean $= 550 - 4.8 = 545.2$.
Ans: ₹ 545.2
Q7 2025
00:00
The mean of the following frequency distribution is 50. The sum of all frequencies is 120. Find the missing frequencies $f_1$ and $f_2$. | Class | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 | |-------|------|-------|-------|-------|--------| | Freq | 17 | $f_1$ | 32 | $f_2$ | 19 |
$f_1+f_2 = 120 - 68 = 52$.
$\sum fx = 10(17) + 30f_1 + 50(32) + 70f_2 + 90(19) = 120 \times 50 = 6000$.
$3480 + 30f_1 + 70f_2 = 6000 \Rightarrow 30f_1 + 70f_2 = 2520 \Rightarrow 3f_1 + 7f_2 = 252$.
Solving $f_1+f_2=52$ and $3f_1+7f_2=252$: $4f_2 = 96 \Rightarrow f_2=24, f_1=28$.
Ans: $f_1=28, f_2=24$
00:00
In a frequency distribution, the mid value of a class is 10 and the width of the class is 6. The lower limit of the class is:
(a)(A) 6
(b)(B) 7
(c)(C) 8
(d)(D) 12
Lower limit $= \text{Mid value} - \text{Width}/2 = 10 - 3 = 7$.
Ans: (B) 7
Q2 2026
00:00
The mean of the first 10 prime numbers is:
(a)(A) 12.9
(b)(B) 15.5
(c)(C) 12.0
(d)(D) 13.4
First 10 primes: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29.
Sum $= 129$. Mean $= 129/10 = 12.9$.
Ans: (A) 12.9
Q3 2021
00:00
The mean of first $n$ natural numbers is:
(a)(A) $n/2$
(b)(B) $(n+1)/2$
(c)(C) $n(n+1)/2$
(d)(D) $n+1$
Mean $= [n(n+1)/2] / n = (n+1)/2$.
Ans: (B)
Q4 2023
00:00
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate. | Literacy Rate (%) | 45-55 | 55-65 | 65-75 | 75-85 | 85-95 | |-------------------|-------|-------|-------|-------|-------| | Number of cities | 3 | 10 | 11 | 8 | 3 |
$\sum f = 35, \sum fx = 2430$. Mean $= 69.43$.
Ans: 69.43%
Q5 2025
00:00
Find the missing frequency $f$ for the following distribution, if the mean of the data is 18. | Class | 11-13 | 13-15 | 15-17 | 17-19 | 19-21 | 21-23 | 23-25 | |-------|-------|-------|-------|-------|-------|-------|-------| | Freq | 7 | 6 | 9 | 13 | $f$ | 5 | 4 |
$18(44+f) = 752 + 20f \Rightarrow f = 20$.
Ans: 20
Q6 2026
00:00
The daily income of 50 workers of a factory is given below. Calculate the mean income using the Assumed Mean Method. | Daily Income (₹) | 500-520 | 520-540 | 540-560 | 560-580 | 580-600 | |------------------|---------|---------|---------|---------|---------| | No. of workers | 12 | 14 | 8 | 6 | 10 |
Let $A=550$. $\sum fd = -240, \sum f = 50$. Mean $= 550 - 4.8 = 545.2$.
Ans: ₹ 545.2
Q7 2025
00:00
The mean of the following frequency distribution is 50. The sum of all frequencies is 120. Find the missing frequencies $f_1$ and $f_2$. | Class | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 | |-------|------|-------|-------|-------|--------| | Freq | 17 | $f_1$ | 32 | $f_2$ | 19 |
$f_1+f_2 = 120 - 68 = 52$.
$\sum fx = 10(17) + 30f_1 + 50(32) + 70f_2 + 90(19) = 120 \times 50 = 6000$.
$3480 + 30f_1 + 70f_2 = 6000 \Rightarrow 30f_1 + 70f_2 = 2520 \Rightarrow 3f_1 + 7f_2 = 252$.
Solving $f_1+f_2=52$ and $3f_1+7f_2=252$: $4f_2 = 96 \Rightarrow f_2=24, f_1=28$.
Ans: $f_1=28, f_2=24$