RS Aggarwal Solutions for Class 10 Maths Chapter 9 Exercise 9.3: The RS Aggarwal Solutions for Class 10 Maths Chapter 9 Exercise 9.3 provide detailed explanations and step-by-step solutions to problems involving the cumulative frequency distribution and its graphical representation.
These solutions prepared by subject experts at Physics Wallah, help students understand how to construct and interpret cumulative frequency curves (ogives). By practicing these exercises, students can enhance their ability to analyze and interpret data trends effectively.RS Aggarwal Solutions for Class 10 Maths Chapter 9 Exercise 9.3 PDF
Solution:
Here, the maximum class frequency is 45. The class corresponding to this frequency is the modal class. ⇒ modal class = 30 - 40 ∴ lower limit of the modal class (l) = 30 Modal class size (h) = 10 Frequency of the modal class (f1) = 45 Frequency of class preceding the modal class (f0) = 35 Frequency of class succeeding the modal (f2) = 25 Mode is given by, Mode = l + ( f1 − f0 2f1 − f0 − f2 ) × h ⇒ Mode = 30 + ( 45−35 2(45)−35−25) × 10 ⇒ Mode = 30 + ( 10 30) × 10 ⇒ Mode = 30 + 3.33 = 33.33 Hence, the mode is 33.33
Solution:
Here, the maximum class frequency is 28. The class corresponding to this frequency is the modal class. ⇒ modal class = 40 - 60 ∴ lower limit of the modal class (l) = 40 Modal class size (h) = 20 Frequency of the modal class (f1) = 28 Frequency of class preceding the modal class (f0) = 16 Frequency of class succeeding the modal (f2) = 20 Mode is given by, Mode = l + ( f1 − f0 2f1 − f0 − f2 ) × h ⇒ Mode = 40 + ( 28−16 2(28)−16−20) × 20 ⇒ Mode = 40 + ( 12 20) × 20 ⇒ Mode = 40 + 12 = 56 Hence, the mode is 56.
Solution:
Here, the maximum class frequency is 20.
The class corresponding to this frequency is the modal class. ⇒
modal class = 160 - 165
∴ lower limit of the modal class (l) = 160
Modal class size (h) = 5
Frequency of the modal class (f1) = 20
Frequency of class preceding the modal class (f0) = 8
Frequency of class succeeding the modal (f2) = 12
Mode is given by,
Mode = l + (f1 − f02f1 − f0 − f2) × h
⇒ Mode = 26 + (58) × 4
⇒ Mode = 26 + 2.5 = 28.5
Hence, the mode is 28.5.
Expenditure done by maximum number of manual workers is
estimated by finding mode.
So here, the maximum class frequency is 40.
