RS Aggarwal Solutions Chapter 9 Mean, Mode and Median Exercise 9F 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 9F Solutions
1. Write the median class of the following distribution:
Class 
0  10 
10  20 
20  30 
30  40 
40  50 
50  60 
60  70 
Frequency 
4 
4 
8 
10 
12 
8 
4 
Solution
To find 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 
8 
16 
30  40 
10 
26 
40  50 
12 
38 
50  60 
8 
46 
60  70 
4 
50 
Total 
N = Î£f_{i} = 50 

Now, N = 50 ⇒ N/2 = 25
The cumulative frequency just greater than 25 is 26, and the corresponding class is 30 – 40.
Thus, the median class is 30 – 40.
Age (in years) 
0  10 
10  20 
20  30 
30  40 
40  50 
50  60 
Number of patients 
16 
13 
6 
11 
27 
18 
Solution
Here, the maximum class frequency is 27, and the class corresponding to this frequency is 40 – 50. So, the modal class is 40 – 50.
Now,
Modal class = 40 – 50, lower limit (l) of modal class = 40.
Thus, lower limit (l) of modal class is 40.
Monthly pocket money (in ₹) 
0  50 
50  100 
100  150 
150  200 
200  250 
250  300 
Number of students 
2 
7 
8 
30 
12 
1 
Solution
Here the maximum class frequency is 30, and the class corresponding to the
frequency is 150 200. So, the modal class is 150  200.
Also, class mark of the modal class is (150 + 200)2 = 175.
4. A data has 25 observations arranged in a descending order. Which
observation represents the median?
Solution
If the number of observations is odd, then the median is (n + 1)/2th
observation.
Thus, (25 + 1)/2 = 13th observation represents the median.
5. For a certain distribution, mode and median were found to be 1000 and
1250 respectively. Find mean for this distribution using an empirical relation.
Solution
There is an empirical relationship between the three measures of central
tendency.
3 median = mode + 2 mean
⇒ Mean = (3Median  Mode)/2
= (3(1250)  1000)/2
= 1375
Thus, the mean is 1375.
6. In a class test, 50 students obtained marks as follows:
Marks obtained 
0  20 
20  40 
40  60 
60  80 
80  100 
Number of students 
4 
6 
25 
10 
5 
Find the modal class and the median class.
Solution
Here the maximum class frequency is 25, and the class corresponding to this frequency is 40 – 60.
So, the modal class is 40 – 60.
Now, to find the median class let us put the data in the table given below:
Marks Obtained 
Number of students (fi) 
Cumulative frequency (cf) 
0  20 
4 
4 
20  40 
6 
10 
40  60 
25 
35 
60  80 
10 
45 
80 – 100 
5 
50 
Total 
N = Î£f_{i} = 50 

Now, N = 50 ⇒ N/2 = 25.
The cumulative frequency just greater than 25 is 35, and the corresponding class is 40 – 60. Thus, the median class is 4060.
7. Find the class marks of classes 10 – 25 and 3555.
Solution
Class mark = (Upper limit + Lower limit)/2
∴ Class mark of 10  25 = (10 + 25)/2
And class mark of 35  55 = (35 + 55)/2
= 45
8. While calculating the mean of a given data by the assumed mean method. the following values were obtained.
A = 25, Î£f_{i}d_{i} = 110, Î£f_{i }= 50
Find the mean.
Solution
According to assumedmean method,
= 25 + 110/50
= 25 + 2.2
= 27.2
Thus, mean is 27.2
9. The distribution X and Y with total number of observations 36 and 64, and mean 4 and 3 respectively are combined. What is the mean of the resulting distribution X + Y?
Solution
According to the question,
4 = X/36 and 3 = Y/64
⇒ X = 4 × 36 and Y = 3 × 64
⇒ X = 144 and Y = 192
Now, X + Y = 144 + 192 = 336
And total number of observations = 36 + 64 = 100
Thus, mean = 336/100 = 3.36.
10. In a frequency distribution table with 12 classes, the classwidth is 2.5 and the lowest class boundary is 8.1, then what is the upper class boundary of the highest class?
Solution
Upper class boundary = Lowest class boundary + width × number of classes
= 8.1 + 2.5×12
= 8.1 + 30
= 38.1
Thus, upper class boundary of the highest class is 38.1.
11. á¹¬he observation 29, 32, 48, 50, x + 2, 72, 78, 84, 95 are arranged in ascending order. What is the value of x if the median of the data is 63?
Solution
If number of observations is even, then the median will be the average of (n/2)th and the (n/2 + 1)th observations.
In the given case, n = 10 ⇒ (n/2)th = 5^{th} and (n/2 + 1)th = 6th observation.
Thus, 63 = x + (x + 2)/2
⇒ 126 = 2x + 2
⇒ 124 = 2x
⇒ x = 62
Thus, the value of x is 62.
12. The median of 19 observations is 30. Two more observation are made and the values of these are 8 and 32. Find the median of the 21 observations taken together.
Hint: Since 8 is less than 30 and 32 is more than 30, so the value of median (middle value) remains unchanged.
Solution
Since, 8 is less than 30 and 32 is more than 30, so the middle value remains unchanged
Thus, the median of 21 observations taken together is 30.
13. If the median of x/5, x/4, x/2, x and x/3, where x > 0, is 8, find the value of x.
Hint: Arranging the observations in ascending order, we have x/5, x/4, x/3, x/2, x Median = x/3 = 8.
Solution
Arranging the observations in ascending order, we have
x/5, x/4, x/3, x/2, x
Thus, the median is x/3
⇒ x/3 = 8
⇒ x = 3 × 8
⇒ x = 24
Thus, the value of x is 24.
14. What is the cumulative frequency of the modal class of the following distribution?
Class 
3  6 
6  9 
9  12 
12  15 
15  18 
18  21 
21  24 
Frequency 
7 
13 
10 
23 
54 
21 
16 
Solution
Here the maximum class frequency is 23, and the class corresponding to this frequency is 1215.
So, the modal class is 12.15.
Now to find the cumulative frequency let us put the data in the table given below:
Class 
Frequency (f_{i}) 
Cumulative frequency (cf) 
3  6 
7 
7 
6  9 
13 
20 
9  12 
10 
30 
12  15 
23 
53 
15  18 
4 
57 
18  21 
21 
78 
21  24 
16 
94 
Total 
N = Î£f_{i }= 94 

Thus, the cumulative frequency of the modal class is 53.
Class interval 
0  20 
20  40 
40  60 
60  80 
Frequency 
15 
6 
18 
10 
Solution
Here the maximum class frequency is 18, 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 = 18,
Frequency (f_{0}) of class preceding the modal class = 6,
Frequency (f_{2}) of class preceding the modal class = 10.
Now, let us substitute these values in the formula:
Mode = l + (f_{1 }– f_{0})/(2f_{1 }– f_{0} – f_{2}) × h
= 40 + (18 – 6)/(36 – 6 – 10) × 20
= 40 + (12/20) × 20
= 40 + 12
= 52.
Hence, the mode is 52.
16. The following are the gas of 300 patients getting medical treatment in a hospital on a particular day:
Age (in years) 
10  20 
20  30 
30  40 
40  50 
50  60 
60  70 
Number of patients 
6 
42 
55 
70 
53 
20 
Form a 'less than type' cumulative frequency distribution.
Solution
A ‘less than type’ cumulative frequency distribution table is given below:
Age (in years) 
Cumulative frequency (cf) 

Less than 20 
60 

Less than 30 
102 

Less than 40 
157 

Less than 50 
227 

Less than 60 
280 

Less than 70 
300 

Class 
Frequency (f) 
Cumulative frequency (cf) 
100  200 
11 
11 
200 – 300 
12 
P 
300  400 
10 
33 
400  500 
Q 
46 
500 – 600 
20 
66 
600  700 
14 
80 
Solution
Here, p = 11 + 12 = 23
And 33 + q = 46
⇒ q = 46 – 33
= 13
Thus, p is 23 and q is 13.
Now,
Here the maximum class frequency is 20, and the class corresponding to this frequency is 500 – 600.
Also, Î£f = N = 80
⇒ N/2 = 40.
The cumulative frequency just greater than 40 is 46, and the corresponding class is 400 – 500.
So, the modal class is 500 – 600.
Also, Î£f = N = 80
⇒ N/2 = 40.
The cumulative frequency just greater than 40 is 46, and the corresponding class is 400 – 500.
Thus, the median class is 400 – 500.
18. The following frequency distribution gives the monthly consumption of electricity of 64 consumers of locality.
Monthly consumption (in units) 
65  85 
85  105 
105  125 
125  145 
145  165 
165  185 
Number of consumers 
4 
5 
13 
20 
14 
8 
Form a ‘more than type’ cumulative frequency distribution.
Solution
The cumulative frequency distribution table of more than type is as follows:
Monthly consumption (in units) (lower class limits) 
Cumulative frequency (cf) 
More than 65 
60 + 4 = 64 
More than 85 
55 + 5 = 60 
More than 105 
42 + 13 = 55 
More than 125 
22 + 20 = 42 
More than 145 
8 + 14 = 22 
More than 165 
8 
19. The following table gives the lifetime (in days) of 100 electric bulbs of a certain brand.
Lifetime (in days) 
Less than 50 
Less than 100 
Less than 150 
Less than 200 
Less than 250 
Less than 300 
Number of Bulbs 
7 
21 
52 
79 
91 
100 
Solution
The frequency distribution is as follows:
Lifetime (in days) 
Frequency (f) 
0  50 
7 
50  100 
14 
100  150 
31 
150  200 
27 
200  250 
12 
250  300 
9 
Marks obtained (in percent) 
11  20 
21  30 
31  40 
41  50 
51  60 
61  70 
71  80 
Number of students 
141 
221 
439 
529 
495 
322 
153 
(a) Convert the given frequency distribution into the continuous form.
(b) Find the median class and write its class mark.
(c) Find the modal class and write its cumulative frequency.
Solution
(a) The frequency distribution into the continuous form is as follows:
Marks obtained (in percent) 
Number of students (f) 
10.5 – 20.5 
141 
20.5 – 30.5 
221 
30.5 – 40.5 
439 
40.5 – 50.5 
529 
50.5 – 60.5 
495 
60.55 – 70.5 
322 
70.5 – 80.5 
153 
(b) Now, to find the median class let us’ put the data in the table given below:
Marks obtained (in percent) 
Number of students (f) 
Cumulative frequency (cf) 
10.5 – 270.5 
141 
141 
20.5 – 30.5 
221 
362 
30.5 – 40.5 
439 
801 
40.5 – 50.5 
529 
1330 
50.5 – 60.5 
495 
1825 
60.5 – 70.5 
322 
2147 
70.5 – 80.5 
153 
2300 
Now, N = 2300
⇒ N/2 = 1150
The cumulative frequency just greater than 1150 is 1330, and the corresponding class is 40.5 – 50.5.
Thus, the median class is 40.5 – 50.5.
Now, class mark = (Upper class limit + lower class limit)/2
= (40.5 + 50.5)/2 = 91/2
= 45.5
Thus, class mark of the median class is 45.5.
(c) Here the maximum class frequency is 529, and the class corresponding to \1this frequency is 40.5 – 50.5.
So, the modal class is 40.5 – 50.5 and its cum4ulative frequency is 1330.
21. If the mean of the following distribution is 27, find the value of p.
Class 
0  10 
10  20 
20  30 
30  40 
40  50 
Frequency 
8 
P 
12 
13 
10 
Solution
The given data is shown as follows:
Class 
Frequency (f) 
Class mark (x_{i}) 
f_{i}x_{i} 
0  10 
8 
5 
40 
10  20 
P 
15 
15p 
20  30 
12 
25 
300 
30  40 
13 
35 
455 
40  50 
10 
45 
450 
Total 
Î£f_{i} = 43 + p 

Î£f_{i}x_{i} = 1245 + 15p 
The mean of given data is given by

⇒ 1161 + 27p = 1245 + 15p
⇒ 27p – 15p = 1245 – 1161
⇒ 12p = 84
⇒ p = 7
Thus, the value 4 of p is 7.
22. Calculate the missing frequency form the following distribution, it being median of the distribution is 24.
Age (in years) 
0  10 
10  20 
20  30 
30  40 
40  50 
Number of persons 
5 
25 
? 
18 
7 
Solution
Let the missing frequency be x.
To find the median let us put data in the table given below:
Age (in years) 
Number of persons (f) 
Cumulative frequency (cf) 
0  10 
5 
5 
10  20 
25 
30 
20  30 
X 
30 + x 
30 – 40 
18 
48 + x 
40  50 
7 
55 + x 
The given median is 24,
∴ The median class is 20 – 30.
∴ l = 20, h = 10, N = 55 + x, f = x and cf = 30
Median = l + ((N/2  cf)/f) × h
⇒ 24 = 20 + ((55 + x)/2 – 30)/x} × 10
⇒ 24 – 20 = (55 + x – 60)/2x × 10
⇒ 4 = (x – 5)/2x × 10
⇒ 8x = 10x – 50
⇒ 2x = 50
⇒ x = 25
Thus, the missing frequency is 25.