In this chapter, we provide RS Aggarwal Solutions for Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9B Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9B Maths pdf, free RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9B Maths book pdf download. Now you will get step by step solution to each question.
| Textbook | NCERT |
| Class | Class 10 |
| Subject | Maths |
| Chapter | Chapter 9 |
| Chapter Name | Mean, Median, Mode of Grouped Data |
| Exercise | 9 B |
| Category | RS Aggarwal Solutions |
RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9B
Question 1:
| Class | Frequency fi | C.F. |
| 0 – 10 | 3 | 3 |
| 10 – 20 | 6 | 9 |
| 20 – 30 | 8 | 17 |
| 30 – 40 | 15 | 32 |
| 40 – 50 | 10 | 42 |
| 50 – 60 | 8 | 50 |
| N=∑ fi=50 |
Now, N=50⟹N2=25
The cumulative frequency just greater than 25 is 32 and corresponding class is 30 – 40.
Thus, the median class is 30 – 40
l = 30, h = 10, f = 15, c = C.F. preceding class 30 – 40 is 17 and N2= 25
Hence the median is 35.33
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Question 2:
We prepare the frequency table, given below
| Class | Frequency fi | C.F. |
| 0 – 7 | 3 | 3 |
| 7 – 14 | 4 | 7 |
| 14 – 21 | 7 | 14 |
| 21 – 28 | 11 | 25 |
| 28 – 35 | 0 | 25 |
| 35 – 42 | 16 | 41 |
| 42 – 49 | 9 | 50 |
| N=∑ fi=50 |
Now,N=50⟹N2=25
The cumulative frequency is 25 and corresponding class is 21 – 28.
Thus, the median class is 21 – 28
l = 21, h = 7, f = 11, c = C.F. preceding class 21 – 28 is 14 and N2= 25
Hence the median is 28.
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Question 3:
We prepare the frequency table given below:
| Daily wages | Frequency fi | C.F. |
| 0 – 100 | 40 | 40 |
| 100 – 200 | 32 | 72 |
| 200 – 300 | 48 | 120 |
| 300 – 400 | 22 | 142 |
| 400 – 500 | 8 | 150 |
| N=∑ fi=150 |
Now, N=150⟹N2=75
The cumulative frequency just greater than 75 is 120 and corresponding class is 200 – 300.
Thus, the median class is 200 – 300
l = 200, h = 100, f = 48
c = C.F. preceding median class = 72 and N2= 75
Hence the median of daily wages is Rs. 206.25.
Question 4:
We prepare the frequency table, given below:
| Class | Frequency fi | C.F. |
| 5 – 10 | 5 | 5 |
| 10 – 15 | 6 | 11 |
| 15 – 20 | 15 | 26 |
| 20 – 25 | 10 | 36 |
| 25 – 30 | 5 | 41 |
| 30 – 35 | 4 | 45 |
| 35 – 40 | 2 | 47 |
| 40 – 45 | 2 | 49 |
| N=∑ fi=49 |
Now, N=49⟹N2=24.5
The cumulative frequency just greater than 24.5 is 26 and corresponding class is 15 – 20.
Thus, the median class is 15 – 20
l = 15, h = 5, f = 15
c = CF preceding median class = 11 and N2=24.5
Median of frequency distribution is 19.5
Question 5:
We prepare the cumulative frequency table as given below:
| Consumption | Frequency fi | C.F. |
| 65 – 85 | 4 | 4 |
| 85 – 105 | 5 | 9 |
| 105 – 125 | 13 | 22 |
| 125 – 145 | 20 | 42 |
| 145 – 165 | 14 | 56 |
| 165 – 185 | 7 | 63 |
| 185 – 205 | 4 | 67 |
| N=∑ fi=67 |
Now, N=67⟹N2=33.5
The cumulative frequency just greater than 33.5 is 42 and the corresponding class 125 – 145.
Thus, the median class is 125 – 145
l = 125, h = 20, and c = CF preceding the median class = 22, N2= 33.5
Hence median of electricity consumed is 136.5
Question 6:
Frequency table is given below:
| Hieght | Frequency fi | C.F. |
| 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=∑ fi=100 |
N=100⟹N2=50
The cumulative frequency just greater than 50 is 56 and the corresponding class is 150 – 155
Thus, the median class is 150 – 155
l = 150, h = 5, f = 22, c = C.F.preceding median class = 34
Hence, Median = 153.64
Question 7:
The frequency table is given below. Let the missing frequency be x.
| Class | Frequency fi | C.F. |
| 0 – 10 | 5 | 5 |
| 10 – 20 | 25 | 30 |
| 20 – 30 | x | 30+x |
| 30 – 40 | 18 | 48+x |
| 40 – 50 | 7 | 55+x |
Median = 24
Median class is 20 – 30
l = 20, h = 10, f = x, c = C.F. preceding median class = 30
Hence, the missing frequency is 25.
Question 8:
Let f1 and f2 be the frequencies of classes 20 – 30 and 40 – 50 respectively, then
Median is 35, which lies in 30 – 40, so the median class is 30 – 40.
l = 30, h = 10, f = 40, N = 170 and c = 10 + 20 +f1 = (30 + f1)
Question 9:
Let f1 and f2 be the frequencies of class intervals 0 – 10 and 40 – 50
Median is 32.5 which lies in 30 – 40, so the median class is 30 – 40
l = 30, h = 10, f = 12, N = 40 and c = f1 +5+9= (14 + f1)
Question 10:
The given series is of inclusive form. Converting it into exclusive form and preparing the cumulative frequency table, we get
| Class | Frequency fi | C.F. |
| 18.5 – 25.5 | 35 | 35 |
| 25.5 – 32.5 | 96 | 131 |
| 32.5 – 39.5 | 68 | 199 |
| 39.5 – 46.5 | 102 | 301 |
| 46.5 – 53.5 | 35 | 336 |
| 53.5 – 60.5 | 4 | 340 |
| N=∑ fi=340 |
N=340⟹N2=170
The cumulative frequency just greater than 170 is 199 and the corresponding class is 32.5 – 39.5.
Median class is 32.5 – 39.5
l = 32.5, h = 7, f = 68, c = C.F. of preceding median class = 131
Hence median is 36.5 years
Question 11:
Given series is in inclusive form converting it into exclusive form and preparing the cumulative frequency table, we get
| Wages per day(in Rs.) | Frequency fi | C.F. |
| 60.5 – 70.5 | 5 | 5 |
| 70.5 – 80.5 | 15 | 20 |
| 80.5 – 90.5 | 20 | 40 |
| 90.5 – 100.5 | 30 | 70 |
| 100.5 – 110.5 | 20 | 90 |
| 110.5 – 120.5 | 8 | 98 |
| N=∑ fi=98 |
N=98⟹N2=49
The cumulative frequency just greater than 49 is 70 and corresponding class is 90.5 – 100.5.
median class is 90.5 – 100.5
l = 90.5, h = 10, f = 30, c = CF preceding median class = 40
Hence, Median = Rs 93.50
Question 12:
The given series is converted from inclusive to exclusive form and preparing the cumulative frequency table, we get
| Marks | Frequency fi | C.F. |
| 10.5 – 15.5 | 2 | 2 |
| 15.5 – 20.5 | 3 | 5 |
| 20.5 – 25.5 | 6 | 11 |
| 25.5 – 30.5 | 7 | 18 |
| 30.5 – 35.5 | 14 | 32 |
| 35.5 -40.5 | 12 | 44 |
| 40.5 -45.5 | 4 | 48 |
| 45.5 -50.5 | 2 | 50 |
| N=∑ fi=50 |
N=50⟹N2=25
The cumulative frequency just greater than 25 is 32.
The corresponding class is 30.5 – 35.5.
Thus, the median class is 30.5 – 35.5
l = 30.5, h = 5, f = 14, c = C.F preceding median class = 18
Hence, Median = 33
Question 13:
The given series is converted from inclusive to exclusive form and preparing the cumulative frequency table, we get
| Marks | Frequency fi | C.F. |
| 0.5 – 5.5 | 7 | 7 |
| 5.5 – 10.5 | 10 | 17 |
| 10.5 – 15.5 | 16 | 33 |
| 15.5 – 20.5 | 32 | 65 |
| 20.5 – 25.5 | 24 | 89 |
| 25.5 – 30.5 | 16 | 105 |
| 30.5 – 35.5 | 11 | 116 |
| 35.5 – 40.5 | 5 | 121 |
| 40.5 – 45.5 | 2 | 123 |
| N=∑ fi=123 |
N=123⟹N2=61.5
The cumulative frequency just greater than 61.5 is 65.
The corresponding median class is 15.5 – 20.5.
Then the median class is 15.5 – 20.5
l = 15.5, h = 5, f = 32, c = C.F. preceding median class = 33
Hence, Median = 19.95
Question 14:
| 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⟹N2=100
The cumulative frequency just greater than 100 is 116 and the corresponding class is 50 – 60.
Thus the median class is 50 – 60
l = 50, h = 10, f = 24, c = C.F. preceding median class = 92, N2= 100
Hence, Median = 53.33
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