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

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 9B

**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 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,

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 = 25

Hence the median is 35.33

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**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,

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 = 25

Hence the median is 28.

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**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,

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 = 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,

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 =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,

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, = 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 |

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 |

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
| 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 |

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 |

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 |

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 |

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, = 100

Hence, Median = 53.33

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

If you have any doubts, please comment below. A Plus Topper try to provide online math tutoring for you.

AGRIM says

Nice

Ananya says

In question no. 2 there’s no greater cf taken. Why??