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 f_{i} | 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=∑ f_{i}=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

**Read More:**

**Question 2:**

We prepare the frequency table, given below

Class | Frequency f_{i} | 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=∑ f_{i}=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.

**More Resources**

**Question 3:**

We prepare the frequency table given below:

Daily wages | Frequency f_{i} | C.F. |

0 – 100 | 40 | 40 |

100 – 200 | 32 | 72 |

200 – 300 | 48 | 120 |

300 – 400 | 22 | 142 |

400 – 500 | 8 | 150 |

N=∑ f_{i}=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 f_{i} | 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=∑ f_{i}=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 f_{i} | 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=∑ f_{i}=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 f_{i} | 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=∑ f_{i}=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 f_{i} | 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 f_{1} and f_{2} 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 +f_{1} = (30 + f_{1})**Question 9:**

Let f_{1} and f_{2 }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 = f_{1} +5+9= (14 + f_{1})**Question 10:**

The given series is of inclusive form. Converting it into exclusive form and preparing the cumulative frequency table, we get

Class | Frequency f_{i} | 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=∑ f_{i}=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 f_{i} | 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=∑ f_{i}=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 f_{i} | 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=∑ f_{i}=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 f_{i} | 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=∑ f_{i}=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 f_{i} | 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=∑ f_{i}=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

**All Chapter RS Aggarwal Solutions For Class 10 Maths**

—————————————————————————–**All Subject NCERT Exemplar Problems Solutions For Class10**

**All Subject NCERT Solutions For Class 10**

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rsaggarwalsolutions.in to your friends.