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RS Aggarwal Solutions for Class 10 Maths Chapter 9 Exercise 9.6

In this article we have provided RS Aggarwal Solutions for Class 10 Maths Chapter 9 Exercise 9.6 prepared by our experts to help students to prepare better for their examinations.
authorImageAnanya Gupta12 Jul, 2024
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RS Aggarwal Solutions for Class 10 Maths Chapter 9 Exercise 9.6

RS Aggarwal Solutions for Class 10 Maths Chapter 9 Exercise 9.6: RS Aggarwal Solutions for Class 10 Maths Chapter 9 Exercise 9.6 focus on cumulative frequency distribution and how to draw cumulative frequency curves, called ogives.

They help students learn how to plot and understand ogives step by step. By practicing these exercises, students can get better at analyzing and visualizing data, which is important for their math exams. These solutions are a great resource for understanding cumulative frequency and its uses in statistics.

RS Aggarwal Solutions for Class 10 Maths Chapter 9 Exercise 9.6 Overview

These solutions for RS Aggarwal Class 10 Maths Chapter 9 Exercise 9.6 were created by subject experts of Physics Wallah. This exercise teaches about cumulative frequency distribution and how to draw cumulative frequency curves, called ogives. The solutions explain each step clearly and simply, helping students learn how to plot and understand ogives. By practicing these exercises, students can improve their skills in analyzing and visualizing data, which is important for their exams. These expert-prepared solutions are a great resource for understanding cumulative frequency and its uses in statistics.

RS Aggarwal Solutions for Class 10 Maths Chapter 9 Exercise 9.6 PDF

The PDF link for RS Aggarwal Solutions for Class 10 Maths Chapter 9 Exercise 9.6 is available below. By using this PDF, students can easily understand how to plot and interpret cumulative frequency curves. This resource is very useful for improving data analysis and visualization skills, helping students prepare effectively for their exams.

RS Aggarwal Solutions for Class 10 Maths Chapter 9 Exercise 9.6 PDF

RS Aggarwal Solutions for Class 10 Maths Chapter 9 Mean Median Mode Of Grouped Data Cumulative Frequency Graph And Ogive Exercise 9.6

Here we have provided RS Aggarwal Solutions for Class 10 Maths Chapter 9 Exercise 9.6 for the ease of students so that they can prepare better for their exams.

Q. Which of the following is not a measure of central tendency?

(a) Mean (b) Mode (c) Median (d) Range

Solution: C

Mean and mode does not require construction of cumulative frequency, but median necessarily requires construction of cumulative frequency, unless it is raw data (in which median is the (n/2)th value, when there are n number of observations; and the average of (n/2)th and (n/2 + 1)th values, when there are n observations).

Q. Which of the following cannot be determined graphically?

(a) Mean (b) Mode (c) Median (d) None of these

Solution: C

Given: mean = 27 and median = 33 We have to find the value of mode. Empirical relationship is given by, Mode = 3(Median) – 2(Mean) ⇒ Mode = 3(33) – 2(27) ⇒ Mode = 99 – 54 = 45

Q. Which of the following measures of central tendency is influenced by extreme values?

(a) Mean (b) Mode (c) Median (d) None of these

Solution: B

Solution:

We prepare the cumulative frequency table, as shown:
Class Frequency(f) Cumulative frequency
135-140 6 6
140-145 10 16
145-150 18 34
150-155 22 56
155-160 20 76
160-165 15 91
165-170 6 97
170-175 3 100
N=Σf=100
N= 100 ⇒N/2=50 The cumulative frequency greater than 50 is 56 and the corresponding class is 150-155. Thus the median class is 150-155. Therefore, l= 150, h = 5, f = 22, c.f. of preceding class = 34 and N/2=50 Median, M=l+((N/2−cf)/f)×h Median=150+(50–34/22)×5 Median=150+3.64 Median=153.64 Hence, median is 153.64.

Question:

The median of the following data is 16. Find the missing frequencies a and b if the total of frequencies is 70.

Class 0 − 5 5 − 10 10 − 15 15 − 20 20 − 25 25 − 30 30 − 35 35 − 40
Frequency 12 a 12 15 b 6 6 4
Solution
We prepare the cumulative frequency table, as shown:
Class Frequency fi Cumulative frequency
0-5 12 12
5-10 a 12+a
10-15 12 24+a
15-20 15 39+a
20-25 b 39+a+b
25-30 6 45+a+b
30-35 6 51+a+b
35-40 4 55+a+b
Total N=Σfi=70
Let a,b be the missing frequencies of class intervals 5-10 and 20-25 respectively. Then, 55 + a + b = 70 ⇒a+b=15 ........(i) Median is 16, which lies in 15-20. So the median class is 15-20 . Therefore, l = 15, h= 5, N= 70, f = 15, and cf = 24+a Now, Median, M=l+((N/2−cf)/f)×h 16=15+(70/2−(24+a)/15)×5 ⇒16=15+(35–24–a)/3 ⇒16–15=(11−a)/3 ⇒1×3=11–a ⇒a=11–3 ⇒a=8 Therefore, b=15–a ( From (i) ) ⇒b=15–8 ⇒b=7 Hence, a= 8 and b = 7 .
Q. The distributions X and Y with total number of observations 36 and 64, and mean 4 and 3 respectively are combined.What is the mean of the resulting distribution X+Y?
Solution:
Number of observations in X=36 and Mean=4 Number of observations in Y=64 and Mean=3 Therefore Mean of the distribution X+Y=(36×4+64×3)/(36+64) =(144+192)/100 =336/100=3.36
Solution:
Using Direct method, the given data is shown as follows:
Literacy rate (%) Number of cities (fi) Class mark (xi) (fixi)
45 - 55 4 50 200
55 - 65 11 60 660
65 - 75 12 70 840
75 - 85 9 80 720
85 - 95 4 90 360
Total ∑fi=40 ∑(fi×xi)=2780
The mean of given data is given by ¯ x = ∑(fi×xi)/∑fi = 2780/40 = 69.5 Thus, the mean literacy rate is 69.5%. Let us choose a = 25, h = 10, then di = xi - 25 and ui = (xi−25)/10 Using Step-deviation method, the given data is shown as follows:
Class Frequency (fi) Class mark (xi) di= xi- 25 ui = (xi−25)/10 fiui
0 - 10 7 5 -20 -2 -14
10 - 20 10 15 -10 -1 -10
20 - 30 15 25 0 0 0
30 - 40 8 35 10 1 8
40 - 50 10 45 20 2 20
Total ∑fi=50 ∑(fi×ui)=4
The mean of given data is given by ¯ x =a+(∑fiui/∑fi)×h = 25 + 4/50 x 10 = 25 + 4/5 = (125+4)/5 = 129/5 = 25.8 Thus, the mean is 25.8
Solution:
Let us choose a = 40, h = 10, then di = xi - 40 and ui = (xi−40)/10 Using Step-deviation method, the given data is shown as follows:
Class Frequency (fi) Class mark (xi) di= xi- 40 ui = xi−40/10 fiui
5 - 15 6 10 -30 -3 -18
15 - 25 10 20 -20 -2 -20
25 - 35 16 30 -10 -1 -16
35 - 45 15 40 0 0 0
45 - 55 24 50 10 1 24
55 - 65 8 60 20 2 16
65 - 75 7 70 30 3 21
Total ∑fi=86 ∑(fi×ui)=7
The mean of given data is given by ¯ x =a+(∑fiui/∑fi)×h = 40 + (7/86) × 10 = 40 + 70/86 = 40 + 0.81 = 40.81 Thus, the mean is 40.81.
Solution:
Let us choose a = 40, h = 10, then di = xi - 40 and ui = (xi−40)/10 Using Step-deviation method, the given data is shown as follows:
Class Frequency (fi) Class mark (xi) di= xi- 40 ui = xi−40/10 fiui
5 - 15 6 10 -30 -3 -18
15 - 25 10 20 -20 -2 -20
25 - 35 16 30 -10 -1 -16
35 - 45 15 40 0 0 0
45 - 55 24 50 10 1 24
55 - 65 8 60 20 2 16
65 - 75 7 70 30 3 21
Total ∑fi=86 ∑(fi×ui)=7
The mean of given data is given by ¯ x =a+(∑fiui/∑fi)×h = 40 + (7/86) × 10 = 40 + 70/86 = 40 + 0.81 = 40.81 Thus, the mean is 40.81.
Solution:
Let us choose A = 201.5 , h = 1 , then d i = x i − 201.5 and u i = ( x i − A ) h = x i − 201.5 1 Using the Step-deviation method, the given data is shown as follows: Weight(in grams) Number of packets ( f i ) Class mark ( x i ) d i = A − x i = 201.5 − x i u i = ( x i − 201.5 ) 1 f i u i 200 − 201 13 200.5 1 1 13 201 − 202 27 201.5 0 0 0 202 − 203 18 202.5 − 1 − 1 − 18 203 − 204 10 203.5 − 2 − 2 − 10 204 − 205 1 204.5 − 3 − 3 − 3 205 − 206 1 205.5 − 4 − 4 − 4 Total ∑ f i = 70 ∑ f i u i = − 22 N = ∑ f i u i = − 22 h = 1 A = 201.5 Mean = A + ∑ f i u i N × h = 201.5 + − 22 70 × 1 = 201.5 − 0.31 = 201.19 Q. In the following data the median of the runs scored by 60 top batsmen of the world in one-day international cricket matches is 5000 . Find the missing frequencies x and y .
Runs scored 2500-3500 3500-4500 4500-5500 5500-6500 6500-7500 7500-8500
Number of batsman 5 x y 12 6 2

Solution:

We prepare the cumulative frequency table, as shown:
Runs scored No. of batsmen Cumulative frequency
2500-3500 5 5
3500-4500 x 5+x
4500-5500 y 5+x+y
5500-6500 12 17+x+y
6500-7500 6 23+x+y
7500-8500 2 25+x+y
Total N=Σfi=60
Let x,y be the missing frequencies of class intervals 3500-4500 and 4500-5500 respectively. Then, 25 + x + y = 60 ⇒x+y=35 ----------(i) Median is 5000, which lies in 4500-5500. So the median class is 4500-5500. Therefore, l= 4500, h= 1000, N= 60, f= y, and cf = 5+x Now, Median,M=l+((N/2−cf) /f)×h ⇒5000–4500 = ((30–5–x)/y)×1000 ⇒500 = ((25−x)/y)×1000 ⇒y = 50–2x ⇒35–x=50–2x ( From (i) ) ⇒2x–x=50–35 ⇒x=15 Therefore, y=35–x ⇒y=35–15 ⇒y=20 Hence, x= 15 and y = 20.
Solution:
The given series is of inclusive form. Converting it into exclusive form and preparing the cumulative frequency table, we get
Marks Frequency fi C.F
0-10 12 12
10-20 20 32
20-30 25 57
30-40 23 80
40-50 12 92
50-60 24 116
60-70 24 116
70-80 36 200
N=∑fi=200
N =200 => N/2 = 100 Now, The cumulative frequency just greater than 100 is 116 and corresponding class is 50-60. Thus, the median class is 50-60 I = 50, h = 10, f = 24, c = C.F. preceding median class = 92 and N/2 = 100 Median = I+[h×(N/2−c)/f]=50+[10×(100−92)/24] =50+[10×84/24]=50+[10×84/24]= 50 + 3.33 = 53.33 Hence, Median = 53.33

Solution:

We prepare the cumulative frequency table, as shown below:
Class Frequency Cumulative frequency
0-10 f1 f1
10-20 5 f1+5
20-30 9 f1+14
30-40 12 f1+26
40-50 f2 f1+f2+26
50-60 3 f1+f2+29
60-70 2 f1+f2+31
N=Σf=40
Now, f1+f2+31 = 40 ⇒f1+f2=9 ⇒f2=9–f1.......(i) The median is 32.5 which lies in 30 – 40. Hence, median class = 30 – 40 Therefore, l= 30,N/2=20, f= 12, and cf = 14+f1 Now, Median=32.5 Also, Median (M)=l+(N/2−cf/f)×h 32.5=30+(20–(14+f1)/12)×10 ⇒32.5=30+6–f1/12×10 ⇒2.5=6–f1/12×10 ⇒60–10f1=30 ⇒10f1=30 ⇒f1=3 From eq. (i), we have f2=9−3⇒f2=6 Q. Assertion-and-Reason Type Each question consists of two statements, namely, Assertion (A) and Reason (R).For selecting the correct answer, use the following code: (a) Both Assertion (A) and Reason (R) are the true and Reason (R) is a correct explanation of Assertion (A). (b) Both Assertion (A) and Reason (R) are the true but Reason (R) is not a correct explanation of Assertion (A). (c) Assertion (A) is true and Reason (R) is false. (d) Assertion (A) is false and Reason (R) is true. A s s e r t i o n ( A ) R e a s o n ( R ) C o n s i d e r t h e f o l l o w i n g f r e q u e n c y T h e v a l u e o f t h e v a r i a b l e w h i c h d i s t r i b u t i o n : o c c u r s m o s t o f t e n i s t h e m o d e C l a s s i n t e r v a l 3 − 6 6 − 9 9 − 12 12 − 15 15 − 18 18 − 21 F r e q u e n c y 2 5 21 23 10 12 T h e m o d e o f t h e a b o v e d a t a i s 12.4.

Solution:

The correct answer is: (b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A). Clearly, Reason (R) is true. But we got the value of mode thus: The maximum frequency is 23 and the modal class is 12-15. Using, M o d e = l + ( f 1 − f 0 2 f 1 − f 0 − f 2 ) ∗ h where l = lower limit of the modal class, h = size of the class interval (assuming all class sizes to be equal), f 1 = frequency of the modal class, f 0 = frequency of the class preceding the modal class, f 2 = frequency of the class succeeding the modal class. M o d e = 12 + ( 23 − 21 2 ∗ 23 − 21 − 10 ) ∗ 3 ⇒ M o d e = 12 + ( 2 15 ) ∗ 3 ⇒ M o d e = 12 + 0.4 ⇒ M o d e = 12.4 ∴ Assertion (A) and Reason (R) are true. However, Reason (R) isn't the correct explanation of Assertion (A). Q. The mean of 2, 7, 6 and x is 15 and the mean of 18, 1, 6, x and y is 10.What is the value of y? (a)5 (b)10 (c)20 (d)30
Solution:
In ( 1 ) case:- The average of 2, 7, 6 and x is 5 ⇒ 2 + 7 + 6 + x 4 = 5 ⇒ 15 + x 4 = 5 ⇒ 15 + x = 20 ⇒ x = 20 − 15 ⇒ x = 5 ------( 1 ) In ( 2 ) case:- the average of 18, 1, 6, x and y is 10 ⇒ 18 + 1 + 6 + x + y 5 = 10 ⇒ 25 + x + y 5 = 10 ⇒ 25 + 5 + y 5 = 10 -----from( 1 ) ⇒ 30 + y 5 = 10 ⇒ 30 + y = 50 ⇒ y = 50 − 30 ⇒ y = 20

Benefits of RS Aggarwal Solutions for Class 10 Maths Chapter 9 Exercise 9.6

  • Improved Data Analysis Skills: By practicing these solutions, students can enhance their ability to analyze and visualize data using cumulative frequency curves.
  • Exam Preparation: These solutions are excellent for exam preparation provide comprehensive practice that helps reinforce understanding and boosts confidence.
  • Conceptual Clarity: Students develop a deeper understanding of cumulative frequency and its applications, which are important for mastering statistics.
  • Expert Guidance: The solutions are prepared by subject experts from Physics Wallah, ensuring accuracy and reliable guidance for students.
  • Accessible Resource: The PDF format makes it easy for students to access and use the solutions anytime, providing a convenient study aid.

RS Aggarwal Solutions for Class 10 Maths Chapter 9 Exercise 9.6 FAQs

What skills can students improve by using these solutions?

Students can improve their data analysis and visualization skills, particularly in constructing and interpreting cumulative frequency curves (ogives).

Who prepares these solutions and why is that important?

The solutions are prepared by subject experts from Physics Wallah, ensuring they are accurate, reliable, and aligned with the curriculum, providing trustworthy guidance for students.

Can these solutions help with other statistical concepts?

Yes, while they focus on cumulative frequency and ogives, the foundational skills learned can be applied to other statistical concepts, enhancing overall mathematical proficiency.

How do RS Aggarwal Solutions help in understanding cumulative frequency distribution?

The solutions provide clear, step-by-step explanations that simplify the process of understanding cumulative frequency distribution and drawing ogives. This helps students grasp the concept more easily.
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