In this chapter, we provide RS Aggarwal Solutions for Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9 D 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 9 D Maths pdf, free RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9 D 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 D |

Category | RS Aggarwal Solutions |

**Question 1:**

Let assumed mean be 35, h = 10, now we have

Class | Frequency f_{i} | Mid value x_{i} | ui=(xi−Ah) | C.F | f_{i }u_{i} |

0 – 1010 – 2020 – 3030 – 4040 – 5050 – 6060 – 70 | 510183020125 | 5152535 = A455565 | -3-2-10123 | 51533638395100 | -15-20-180202415 |

N=100 | ∑(f_{i}u_{i})=6 |

(i) Mean

=

(ii) N = 100, N2= 50

Cumulative frequency just after 50 is 63

Median class is 30 – 40

l = 30, h = 10, N = 100, c = 33, f = 30

(iii) Mode = 3 × median – 2 × mean

= 3 × 35.67 – 2 × 35.6 = 107.01 – 71.2

= 35.81

Thus, Mean = 35.6, Median = 35.67 and Mode = 35.81**Question 2:**

Let assumed mean A be 8.5. Class interval h = 3

Class | Frequency f_{i} | Mid value x_{i} | ui=(xi−Ah) | f_{i }u_{i} | C.F |

1-44-77-1010-1313-1616-19 | 630401644 | 2.55.58.5 = A11.514.517.5 | -2-10123 | -12-30016812 | 636769296100 |

N=100 | ∑(f_{i}u_{i})=-6 |

N = total frequency = 100

(i) Mean

=

(ii) N2= 50, Cumulative frequency just after 50 is 76

Median class is 7 – 10

l = 7, h = 3, N = 100, f = 40, c = 36

(iii) Mode = 3 × Median – 2 × Mean

= 3 × 8.05 – 2 × 8.32 = 24.15 – 16.64

= 7.51

Thus, mean = 8.32, Median = 8.05, Mode = 7.51**Question 3:**

Let the assumed mean A be 145. Class interval h = 10.

Class | Frequency f_{i} | Mid value x_{i} | ui=(xi−Ah) | f_{i }u_{i} | C.F |

120-130130-140140-150150-160160-170 | 2812208 | 125135145=A155165 | -2-1012 | -4-802016 | 210224250 |

N=50 | ∑(f_{i}u_{i})=24 |

(i) Mean

=

= 145 + 4.8 = 149.8

(ii) N = 50, N2= 25

Cumulative frequency just after 25 is 42

Corresponding median class is 150 – 160

Cumulative frequency before median class, c = 22

Median class frequency f = 20

(iii) Mode = 3 × median – 2 × mean

= 3 × 151.5 – 2 × 149.8 = 454.5 – 299.6

= 154.9

Thus, Mean = 149.8, Median = 151.5, Mode = 154.9**Question 4:**

Let assumed mean A = 150 and h = 20

Class | Frequency f_{i} | Mid value x_{i} | ui=(xi−Ah) | f_{i }u_{i} | C.F |

100-120120-140140-160160-180180-200 | 12148610 | 110130150= A170190 | -2-1012 | -24-140620 | 1226344050 |

N=50 | ∑(f_{i}u_{i})=-12 |

(i) Mean

(ii) N = 50, N2= 25

Cumulative frequency just after 25 is 26

Corresponding frequency median class is 120 – 140

So, l = 120, f = 14, h = 20, c = 12

(iii) Mode = 3 × Median – 2 × Mode

= 3 × 138.6 – 2 × 145.2

= 415.8 – 190.4

= 125.4

Hence, Mean = 145.2, Median = 138.6 and Mode = 125.4**Question 5:**

Let assumed mean = 225 and h = 50

Class | Frequency f_{i} | Mid value x_{i} | ui=(xi−Ah) | f_{i }u_{i} | C.F |

100-150150-200200-250250-300300-350 | 671232 | 125175225275325 | -2-1012 | -12-7034 | 613252830 |

N=30 | ∑(f_{i}u_{i})=-12 |

(i) Mean

=

(ii) N = 30, N2= 15

Cumulative frequency just after 15 is 25

corresponding class interval is 200 – 250

Median class is 200 – 250

Cumulative frequency c just before this class = 13

So I=200, f=12, N2= 15, h=50, c=13

Hence, Mean = 205 and Median = 208.33

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