RD Sharma Class 9 Solutions Chapter 24 Exercise 24.2: You can get the PDF for RD Sharma Class 9 Solutions Maths Chapter 24 Exercise 24.2 for free. This PDF is created by our expert team to provide students the detailed solutions. These solutions will be very beneficial for your Class 9 exams.
Download RD Sharma Class 9 Solutions Maths Chapter 24 Exercise 24.2
RD-SHARMA-Solutions-Class-9-Maths-Chapter-24-Ex-24.2
Access RD Sharma Class 9 Solutions Maths Chapter 24 Exercise 24.2
Question 1: Calculate the mean for the following distribution:
Solution:
Formula to calculate mean:
= 281/40
= 7.025
⇒ Mean for the given distribution is 7.025.
Question 2: Find the mean of the following data:
Solution:
Formula to calculate mean:
= 2650/106
= 25
⇒ Mean for the given data is 25.
Question 3: The mean of the following data is 20.6 . Find the value of p.
Solution:
Formula to calculate mean:
= (25p + 530)/50
Mean = 20.6 (Given)
So,
20.6 = (25p + 530)/50
25p + 530 = 1030
25p = 1030 − 530 = 500
or p = 20
⇒ The value of p is 20.
Question 4: If the mean of the following data is 15, find p.
Solution:
Formula to calculate mean:
= (10p + 445)/(p + 27)
Mean = 15 (Given)
So, (10p + 445)/(p + 27) = 15
10p + 445 = 15(p + 27)
10p – 15p = 405 – 445 = -40
-5p = -40
or p = 8
⇒ The value of p is 8.
Question 5: Find the value of p for the following distribution whose mean is 16.6.
Solution:
Formula to calculate mean:
= (24p + 1228)/100
Mean = 16.6 (given)
So, (24p + 1228)/100 = 16.6
24p + 1228 = 1660
24p = 1660 – 1228 = 432
p = 432/24 = 18
⇒ The value of p is 18.
Question 6: Find the missing value of p for the following distribution whose mean is 12.58.
Solution:
Formula to calculate mean:
= (7p + 524)/50
Mean = 12.58 (given)
So, (7p + 524)/50 = 12.58
7p + 524 = 12.58 x 50
7p + 524 = 629
7p = 629 – 524 = 105
p = 105/7 = 15
⇒ The value of p is 15.
Question 7: Find the missing frequency (p) for the following distribution whose mean is 7.68.
Solution:
Formula to calculate mean:
= (9p + 303)/(p+41)
Mean = 7.68 (given)
So, (9p + 303)/(p+41) = 7.68
9p + 303 = 7.68 (p + 41)
9p + 303 = 7.68p + 314.88
9p − 7.68p = 314.88 − 303
1.32p = 11.88
or p = (11.881)/(1.32) = 9
⇒ The value of p is 9.
We have included all the information regarding CBSE RD Sharma Class 9 Solutions Chapter 24 Exercise 24.2. If you have any queries feel free to ask in the comment section.
FAQ: RD Sharma Class 9 Solutions Chapter 24 Exercise 24.2
Can I open RD Sharma Class 9 Solutions Chapter 24 Exercise 24.2 PDF on my smartphone?
Yes, you can open RD Sharma Class 9 Solutions Chapter 24 Exercise 24.2 PDF on any device.
Is RD Sharma enough for Class 12 Maths?
RD Sharma is a good book that gives you thousands of questions to practice.
Can I download RD Sharma Class 9 Solutions Chapter 24 Exercise 24.2 PDF free?
Yes, you can download RD Sharma Class 9 Solutions Chapter 24 Exercise 24.2 PDF free.
What are the benefits of studying from RD Sharma Solutions Class 12?
By practicing these solutions, students can earn higher academic grades. Our experts solve these solutions with utmost accuracy to help students in their studies.