RS Aggarwal Solutions Chapter 9 Mean, Mode and Median Exercise 9D 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 9D Solutions
1. Find the mean, median and mode of the following data:
Class 
0  10 
10  20 
20  30 
30  40 
40  50 
50  60 
60  70 
Frequency 
4 
4 
7 
10 
12 
8 
5 
Solution
To find the mean let us put the data in the table given below:
Class 
Frequency (f_{i}) 
Class mark (x_{i}) 
f_{i} x_{i} 
0  10 
4 
5 
20 
10  20 
4 
15 
60 
20  30 
7 
25 
175 
30  40 
10 
35 
350 
40  50 
12 
45 
540 
50 – 60 
8 
55 
440 
60 70 
5 
65 
325 
Total 
Î£f_{i }= 50 

Î£f_{i} x_{i }= 1910 
= 1910/50
= 38.2
Thus, the mean of the given data is 38.2.
Now, to find the median let us put the data in the table given below:
Class 
Frequency (f_{i}) 
Cumulative Frequency (cf) 
0  10 
4 
4 
10  20 
4 
8 
20  30 
7 
15 
30  40 
10 
25 
40  50 
12 
37 
50 – 60 
8 
45 
60 – 70 
5 
50 
Total 
N = Î£f_{i} = 50 

Now, N = 50
⇒ N/2 = 25
The cumulative frequency just greater than 25 is 37 and the corresponding class is 40 – 50.
Thus, the median class is 40 – 50.
∴ l = 40, h = 10, N = 50, f = 12 and cf = 25.
Now,
Median = l + ((N/2 – cf)/f) × h
= 40 + (25 – 25)/12 × 10
= 40.
Thus, the median is 40.
We know that,
Mode = 3(Median) – 2(Mean)
= 3 × 40 – 2 × 38.2
= 120 – 76.4
= 43.6
Hence, Mean = 38.2, Median = 40 and Mode = 43.6
2. Find the mean, median and mode of the following data:
Class 
0  20 
20  40 
40  60 
60  80 
80  100 
100  120 
120  140 
Frequency 
6 
8 
10 
12 
6 
5 
3 
Solution
To find the mean let us put the data in the table given below:
Class 
Frequency (f_{i}) 
Class mark (x_{i}) 
f_{i }x_{i} 
0  20 
6 
10 
60 
20  40 
8 
30 
240 
40  60 
10 
50 
500 
60  80 
12 
70 
840 
80  100 
6 
90 
540 
100  120 
5 
110 
550 
120  140 
3 
130 
390 
Total 
Î£f_{i }= 50 

Î£f_{i} x_{i} = 3120 
= 3120/50
= 62.4
Thus, the mean of the given data is 62.4.
Now, to find the median let us put the data in the table given below:
Class 
Frequency (f_{i}) 
Cumulative Frequency (cf) 
0  20 
6 
6 
20  40 
8 
14 
40  60 
10 
24 
60  80 
12 
36 
80  100 
6 
42 
100 – 120 
5 
47 
120 – 140 
3 
50 
Total 
N = Î£f_{i }= 50 

Now, N = 50
⇒ N/2 = 25
The cumulative frequency just greater than 25 is 36 and the corresponding class is 60 – 80.
Thus, the median class is 60 – 80.
∴ l = 60, h = 20, N = 50, f = 12 and cf = 24.
Now,
Median = l + ((N/2 – cf)/f) × h
= 60 + (25 – 24)/12 × 20
= 60 + 1.67
= 61.67
Thus, the median is 61.67.
We know that,
Mode = 3(Median) – 2(mean)
= 3 × 61.67 – 2 × 62.4
= 185.01 – 124.8
= 60.21
Hence, Mean = 62.4, Median = 61.67 and Mode = 60.21
3. Find the mean, median and mode of the following data:
Class 
0  50 
5  100 
100  150 
150  200 
200  250 
250  300 
300  350 
Frequency 
2 
3 
5 
6 
5 
3 
1 
Solution
To find the mean let us put the data in the table given below:
Class 
Frequency (f_{i}) 
Class mark (x_{i}) 
f_{i} x_{i} 
0  50 
2 
25 
50 
50  100 
3 
75 
225 
100  150 
5 
125 
625 
150  200 
6 
175 
1050 
200  250 
5 
225 
1125 
250  300 
3 
275 
825 
300 – 350 
1 
325 
325 
Total 
Î£f_{i } = 25 

Î£f_{i} x_{i} = 4225 
= 4225/25
= 169
Thus, mean of the given data is 169.
Now, to find the median let us put the data in the table given below:
Class 
Frequency (f_{i}) 
Cumulative frequency (cf) 
0  50 
2 
2 
50  100 
3 
5 
100  150 
5 
10 
150  200 
6 
16 
200 – 250 
5 
21 
250 – 300 
3 
24 
300 – 350 
1 
25 
Total 
N = Î£f_{i} = 25 

Now, N = 25, ⇒ N/2 = 12.5
The cumulative frequency just greater than 12.5 is 16 and the corresponding class is 150 – 200.
Thus, the median class is 150 – 200.
∴ l = 150, h = 50, N = 25, f = 6 and cf = 10.
Now,
Median = l + ((N/2 – cf)/f) × h
= 150 + (12.5 – 10)/6 × 50
= 150 + 20.83
= 170.83
Thus, the median is 170.83
We know that,
Mode = 3(Median) – 2(mean)
= 3 × 170.83 – 2 × 169
= 512.49 – 33.8
= 174.49
Hence, mean = 169, Median = 170.83 and Mode = 174.49
4. Find the mean, median and mode of the following data:
Marks Obtained 
25  35 
35  45 
45  55 
55  65 
65  75 
75  85 
No. of students 
7 
31 
33 
17 
11 
1 
Solution
To find the mean let us put the data in the table given below:
Marks Obtained 
Number of students (f_{i}) 
Class mark (x_{i}) 
f_{i }x_{i} 
25  35 
7 
30 
210 
35  45 
31 
40 
1240 
45  55 
33 
50 
1650 
55  65 
17 
60 
1020 
65  75 
11 
70 
770 
75  85 
1 
80 
80 
Total 
Î£f_{i} = 100 

Î£f_{i} x_{i} = 4970 
= 49.7
Thus, mean of the given data is 49.7.
Now, to find the median let us put the data in the table given below:
Class 
Frequency (f_{i}) 
Cumulative frequency (cf) 
25  35 
7 
7 
35  45 
31 
38 
45  55 
33 
71 
55  65 
17 
88 
65  75 
11 
99 
75  85 
1 
100 
Total 
N = Î£f_{i} = 100 

Now, N = 100 ⇒ N/2 = 50.
The cumulative frequency just greater than 50 is 71 and the corresponding class is 45 – 55.
Thus, the median class is 45 – 55.
∴ l = 45, h = 10, N = 100, f = 33 and cf = 38.
Now,
Median = l + ((N/2 – cf)/f) × h
= 45 + (50 – 38)/33 × 10
= 45 + 3.64
= 48.64
Thus, the median is 48.64
We know that,
Mode = 3(Median) – 2(mean)
= 3 × 48.64 – 2 × 49.70
= 145.92 – 99.4
= 46.52
Hence, Mean = 49.70, Median = 48.64 and Mode = 46.52
5. A survey regarding the heights (in cm) of 50 girls of a class was conducted and the following data was obtained:
Height in cm 
120  130 
130  140 
140  150 
150  160 
160  170 
No. of girls 
2 
8 
12 
20 
8 
Find the mean, median and mode of the above data.
Solution
We have the following:
Height in cm 
Mid value (x_{i}) 
Frequency (f_{i}) 
Cumulative frequency 
(f_{i} × x_{i}) 
120  130 
125 
2 
2 
250 
130  140 
135 
8 
10 
1080 
140  150 
145 
12 
22 
1740 
150  160 
155 
20 
42 
3100 
160  170 
165 
8 
50 
1320 


Î£f_{i }= 50 

Î£f_{i} × x_{i} = 7490 
= 149.8
Now, N = 50
⇒ N/2 = 25.
The cumulative frequency just greater than 25 is 42 and the corresponding class is 150 – 160.
Thus, the median class is 150 – 160.
∴ l = 150, h = 10, f = 20, c = cf of preceding class=22 and N/2 = 25
Now,
Median, Me = l + {h × ((N/2 – c)/f)}
= 150 + {10 × ((25 – 22)/20)}
= (150 + 10 × 3/20)
= 151.5
Mode = 3(median) – 2(Mean)
= 3 × 151.5 – 2 × 149.8
= 154.9
6. The following table gives the daily income of 50 workers of a factory:
Daily income (in Rs) 
100  120 
120  140 
140  160 
160  180 
180  200 
No. of workers 
12 
14 
8 
6 
10 
Find the mean, median and mode of the above data.
Solution
We have the following:
Daily income 
Mid values (x_{i}) 
Frequency (f_{i}) 
Cumulative frequency 
(f_{i} × x_{i}) 
100  120 
110 
12 
12 
1320 
120  140 
130 
14 
26 
1820 
140  160 
150 
8 
34 
1200 
160  180 
170 
6 
40 
1020 
180  200 
190 
10 
50 
1900 


Î£f_{i} = 50 

Î£f_{i} × x_{i} = 7260 
= 7260/50
= 145.2
Now, N = 50
⇒ N/2 = 25
The cumulative frequency just greater than 25 is 26 and the corresponding class is 120 – 140.
Thus, the median class is 120 – 140.
∴ l = 120, h = 20, f = 14, c = cf of preceding class=12 and N/2 = 25
Now,
Median, M_{e }= l + {h × ((N/2 – c)/f)}
= 120 + {20 × (25 – 12)/14}
= (120 + 20 × 13/14)
= 138.57
Mode = 3(median) – 2(mean)
= 3 × 138.57 – 2 × 145.2
= 125.31
7. The table shows the daily expenditure on food of 30 households in a locality:
Daily expenditure (in Rs) 
Number of households 
100  150 
6 
150  200 
7 
200  250 
12 
250  300 
3 
300  350 
2 
Find the mean and median daily expenditure on food.
Solution
We have the following:
Daily expenditure (in Rs) 
Mid value (x_{i}) 
Frequency (f_{i}) 
Cumulative frequency 
(f_{i }× x_{i}) 
100  150 
125 
6 
6 
750 
150  200 
175 
7 
13 
1225 
200  250 
225 
12 
25 
2700 
250  300 
275 
3 
28 
825 
300  350 
325 
2 
30 
650 


Î£f_{i} = 30 

Î£f_{i} × x_{i} = 6150 
⇒ N/2 = 15
The cumulative frequency just
greater than 15 is 25 and the corresponding class is 200 – 250.
Thus, the median class is 200 –
250.
∴ l = 200, h = 50, f = 12, c = cf of preceding class = 13
and N/2 = 15
Now,
Median, M_{e} = l + {h ×
(N/2 – c)/f)}
= 200 + {50 + (15 – 13)/12)}
= (200 + 50 × 2/12)
= 200 + 8.33
= 208.33