RS Aggarwal Solutions Chapter 9 Mean, Mode and Median Exercise 9C Class 10 Maths
Chapter Name  RS Aggarwal Chapter 9 Mean, Mode and Median 
Book Name  RS Aggarwal Mathematics for Class 10 
Other Exercises 

Related Study  NCERT Solutions for Class 10 Maths 
Exercise 9C Solutions
1. Find the mode of the following distribution:
Marks 
10  20 
20  30 
30  40 
40  50 
50  60 
Frequency 
12 
35 
45 
25 
13 
Solution
Here, the maximum class frequency is 45, and the class corresponding to this frequency is 30 – 40. So, the modal class is 30 – 40.
Now,
Modal class=30 – 40, lower limit (l) of modal class=30, class size (h) = 10,
Frequency (f_{1}) of the modal class=45,
Frequency (f_{0}) of class preceding the modal class=35,
frequency (f_{2}) of class succeeding the modal class=25
Now, let us substitute these values in the formula:
Mode = l + (f_{1} – f_{0})/(2f_{1} – f_{0} – f_{2}) × h
= 30 + (45 – 35)/(90 – 35 – 45) × 10
= 30 + (10/30) × 10
= 30 + 3.33
= 33.33
Hence, the mode is 33.33.
2. Compute the mode of the following data:
Class 
0  20 
20  40 
40  60 
60  80 
80  100 
Frequency 
25 
16 
28 
20 
5 
Solution
Here, the maximum class frequency is 28, and the class corresponding to this frequency is 40 – 60. So, the modal class is 40 – 60.
Now,
Modal class = 40 – 60, lower limit (l) of modal class=40, Class size (h) = 20,
frequency (f_{1}) of the modal class = 28,
frequency (f_{0}) of class preceding the modal class = 16,
frequency (f_{2}) of class succeeding the modal class = 20
Now, let us substitute these values in the formula:
Mode = l + (f_{1} – f_{0})/(2f_{1} – f_{0} – f_{2}) × h
= 40 + (28 – 16)/(56 – 16 – 20) × 20
= 40 + (12/20) × 20
= 40 + 12
= 52.
Hence, the mode is 52.
3. Heights of students of class X are given in the flowing frequency distribution
Height (in cm) 
150  155 
155  160 
160  165 
165  170 
170  175 
Number of students 
15 
8 
20 
12 
5 
Find the modal height.
Also, find the mean height. compared and intercept the two measures of central tendency.
Solution
Here, the maximum class frequency is 20, and the class corresponding to this frequency is 160 – 165. So, the modal class is 160 – 165.
Now,
Modal class=160 – 165, lower limit (l) of modal class=160, class size (h) = 5,
frequency (f_{1}) of the modal class=20,
frequency (f_{0}) of class preceding the modal class=8
frequency (f_{2}) of class succeeding the modal class=12
Now, let us substitute these values in the formula:
Mode = l + (f_{1 }– f_{0})/(2f_{1} – f_{0} – f_{2}) × h
= 160 + (20 – 8)/(40 – 8 – 12) × 5
= 160 + (12/20) × 5
= 160 + 3
= 163
Hence, the mode is 163.
It represents that the height of maximum number of students is 163 cm.
Now, to find the mean let us put the data in the table given below:
Height (in cm) 
Number of students (f_{1}) 
Class mark (x_{i}) 
f_{i} x_{i} 
150  155 
15 

2287.5 
155  160 
8 

1260 
160  165 
20 

3250 
165  170 
12 

2010 
170  175 
5 

862.5 
Total 
Î£f_{i} = 60 

Î£f_{i} x_{i} = 9670 
= 161.17
Thus, mean of the given data is 161.17.
It represents that on an average, the height of a student is 161.17 cm.
4. Find the mode of the following distribution:
Class interval 
10  14 
14  18 
18  22 
22  26 
26  30 
30  34 
34  38 
38  42 
Frequency 
8 
6 
11 
20 
25 
22 
10 
4 
Solution
As the class 26 – 30 has the maximum frequency, it is the modal class.
Now, x_{k} = 26, h = 4, f_{k }= 25, f_{k+1} = 22
∴ Mode, M_{0} = x_{k} + {h × (f _{k} – f_{k – 1})/(2f_{k} – f_{k  1} – f_{k+1})}
= 26 + {4 × (25 – 20)/(2 × 25 – 20 – 22)}
= 26 + {4 × 5/8}
= (26 + 2.5)
= 28.5
5. Given below is the distribution of total household expenditure of 200 manual workers in a city:
Expenditure (in Rs) 
Number of manual workers 
1000  1500 
24 
1500  2000 
40 
2000  2500 
31 
2500  3000 
28 
3000 – 3500 
32 
3500 – 4000 
23 
4000 – 4500 
17 
4500  5000 
5 
Solution
As the class 1500 – 2000 has the maximum frequency, it is the modal class.
Now, x_{k} = 1500, h = 500, f_{k} = 40, f_{k1} = 24, f_{k+1} = 31
∴ Mode, M_{0} = x_{k} + {h × (f_{k} – f_{k1})/(2f_{k} – f_{k1} – f_{k+1})}
= 1500 + {500 × (40 – 24)/(2 × 40 – 24 – 31)}
= 1500 + {500 × 16/25}
= (1500 + 320)
= 1820
Hence, mode = Rs 1820
6. Calculate the mode from the following data:
Monthly salary (in Rs) 
No of employees 
0  5000 
90 
5000  10000 
150 
10000  15000 
100 
15000  20000 
80 
20000  25000 
70 
25000  30000 
10 
Solution
As the class 5000 – 10000 has the maximum frequency, it is the modal class.
Now, x_{k} = 5000, h = 5000, f_{k} = 150, f_{k+1} = 100
∴ Mode, M_{0} = x_{k} + {h × (f_{k} – f_{k 1})/(2f_{k} – f_{k1} – f_{k+1})}
= 5000 + {5000 × (150 – 90)/(2 × 150 – 90 100)}
= 5000 + {5000 × 60/110}
= (5000 + 2727.27)
= 7727.27
Hence, mode = Rs 7727.27
7. Compute the mode from the following data:
Age (in years) 
0  5 
5  10 
10  15 
15  20 
20  25 
25  30 
30  35 
No. of patients 
6 
11 
18 
24 
17 
13 
5 
Solution
As the class 15 – 20 has the maximum frequency, it is the modal class
Now, x_{k} = 15, h = 5, f_{k} = 24, f_{k1} = 18, f_{k + 1} = 17
∴ Mode, M_{0} = x_{k} + {h × (f_{k} – f_{k 1})/(2f_{k} – f_{k1} – f_{k+1})}
= 15 + {5 × (24 – 18)/(2 × 24 – 18 – 17)}
= 15 + {5 × 6/13}
= (15 + 2.3)
= 17.3
Hence, mode = 17.3 years
8. Compute the mode from the following series:
Size 
45  55 
55  65 
65  75 
75  85 
85  95 
95  105 
105  115 
Frequency 
7 
12 
17 
30 
32 
6 
10 
Solution
As the class 85 – 95 has the maximum frequency, it is modal class.
Now, x_{k} = 85, h = 10, f_{k }= 32, f_{k+1} = 6
∴ Mode, M_{0} = x_{k} + {h × (f_{k} – f_{k1})/(2f_{k} – f_{k1} – f_{k+1})}
= 85 + {10 × (32 – 30)/(2 × 32 – 30 – 6)}
= 85 + {10 × 2/28}
= (85 + 0.71)
= 85.71
Hence, mode = 85.71
9. Compute the mode from the following data:
Class interval 
1  5 
6  10 
11  15 
16  20 
21  25 
26  30 
31  35 
36  40 
41  45 
46  50 
Frequency 
3 
8 
13 
18 
28 
20 
13 
8 
6 
4 
Solution
Clearly, we have to find the mode of the data. The given data is an inclusive series. So, we will convert it to an exclusive form as given below:
Class Interval 
0.5 – 5.5 
5.5 – 10.5 
10.5 – 15.5 
15.5  20.5 
20.5 – 25.5 
25.5 – 30.5 
30.5 – 35.5 
35.5 – 40.5 
40.5 – 45.5 
45.5 – 50.5 
Frequency 
3 
8 
13 
18 
28 
20 
13 
8 
6 
4 
As the class 20.5 – 25.5 has the maximum frequency, it is modal class.
Now, x_{k} = 20.5, h = 5, f_{k} = 28, f_{k1} = 18, f_{k+1} = 20
∴ Mode, M_{0} = x_{k} + {h × (f_{k} – f_{k1})/(2f_{k} – f_{k1} – f_{k+1})}
= 20.5 + {5 × (28 – 18)/(2 × 28 – 18 – 20)}
= 20.5 + {5 × 10/18}
= (20.5 + 2.78)
= 23.78
Hence, mode = 23.28.
10. The age wise participation of students in the annual function of a school is shown in the following distribution.
Age (in years) 
5  7 
7  9 
9  11 
11  13 
13  15 
15  17 
17  19 
Number of students 
x 
15 
18 
30 
50 
48 
x 
Find the missing frequencies when the sum of frequencies is 181. Also find the mode of the data.
Solution
It is given that the sum of frequencies is 181.
∴ x + 15 + 18 + 30 + 50 + 48 + x = 181
⇒ 2x + 161 = 181
⇒ 2x = 181 – 161
⇒ 2x = 20
⇒ x = 10
Thus, x = 10
Here, the maximum class frequency is 50, and the class corresponding to this frequency is 13 – 15. So, the modal class is 13 – 15.
Now,
Modal class=13 – 15, lower limit (l) of modal Class = 13, class size (h) = 2,
frequency (f_{1}) of the modal class=50,
frequency (f_{0}) of class preceding the modal class=30,
frequency (f_{2}) of class succeeding of modal class=48
Now, let us substitute these values in the formula:
Mode = l + (f_{1} – f_{0})/(2f_{1} – f_{0} – f_{2}) × h
= 13 + (50 – 30)/(100 – 30 – 48) × 2
= 13 + (20/22) × 2
= 13 + 1.82
= 14.82
Hence, the mode is 14.82.