RD Sharma Solutions for Class 9 Maths Chapter 24 Measure of Central Tendency

Infinity Learn offers detailed solutions for RD Sharma Solutions for Class 9 Maths  Chapter 24 – Measures of Central Tendency to help students understand the topic thoroughly. Each exercise is solved step-by-step in a student-friendly format, making it easier to follow and learn effectively. These solutions are created by expert educators at IL and are fully aligned with the latest CBSE curriculum. Regular practice using these solutions can enhance conceptual clarity and build a strong foundation, especially since many of these concepts appear again in higher classes.

Measures of Central Tendency are an essential part of statistics and are used to represent a dataset with a single, central value. This value typically gives an overall idea of the data distribution and is sometimes referred to as a measure of central location. The most common measure students are introduced to is the mean, which represents the average of a data set. Along with the mean, the median and mode are also key measures of central tendency. For thorough exam preparation, teachers often suggest that students refer to RD Sharma Class 9 Solutions provided by Infinity Learn to reinforce their understanding and improve performance.

RD Sharma Solution for Class 9 Chapter 24 - Download PDF

Here are the RD Sharma Solutions Class 9 Maths Chapter 24 Measure of Central Tendency Solutions, designed to help students prepare effectively for their exams. By referring to these solutions and practicing the problems, students can boost their confidence and improve their scores.

RD Sharma Solution for Class 9 Chapter 24 - Question with Answers

Part A: Mean of Ungrouped Data

Q: Find the mean of 3, 5, 7, 9, 11.
A: Mean = (3+5+7+9+11)/5 = 35/5 = 7

Q: Mean of 5 observations is 12. If four of them are 10, 15, 8, 14, find the fifth observation.
A: Total = 12×5 = 60; Fifth = 60 - (10+15+8+14) = 13

Q: Mean of 6 numbers is 18. If one number is 30, find the mean of remaining.
A: Total = 108; Remaining = 108 - 30 = 78; Mean = 78/5 = 15.6

Q: Find x if mean of 6, 8, 10, x, 14 is 10.
A: (6+8+10+x+14)/5 = 10 → 38+x = 50 → x = 12

Q: Find mean of first five even numbers.
A: (2+4+6+8+10)/5 = 30/5 = 6

Part B: Median of Ungrouped Data

Q: Find median of 2, 5, 7, 9, 11.
A: Median = middle = 7

Q: Find median of 6, 8, 3, 4, 7, 2
A: Ordered = 2,3,4,6,7,8 → Median = (4+6)/2 = 5

Q: Find median of first 10 odd numbers.
A: List: 1 to 19 → Median = (5th + 6th)/2 = (9+11)/2 = 10

Q: Find median of 15, 18, 10, 20, 12, 16, 14.
A: Ordered = 10,12,14,15,16,18,20 → Median = 15

Q: If median of 5, x, 12 is 9, find x.
A: Ordered = 5, x, 12 → Median = x = 9

Part C: Mode of Ungrouped Data

Q: Find mode of 2, 3, 4, 4, 5, 6
A: Mode = 4

Q: Find mode of 1, 2, 2, 3, 3, 4
A: Bimodal → 2 and 3

Q: Find mode of 7, 9, 9, 10, 11, 12
A: Mode = 9

Q: If mode = 15 in 9, 15, 13, 15, x, find x if x repeats most
A: x must not occur more than 2 times → Any number ≠ 15

Q: Find mode of 6, 8, 6, 10, 8, 8, 6, 6
A: 6 occurs 4 times → Mode = 6

Part D: Mean of Grouped Data

Q: Find mean of: | x | 10 | 20 | 30 | | f | 3 | 4 | 3 |
A: Mean = (10×3 + 20×4 + 30×3)/10 = (30+80+90)/10 = 200/10 = 20

Q: Calculate mean: | x | 5 | 10 | 15 | | f | 2 | 3 | 5 |
A: Mean = (5×2 + 10×3 + 15×5)/10 = (10+30+75)/10 = 115/10 = 11.5

Q: Find mean using assumed mean method (A=20): | x | 10 | 20 | 30 | | f | 5 | 7 | 8 |
A: Mean = A + (Σfd / Σf) = 20 + ((-10×5 + 0×7 + 10×8)/20) = 20 + (30/20) = 21.5

Q: Mean of marks scored by students is 50. There are 25 students. Find total marks.
A: 25 × 50 = 1250

Q: Find mean from: | Class | 0–10 | 10–20 | 20–30 | | f | 5 | 10 | 5 |
A: Midpoints = 5, 15, 25; Mean = (5×5 + 10×15 + 5×25)/20 = (25+150+125)/20 = 15

Part E: Median of Grouped Data

Q: Find median class: | Class | 0–10 | 10–20 | 20–30 | | f | 5 | 15 | 10 |
A: Total = 30, Median class = 10–20 (cumulative > 15)

Q: Median formula used?
A: Median = $l + \frac{\frac{N}{2} - CF}{f} \times h$

Q: What is l in median formula?
A: Lower boundary of the median class

Q: Calculate median: | Class | 0–10 | 10–20 | 20–30 | | f | 5 | 7 | 8 |
A: N=20, N/2=10 → CF=5 → Median Class=10–20
l=10, f=7, CF=5, h=10
Median = 10 + (5/7)×10 ≈ 17.14

Q: Find cumulative frequency for: | f | 4 | 6 | 10 |
A: CF: 4, 10, 20

Part F: Mode of Grouped Data

Q: Mode formula:
A: Mode = $l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$

Q: Mode class is the class with:
A: Maximum frequency

Q: Find modal class: | Class | 0–10 | 10–20 | 20–30 | | f | 3 | 9 | 6 |
A: Modal class = 10–20

Q: What is $f_0, f_1, f_2$?
A: $f_1$=modal class freq, $f_0$=freq before, $f_2$=freq after

Q: Calculate mode: | Class | 0–10 | 10–20 | 20–30 | | f | 5 | 10 | 5 |
A: l=10, h=10, f1=10, f0=5, f2=5
Mode = 10 + (5/(20–5–5))×10 = 10 + (5/10)×10 = 15