## RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9C & 9D

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9C & 9D

**Other Exercises**

- RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9A
- RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9B
- RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9C & 9D
- RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9E
- Mean, Median, Mode of Grouped Dat

**Exercise 9C**

**Question 1:**

As the class 50 – 60 has maximum frequency, so it is the modal class:

Hence, mode = 53.33

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**Question 2:**

As the class 40 – 50 as maximum frequency, so it s modal class.

Hence, mode = 43.75 years

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**Question 3:**

As the class 26 – 30 has maximum frequency so it is modal class

Hence, mode = 28.5

**Question 4:**

As the class 1500 – 2000 has maximum frequency, so it os modal class

Hence the average expenditure done by maximum number of workers = Rs. 1820

**Question 5:**

As the class 5000 – 10000 has maximum frequency, so it is modal class

Hence, mode = Rs. 7727.27

**Question 6:**

As the class 15 – 20 has maximum frequency so it is modal class.

Hence mode = 17.3 years

**Question 7:**

As the class 85 – 95 has the maximum frequency it is modal class

Hence, mode = 85.71

**Question 8:**

The given series is converted from inclusive to exclusive form and on preparing the frequency table, we get

Class | Frequency |

0.5 – 5.5 | 3 |

5.5 – 10.5 | 8 |

10.5 – 15. 5 | 13 |

15.5 – 20.5 | 18 |

20.5 – 25. 5 | 28 |

25.5 – 30.5 | 20 |

30.5 – 35.5 | 13 |

35.5 – 40.5 | 8 |

40.5 – 45.5 | 6 |

45.5 – 50.5 | 3 |

As the class 20.5 – 25.5 has maximum frequency, so it is modal class

Hence, mode = 23.28

**Exercise 9D**

**Question 1:**

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

Class | Frequency f_{i} | Mid value x_{i} | \({ u }_{ i }=\left( \frac { { x }_{ i }-A }{ h } \right) \) | C.F | f_{i }u_{i} |

0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 | 5 10 18 30 20 12 5 | 5 15 25 35 = A 45 55 65 | -3 -2 -1 0 1 2 3 | 5 15 33 63 83 95 100 | -15 -20 -18 0 20 24 15 |

N=100 | ∑(f_{i}_{ }u_{i})=6 |

(i) Mean

=

(ii) N = 100, \(\frac { N }{ 2 } \)= 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} | \({ u }_{ i }=\left( \frac { { x }_{ i }-A }{ h } \right) \) | f_{i }u_{i} | C.F |

1-4 4-7 7-10 10-13 13-16 16-19 | 6 30 40 16 4 4 | 2.5 5.5 8.5 = A 11.5 14.5 17.5 | -2 -1 0 1 2 3 | -12 -30 0 16 8 12 | 6 36 76 92 96 100 |

N=100 | ∑(f_{i}_{ }u_{i})=-6 |

N = total frequency = 100

(i) Mean

=

(ii) \(\frac { N }{ 2 } \)= 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} | \({ u }_{ i }=\left( \frac { { x }_{ i }-A }{ h } \right) \) | f_{i }u_{i} | C.F |

120-130 130-140 140-150 150-160 160-170 | 2 8 12 20 8 | 125 135 145=A 155 165 | -2 -1 0 1 2 | -4 -8 0 20 16 | 2 10 22 42 50 |

N=50 | ∑(f_{i}_{ }u_{i})=24 |

(i) Mean

=

= 145 + 4.8 = 149.8

(ii) N = 50, \(\frac { N }{ 2 } \)= 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} | \({ u }_{ i }=\left( \frac { { x }_{ i }-A }{ h } \right) \) | f_{i }u_{i} | C.F |

100-120 120-140 140-160 160-180 180-200 | 12 14 8 6 10 | 110 130 150= A 170 190 | -2 -1 0 1 2 | -24 -14 0 6 20 | 12 26 34 40 50 |

N=50 | ∑(f_{i}_{ }u_{i})=-12 |

(i) Mean

(ii) N = 50, \(\frac { N }{ 2 } \)= 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} | \({ u }_{ i }=\left( \frac { { x }_{ i }-A }{ h } \right) \) | f_{i }u_{i} | C.F |

100-150 150-200 200-250 250-300 300-350 | 6 7 12 3 2 | 125 175 225 275 325 | -2 -1 0 1 2 | -12 -7 0 3 4 | 6 13 25 28 30 |

N=30 | ∑(f_{i}_{ }u_{i})=-12 |

(i) Mean

=

(ii) N = 30, \(\frac { N }{ 2 } \)= 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, \(\frac { N }{ 2 } \)= 15, h=50, c=13

Hence, Mean = 205 and Median = 208.33

Hope given RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9C & 9D are helpful to complete your math homework.

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