Statistics: Class 10 Important Questions with Answers

13.7 Quick board MCQs

CBSE 2024 · 30/5/1 For the data 2, 9, x + 6, 2x + 3, 5, 10, 5; if the mean is 7, then the value of x is: (A) 9 (B) 6 (C) 5 (D) 3

Sum $=40+3x=49\Rightarrow x=$ 3 (D)

CBSE 2023 · 30/1/1 The distribution below gives the marks obtained by 80 students: less than 10: 3, less than 20: 12, less than 30: 27, less than 40: 57, less than 50: 75, less than 60: 80. The modal class of this distribution is: (A) 10–20 (B) 20–30 (C) 30–40 (D) 50–60

Convert: 0–10: 3, 10–20: 9, 20–30: 15, 30–40: 30, 40–50: 18, 50–60: 5 ⇒ (C) 30–40

CBSE 2024 · 30/2/1 After an examination, a teacher wants to know the marks obtained by maximum number of the students in her class. She requires to calculate ___ of marks. (A) median (B) mode (C) mean (D) range

"Maximum number of students" = most frequent ⇒ (B) mode

CBSE 2025 · 30/5/1 Following data shows the marks obtained by 100 students in a class test (marks 20, 29, 28, 33, 42, 38, 43, 25 with 6, 28, 24, 15, 2, 4, 1, 20 students). The median will be the average of which two observations? (A) 29 and 33 (B) 25 and 28 (C) 28 and 29 (D) 33 and 38

Sorted: 20 (6), 25 (20), 28 (24), 29 (28) … cf 6, 26, 50, 78. The 50th value is 28 and the 51st is 29 ⇒ (C)

20 MCQs (1 mark each)

1. The class mark of 10–20 is
   (a) 10 (b) 15 (c) 20 (d) 30
2. The modal class is the class with the
   (a) highest cf (b) highest frequency (c) middle position (d) largest class mark
3. If mode = 24 and mean = 30, median is
   (a) 27 (b) 28 (c) 26 (d) 22
4. For n = 50, the median class has cf just ≥
   (a) 50 (b) 25 (c) 25.5 (d) 24
5. In the median formula, cf stands for the cf of the class
   (a) itself (b) after (c) before (d) first
6. If $\sum f_i{u}_i=-10$, $\sum f_i=50$, a = 45, h = 10, mean is
   (a) 43 (b) 47 (c) 44 (d) 45
7. Which measure is found graphically from ogives?
   (a) mean (b) median (c) mode (d) range
8. Mean of first 10 natural numbers
   (a) 5 (b) 5.5 (c) 10 (d) 55
9. If each observation is multiplied by 3, the mean is
   (a) same (b) +3 (c) × 3 (d) ÷ 3
10. The width of class 25–35 is
    (a) 10 (b) 25 (c) 35 (d) 30

1 (b) · 2 (b) · 3 (b) · 4 (b) · 5 (c) · 6 (a) · 7 (b) · 8 (b) · 9 (c) · 10 (a)

Q3: 3 Median = 24 + 60 = 84 ⇒ 28. Q6: 45 + 10(−0.2) = 43.

11. For 0–10: 5, 10–20: 8, 20–30: 12, 30–40: 6, the modal class is
    (a) 10–20 (b) 20–30 (c) 30–40 (d) 0–10
12. For the data in Q11, $l$, $f_0$, $f_1$, $f_2$ for the mode are
    (a) 20, 8, 12, 6 (b) 20, 6, 12, 8 (c) 10, 5, 8, 12 (d) 30, 12, 6, 0
13. Mode of Q11 data is
    (a) 24 (b) 25 (c) 23.2 (d) 26
14. For Q11 data, n/2 and the median class are
    (a) 15.5, 20–30 (b) 15.5, 10–20 (c) 31, 30–40 (d) 13, 10–20
15. The relation used to find mode from mean and median is called
    (a) exact (b) empirical (c) assumed (d) derived
16. If mean = median = mode, the distribution is
    (a) skewed (b) symmetric (c) bimodal (d) cumulative
17. $u_i=\frac{x_i-a}{h}$ is used in the
    (a) direct (b) assumed mean (c) step-deviation (d) median method
18. A "less than 40" cf of 57 and "less than 30" cf of 27 give the frequency of 30–40 as
    (a) 57 (b) 27 (c) 30 (d) 84
19. Mean of 2, 4, 6, 8, 10 is
    (a) 5 (b) 6 (c) 30 (d) 4
20. Median of 3, 8, 5, 9, 1 is
    (a) 5 (b) 8 (c) 3 (d) 9

11 (b) · 12 (a) · 13 (a) · 14 (a) · 15 (b) · 16 (b) · 17 (c) · 18 (c) · 19 (b) · 20 (a)

Q13: $20+\frac{12-8}{24-8-6}\times 10=24$. Q14: n = 31.

Assertion–Reason (10)

(a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true.

1. A: If mean = 20 and median = 26, mode = 38. R: Mode = 3 Median − 2 Mean.
2. A: The modal class has the highest cumulative frequency. R: The last class always has cf = n.
3. A: Adding 5 to each observation increases the mean by 5. R: Mean = sum ÷ number of observations.
4. A: All three methods of finding the mean give the same value. R: They differ only in arithmetic, not in the formula's meaning.
5. A: The median class is the class with the highest frequency. R: The median divides the data into two equal halves.
6. A: Class mark of 40–50 is 45. R: Class mark = (lower + upper) ÷ 2.
7. A: The mode of 2, 3, 3, 5, 5, 5, 7 is 5. R: Mode is the most frequent value.
8. A: In the mode formula f₀ is the frequency of the class after the modal class. R: f₂ is the frequency of the class after the modal class.
9. A: For inclusive classes 10–19, 20–29 we use 9.5–19.5 etc. for median. R: Classes must be continuous for the median formula.
10. A: The mean of the first n natural numbers is $\frac{n+1}{2}$. R: Their sum is $\frac{n(n+1)}{2}$.

Key: 1 (a) · 2 (d) · 3 (a) · 4 (a) · 5 (d) · 6 (a) · 7 (a) · 8 (d) · 9 (a) · 10 (a)

Very short answer (1 mark each)

1. Write the formula of the mode of grouped data. Ans. $l+\frac{f_1-f_0}{2f_1-f_0-f_2}h$
2. If median = 15 and mean = 12, find the mode. Ans. 21
3. Find the class mark of 99.5–109.5. Ans. 104.5
4. Which measure is affected most by extreme values? Ans. mean
5. If $\sum f_i{x}_i=1860$ and $\sum f_i=30$, find the mean. Ans. 62
6. Convert "less than 20: 12, less than 30: 27" into the frequency of 20–30. Ans. 15

Short answer practice (3 marks each)

1. Find the mean of: 0–20: 6, 20–40: 8, 40–60: 10, 60–80: 12, 80–100: 6, 100–120: 5, 120–140: 3.

2. Find the mode: 0–10: 3, 10–20: 8, 20–30: 9, 30–40: 10, 40–50: 3, 50–60: 0, 60–70: 0, 70–80: 2.

Long answer practice (5 marks each)

3. The mean of the following distribution is 50; the total frequency is 120. Find $f_1$ and $f_2$: 0–20: 17, 20–40: $f_1$, 40–60: 32, 60–80: $f_2$, 80–100: 19.

4. Find the median and mode of: 10–20: 5, 20–30: 9, 30–40: 15, 40–50: 13, 50–60: 8. Then estimate the mean from the empirical relation and check it directly.