RS Aggarwal Solutions for Class 9 Maths Exercise 18B PDF
The students can use RS Aggarwal Solutions as a guide to solve the exercise wise problems based on latest CBSE guidelines. The solutions are accurate and explained in a stepwise manner for a better understanding of concepts among students. This exercise mainly contains the formula which can be used in finding means of ungrouped frequency distribution. The students can download PDF and use them as a source of reference for exam preparation. RS Aggarwal Solutions for Class 9 Maths Chapter 18 Mean, Median and Mode of Ungrouped Data Exercise 18B are provided here.
RS Aggarwal Solutions for Class 9 Chapter 18: Mean, Median and Mode of Ungrouped Data Exercise 18B Download PDF
Access RS Aggarwal Solutions for Class 9 Chapter 18: Mean, Median and Mode of Ungrouped Data Exercise 18B
1. Obtain the mean of the following distribution:
Variable (x_{i}) |
4 |
6 |
8 |
10 |
12 |
Frequency (f_{i}) |
4 |
8 |
14 |
11 |
3 |
Solution:
Variable (x_{i}) |
Frequency (f_{i}) |
f_{i} x_{i} |
4 |
4 |
16 |
6 |
8 |
48 |
8 |
14 |
112 |
10 |
11 |
110 |
12 |
3 |
36 |
Â |
Â Î£Â f_{i} = 40 |
Â Î£Â f_{i} x_{i} = 322 |
We know that
Mean = Î£Â f_{i} x_{i}/ Î£Â f_{i}
By substituting the values
Mean = 322/40 = 8.05
2. The following table shows the weights of 12 workers in a factory:
Weight (in kg) |
60 |
63 |
66 |
69 |
72 |
No. of workers |
4 |
3 |
2 |
2 |
1 |
Find the mean weight of the workers.
Solution:
Weight (in kg) (x_{i}) |
No. of workers (f_{i}) |
f_{i} x_{i} |
60 |
4 |
240 |
63 |
3 |
189 |
66 |
2 |
132 |
69 |
2 |
138 |
72 |
1 |
72 |
Â |
Î£Â f_{i} = 12 |
Î£Â f_{i} x_{i} =771 |
We know that
Mean weight of the workers = Î£Â f_{i} x_{i}/ Î£Â f_{i}
By substituting the values
Mean weight of the workers = 771/12 = 64.250kg
Therefore, the mean weight of the workers is 64.250kg.
3. The measurements (in mm) of the diameters of the heads of 50 screws are given below:
Diameter (in mm) (x_{i}) |
34 |
37 |
40 |
43 |
46 |
Number of screws (f_{i}) |
5 |
10 |
17 |
12 |
6 |
Calculate the mean diameter of the heads of the screws.
Solution:
Diameter (in mm) (x_{i}) |
Number of screws (f_{i}) |
Â f_{i} x_{i} |
34 |
5 |
170 |
37 |
10 |
370 |
40 |
17 |
680 |
43 |
12 |
516 |
46 |
6 |
276 |
Â |
Î£Â f_{i} = 50 |
Î£Â f_{i} x_{i} = 2012 |
We know that
Mean diameter of the heads of the screws = Î£Â f_{i} x_{i}/ Î£Â f_{i}
By substituting the values
Mean diameter of the heads of the screws = 2012/50 = 40.24mm
Therefore, the mean diameter of the heads of the screws is 40.24mm.
4. The following data give the number of boys of a particular age in a class of 40 students.
Age (in years) |
15 |
16 |
17 |
18 |
19 |
20 |
Frequency (f_{i}) |
3 |
8 |
9 |
11 |
6 |
3 |
Calculate the mean age of the students.
Solution:
Age (in years) (x_{i}) |
Frequency (f_{i}) |
Â f_{i} x_{i} |
15 |
3 |
45 |
16 |
8 |
128 |
17 |
9 |
153 |
18 |
11 |
198 |
19 |
6 |
114 |
20 |
3 |
60 |
Â |
Î£Â f_{i} = 40 |
Î£Â f_{i} x_{i} = 698 |
We know that
Mean age of the students = Î£Â f_{i} x_{i}/ Î£Â f_{i}
By substituting the values
Mean age of the students = 698/40 = 17.45 years
Therefore, the mean age of the students is 17.45 years.
5. Find the mean of the following frequency distribution:
Variables (x_{i}) |
10 |
30 |
50 |
70 |
89 |
Frequency (f_{i}) |
7 |
8 |
10 |
15 |
10 |
Solution:
Variables (x_{i}) |
Frequency (f_{i}) |
Â f_{i} x_{i} |
10 |
7 |
70 |
30 |
8 |
240 |
50 |
10 |
500 |
70 |
15 |
1050 |
89 |
10 |
890 |
Â |
Î£Â f_{i} = 50 |
Î£Â f_{i} x_{i} = 2750 |
We know that
Mean = Î£Â f_{i} x_{i}/ Î£Â f_{i}
By substituting the values
Mean = 2750/50 = 55
6. Find the mean of daily wages of 40 workers in a factory as per data given below:
Daily wages (in â‚¹ ) (x_{i}) |
250 |
300 |
350 |
400 |
450 |
Number of workers (f_{i}) |
8 |
11 |
6 |
10 |
5 |
Solution:
Daily wages (in â‚¹ ) (x_{i}) |
Number of workers (f_{i}) |
Â Â f_{i} x_{i} |
250 |
8 |
2000 |
300 |
11 |
3300 |
350 |
6 |
2100 |
400 |
10 |
4000 |
450 |
5 |
2250 |
Â |
Î£Â f_{i} = 40 |
Î£Â f_{i} x_{i} = 13650 |
We know that
Mean of daily wages of 40 workers in a factory = Î£Â f_{i} x_{i}/ Î£Â f_{i}
By substituting the values
Mean of daily wages of 40 workers in a factory = 13650/40 = â‚¹ 341.25
Therefore, the mean of daily wages of 40 workers in a factory is â‚¹ 341.25.
7. If the mean of the following data is 20.2, find the value of p.
Variable (x_{i}) |
10 |
15 |
20 |
25 |
30 |
Frequency (f_{i}) |
6 |
8 |
p |
10 |
6 |
Solution:
Variable (x_{i}) |
Frequency (f_{i}) |
Â Â Â f_{i} x_{i} |
10 |
6 |
60 |
15 |
8 |
120 |
20 |
p |
20p |
25 |
10 |
250 |
30 |
6 |
180 |
Â |
Î£Â f_{i} = 30 + p |
Î£Â f_{i} x_{i} = 610 + 20p |
It is given that
Mean = 20.2
We know that
Mean = Î£Â f_{i} x_{i}/ Î£Â f_{i}
So we get
(610 + 20p)/ (30 + p) = 20.2
On further calculation
610 + 20p = 606 + 20.2p
So we get
0.2p = 4
By division
p = 20
Therefore, the value of p is 20.
8. If the mean of the following data is 8, find the value of p.
x |
3 |
5 |
7 |
9 |
11 |
13 |
f |
6 |
8 |
15 |
p |
8 |
4 |
Solution:
x |
f |
Â Â Â Â f_{i} x_{i} |
3 |
6 |
18 |
5 |
8 |
40 |
7 |
15 |
105 |
9 |
p |
9p |
11 |
8 |
88 |
13 |
4 |
52 |
Â |
Î£Â f_{i} = 41 + p |
Î£Â f_{i} x_{i} = 303 + 9p |
It is given that
Mean = 8
We know that
Mean = Î£Â f_{i} x_{i}/ Î£Â f_{i}
So we get
(303 + 9p)/ (41 + p) = 8
On further calculation
303 + 9p = 8 (41 + p)
So we get
303 + 9p = 328 + 8p
It can be written as
9p â€“ 8p = 328 â€“ 303
We get
p = 25
Therefore, the value of p is 25.
9. Find the missing frequency p for the following frequency distribution whose mean is 28.25.
x |
15 |
20 |
25 |
30 |
35 |
40 |
f |
8 |
7 |
p |
14 |
15 |
6 |
Solution:
x |
f |
Â Â Â Â f_{i} x_{i}Â |
15 |
8 |
120 |
20 |
7 |
140 |
25 |
p |
25p |
30 |
14 |
420 |
35 |
15 |
525 |
40 |
6 |
240 |
Â |
Î£Â f_{i} = 50 + p |
Î£Â f_{i} x_{i} = 1445 + 25p |
It is given that
Mean = 28.25
We know that
Mean = Î£Â f_{i} x_{i}/ Î£Â f_{i}
So we get
(1445 + 25p)/ (50 + p) = 28.25
On further calculation
1445 + 25p = 1412.50 + 28.25p
So we get
-28.25p + 25p = -1445 + 1412.50
It can be written as
-3.25p = -32.5
By division
p = 10
Therefore, the value of p is 10.
10. Find the value of p for the following frequency distribution whose mean is 16.6.
x |
8 |
12 |
15 |
p |
20 |
25 |
30 |
f |
12 |
16 |
20 |
24 |
16 |
8 |
4 |
Solution:
x |
f |
Â Â Â Â Â f_{i} x_{i}Â |
8 |
12 |
96 |
12 |
16 |
192 |
15 |
20 |
300 |
p |
24 |
24p |
20 |
16 |
320 |
25 |
8 |
200 |
30 |
4 |
120 |
Â |
Î£Â f_{i} = 100 |
Î£Â f_{i} x_{i} = 1228 + 24p |
It is given that
Mean = 16.6
We know that
Mean = Î£Â f_{i} x_{i}/ Î£Â f_{i}
So we get
(1228 + 24p)/ 100 = 16.6
On further calculation
1228 + 24p = 1660
So we get
24p = 1660 â€“ 1228
By subtraction
24p = 432
We get
p = 18
Therefore, the value of p is 18.
11. Find the missing frequencies in the following frequency distribution whose mean is 34.
x |
10 |
20 |
30 |
40 |
50 |
6 |
Total |
f |
4 |
f_{1} |
8 |
f_{2} |
3 |
4 |
35 |
Solution:
x |
f |
Â f_{i} x_{i}Â |
10 |
4 |
40 |
20 |
f_{1} |
20f_{1} |
30 |
8 |
240 |
40 |
f_{2} |
40f_{2} |
50 |
3 |
150 |
6 |
4 |
240 |
Total |
35 = Î£Â f_{i} =19 + f_{1} + f_{2} |
Î£Â f_{i} x_{i} = 670 + 20f_{1} + 40f_{2} |
We know that
Î£Â f_{i} =19 + f_{1} + f_{2}
It can be written as
35 = 19 + f_{1} + f_{2}
So we get
f_{1} + f_{2} = 16 â€¦â€¦ (1)
It is given that
Mean = 34
We know that
Mean = Î£Â f_{i} x_{i}/ Î£Â f_{i}
So we get
(670 + 20f_{1} + 40f_{2})/ 35 = 34
On further calculation
(335 + 10f_{1} + 20f_{2})/ 35 = 17
So we get
335 + 10f_{1} + 20f_{2} = 595
It can be written as
10f_{1} + 20f_{2} = 260
Dividing the equation by 10
f_{1} + 2f_{2} = 26 â€¦â€¦.. (2)
By subtracting equation (1) from (2)
f_{2} = 10
Substituting in equation (1)
f_{1} + 10 = 16
So we get
f_{1} = 6
Therefore, the missing frequencies are 6 and 10.
12. Find the missing frequencies in the following frequency distribution whose mean is 50.
x |
10 |
30 |
50 |
70 |
90 |
Total |
f |
17 |
f_{1} |
32 |
f_{2} |
19 |
120 |
Solution:
x |
f |
Â Â f_{i} x_{i}Â |
10 |
17 |
170 |
30 |
f_{1} |
30f_{1} |
50 |
32 |
1600 |
70 |
f_{2} |
70f_{2} |
90 |
19 |
1710 |
Total |
120 = Î£Â f_{i} =68 + f_{1} + f_{2} |
Î£Â f_{i} x_{i} = 3480 + 30f_{1} + 70f_{2} |
We know that
Î£Â f_{i} =68 + f_{1} + f_{2}
It can be written as
120 = 68 + f_{1} + f_{2}
So we get
f_{1} + f_{2} = 52
f_{2} = 52 – f_{1} â€¦â€¦. (1)
It is given that
Mean = 50
We know that
Mean = Î£Â f_{i} x_{i}/ Î£Â f_{i}
So we get
(3480 + 30f_{1} + 70f_{2})/ 120 = 50
Substituting equation (1)
3480 + 30f_{1} + 70(52 – f_{1}) / 120 = 50
On further calculation
(3480 + 30f_{1} + 3640 â€“ 70f_{1})/ 120 = 50
So we get
(7120 – 40f_{1})/ 120 = 50
It can be written as
6000 = 7120 – 40f_{1}
By subtraction
40f_{1} = 1120
By division
f_{1} = 28
By substituting it in equation (1)
f_{2} = 52 â€“ 28 = 24
Therefore, the missing frequencies are 28 and 24.
13. Find the value of p, when the mean of the following distribution is 20.
x |
15 |
17 |
19 |
20 + p |
23 |
f |
2 |
3 |
4 |
5p |
6 |
Solution:
x |
f |
Â Â Â f_{i} x_{i}Â |
15 |
2 |
30 |
17 |
3 |
51 |
19 |
4 |
76 |
20 + p |
5p |
100p + 5p^{2} |
23 |
6 |
138 |
Â |
Î£Â f_{i} = 15 + 5p |
Î£Â f_{i} x_{i} = 5p^{2} + 100p + 295 |
It is given that
Mean = 20
We know that
Mean = Î£Â f_{i} x_{i}/ Î£Â f_{i}
So we get
(5p^{2} + 100p + 295)/ (15 + 5p) = 20
On further calculation
5p^{2} + 100p + 295 = 300 + 100p
So we get
5p^{2} = 5
By division
p^{2} = 1
By taking square root
p = 1
Therefore, the value of p is 1.
14. The mean of the following distribution is 50.
x |
10 |
30 |
50 |
70 |
90 |
f |
17 |
5a + 3 |
32 |
7a – 11 |
19 |
Find the value of a and hence the frequencies of 30 and 70.
Solution:
x |
f |
Â f_{i} x_{i}Â |
10 |
17 |
170 |
30 |
5a + 3 |
150a + 90 |
50 |
32 |
1600 |
70 |
7a – 11 |
490a – 770 |
90 |
19 |
171 |
Â |
Î£Â f_{i} = 60 + 12a |
Î£Â f_{i} x_{i} = 2800 + 640a |
It is given that
Mean = 50
We know that
Mean = Î£Â f_{i} x_{i}/ Î£Â f_{i}
So we get
(2800 + 640a)/ (60 + 12a) = 50
On further calculation
2800 + 640a = 3000 + 600a
So we get
40a = 200
By division
a = 5
We know that
Frequency of 30 = 5a + 3
By substituting the values
Frequency of 30 = 5 (5) + 3
So we get
Frequency of 30 = 28
We know that
Frequency of 70 = 7a – 11
By substituting the values
Frequency of 70 = 7 (5) – 11
So we get
Frequency of 70 = 24
Therefore, the value of a is 5 and the frequencies of 30 and 70 is 28 and 24.
s 9 Maths Chapter 18: Mean, Median and Mode of Ungrouped Data
Exercise 18A Solutions 31 Questions
Exercise 18C Solutions 10 Questions
Exercise 18D Solutions 8 Questions
RS Aggarwal Solutions Class 9 Maths Chapter 18 – Mean, Median and Mode of Ungrouped Data Exercise 18B
RS Aggarwal Solutions Class 9 Maths Chapter 18 Mean, Median and Mode of Ungrouped Data Exercise 18B explains the method of finding means of an ungrouped frequency distribution otherwise known as the direct method.
Key features of RS Aggarwal Solutions for Class 9 Maths Chapter 18: Mean, Median and Mode of Ungrouped Data Exercise 18B
- The students can use PDF of solutions while solving the RS Aggarwal textbook to find the other methods of answering problems.
- Time management abilities are improved among students, which is one of the important aspects from the exam point of view.
- The answers are explained in simple language which match the understanding capacity of students.
- The subject experts prepare answers based on CBSE guidelines to help students obtain better academic scores.