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Class 10 Exam  >  Class 10 Tests  >  Measures of Central Tendency - Mean, Median and Mode - Class 10 MCQ

Measures of Central Tendency - Mean, Median and Mode - Class 10 MCQ


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15 Questions MCQ Test - Measures of Central Tendency - Mean, Median and Mode

Measures of Central Tendency - Mean, Median and Mode for Class 10 2025 is part of Class 10 preparation. The Measures of Central Tendency - Mean, Median and Mode questions and answers have been prepared according to the Class 10 exam syllabus.The Measures of Central Tendency - Mean, Median and Mode MCQs are made for Class 10 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Measures of Central Tendency - Mean, Median and Mode below.
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Measures of Central Tendency - Mean, Median and Mode - Question 1
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The marks of 6 students are: 12, 15, 17, 20, 22, 25. Find the median.

Detailed Solution for Measures of Central Tendency - Mean, Median and Mode - Question 1

n = 6 (even).
Median = [(n/2)th term + ((n/2)+1)th term] ÷ 2.
(6/2)th term = 3rd term = 17.
(6/2 + 1)th = 4th term = 20.
Median = (17 + 20) ÷ 2 = 18.5.
Hence Option B is correct.

Measures of Central Tendency - Mean, Median and Mode - Question 2
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The marks of 5 students are: 10, 15, 20, 25, 30. Find the arithmetic mean.

Detailed Solution for Measures of Central Tendency - Mean, Median and Mode - Question 2

Formula: Mean = (Σx) ÷ n.
Σx = 10 + 15 + 20 + 25 + 30 = 100.
n = 5.
Mean = 100 ÷ 5 = 20.
Hence Option A is correct.

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Measures of Central Tendency - Mean, Median and Mode - Question 3
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Find the mean of first 5 natural numbers.

Detailed Solution for Measures of Central Tendency - Mean, Median and Mode - Question 3

Numbers are 1, 2, 3, 4, 5.
Σx = 1 + 2 + 3 + 4 + 5 = 15.
n = 5.
Mean = 15 ÷ 5 = 3.
Hence Option C is correct.

Measures of Central Tendency - Mean, Median and Mode - Question 4
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The ages of 7 students are: 10, 12, 13, 14, 15, 16, 20. Find the median age.
Detailed Solution for Measures of Central Tendency - Mean, Median and Mode - Question 4

n = 7 (odd). Median = (n+1)/2 th term = 4th term. 4th term = 14. Hence median = 14. Option B is correct.

Measures of Central Tendency - Mean, Median and Mode - Question 5
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The weights (in kg) of 5 persons are 67, 65, 71, 57, and 45. What is the arithmetic mean of their weights?
Detailed Solution for Measures of Central Tendency - Mean, Median and Mode - Question 5

Step 1: Sum of weights Σx = 67 + 65 + 71 + 57 + 45 = 305 kg.
Step 2: Number of persons n = 5.
Step 3: Mean = Σx / n = 305 / 5 = 61 kg.
Conclusion: The arithmetic mean is 61 kg, making Option C correct.

Measures of Central Tendency - Mean, Median and Mode - Question 6
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The marks of 7 students are 40, 45, 50, 55, 60, 65, 70. What is the median?
Detailed Solution for Measures of Central Tendency - Mean, Median and Mode - Question 6

Step 1: Arrange in order: 40, 45, 50, 55, 60, 65, 70 (already ordered).
Step 2: n = 7 (odd), median at (7+1)/2 = 4th term.
Step 3: 4th term = 55.
Conclusion: The median is 55, So Option B correct.

Measures of Central Tendency - Mean, Median and Mode - Question 7
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For the grouped data below, what is the mean using direct method?
Detailed Solution for Measures of Central Tendency - Mean, Median and Mode - Question 7

Step 1: Calculate fx: 5 × 4 = 20, 15 × 6 = 90, 25 × 8 = 200, 35 × 2 = 70.
Step 2: Σfx = 20 + 90 + 200 + 70 = 380, Σf = 4 + 6 + 8 + 2 = 20.
Step 3: Mean = Σfx / Σf = 380 / 20 = 19.

Measures of Central Tendency - Mean, Median and Mode - Question 8
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The daily wages of 6 workers are: 120, 150, 160, 180, 200, 250. Find the median.
Detailed Solution for Measures of Central Tendency - Mean, Median and Mode - Question 8

n = 6 (even). Median = Average of (n/2)th term and ((n/2)+1)th term. (6/2)th term = 3rd term = 160. (6/2 + 1)th = 4th term = 180. Median = (160 + 180) ÷ 2 = 170. Hence Option B is correct.

Measures of Central Tendency - Mean, Median and Mode - Question 9
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The marks obtained are: 12, 14, 18, 20, 20, 22, 25. Find the mode.
Detailed Solution for Measures of Central Tendency - Mean, Median and Mode - Question 9

Mode is the value occurring most frequently. Here 20 occurs twice, all others occur once. So Mode = 20. Hence Option A is correct.

Measures of Central Tendency - Mean, Median and Mode - Question 10
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The frequency distribution is:

Find the mean.
Detailed Solution for Measures of Central Tendency - Mean, Median and Mode - Question 10

Mean = Σfx ÷ Σf.

= 830 ÷ 30 = 27.67 ≈ 28.

Hence Option A is correct.

Measures of Central Tendency - Mean, Median and Mode - Question 11
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The ages of 9 students are: 8, 10, 12, 12, 13, 14, 14, 15, 16. Find the mode.
Detailed Solution for Measures of Central Tendency - Mean, Median and Mode - Question 11

Count frequencies:

  • 12 occurs 2 times.
  • 14 occurs 2 times.
  • Others occur once.

When there are two values with the same highest frequency, the data is bimodal.

Modes are 12 and 14. Since both appear in options, the higher (14) is chosen.

Hence Option C is correct.

Measures of Central Tendency - Mean, Median and Mode - Question 12
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The following distribution is given:

Find the median class.
Detailed Solution for Measures of Central Tendency - Mean, Median and Mode - Question 12

Total frequency Σf = 5 + 7 + 10 + 6 + 2 = 30.

n/2 = 30/2 = 15.

Cumulative frequency table:

The 15th observation lies in 20-30.

So median class = 20-30.

Hence Option C is correct.

Measures of Central Tendency - Mean, Median and Mode - Question 13
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The marks of 8 students are: 5, 7, 8, 10, 12, 12, 15, 18. Find the mean.
Detailed Solution for Measures of Central Tendency - Mean, Median and Mode - Question 13

Σx = 5+7+8+10+12+12+15+18 = 87.
n = 8.
Mean = 87 ÷ 8 = 10.875.
Hence Option A is correct.

Measures of Central Tendency - Mean, Median and Mode - Question 14
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In a frequency distribution, the modal class is 20-30 with frequency 10. The preceding class has frequency 7 and succeeding class has frequency 6. Class width = 10. Find the mode.
Detailed Solution for Measures of Central Tendency - Mean, Median and Mode - Question 14

Formula: Mode = L + [(f₁ − f₀) ÷ (2f₁ − f₀ − f₂)] × h.
L = lower limit of modal class = 20.
f₁ = frequency of modal class = 10.
f₀ = frequency of preceding class = 7.
f₂ = frequency of succeeding class = 6.
h = 10.
Mode = 20 + [(10−7)/(20−7−6)] × 10.
= 20 + (3/7) × 10.
= 20 + 30/7 = 24.3.
Hence Option C is correct.

Measures of Central Tendency - Mean, Median and Mode - Question 15
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The following data shows marks of students:

Find the mean marks.
Detailed Solution for Measures of Central Tendency - Mean, Median and Mode - Question 15

Mean = Σfx ÷ Σf.
= 880 ÷ 30 = 29.3.
Rounded = 29.
Hence Option B is correct.

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