RS Aggarwal Solutions Class 10 Mean Median Mode

RS Aggarwal Solutions Class 10 Chapter 9

All these RS Aggarwal class 10 solutions Chapter 9 Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive are solved by Byju's top ranked professors as per CBSE guidelines.

Mean, Median, Mode of Grouped Data Exercise 9.1
Mean, Median, Mode of Grouped Data Exercise 9.2
Mean, Median, Mode of Grouped Data Exercise 9.3
Mean, Median, Mode of Grouped Data Exercise 9.4

Question 1:

If the mean of 5 observations x, x + 2, x + 4, x + 6 and x + 8 is 11, find the value of x.

Solution:

Mean of given observations = \(\frac{Sum\, of\, given\, observations}{Total\, number\, of\, observations}\)

Therefore, 11 = \(\frac{x+(x+2)+(x+4)+(x+6)+(x+8)}{5}\)

\(\Rightarrow\) 55 = 5x + 20

\(\Rightarrow\) 5x = 55 ??? 20

\(\Rightarrow\) 5x = 35

\(\Rightarrow\) x = \(\frac{35}{5}\)

\(\Rightarrow\) x = 7

Hence, the value of x is 7.

Question 2:

If the mean of 25 observations is 27 and each observation is decreased by 7, what will be the new mean?

Solution:

Mean of given observations = \(\frac{Sum\, of\, given\, observations}{Total\, number\, of\, observations}\)

Mean of 25 observations = 27

Therefore, Sum of 25 observations = 27 x 25 = 675

If 7 is subtracted from every number, then the sum = 675 ??? (25 x 7) = 675 ??? 175 = 500

Then, new mean = \(\frac{500}{25}\) = 20

Thus, the new mean will be 20.

Question 3:

Compute the mean of the following data:

Class 1 ??? 3 3 ??? 5 5 ??? 7 7 ??? 9
Frequency 12 22 27 19

Solution:

The given data is shown as follows:

Class Frequency (f1) Class mark (x1) f1x1
1 ??? 3 12 2 24
3 ??? 5 22 4 88
5 ??? 7 27 6 162
7 ??? 9 19 8 152
Total \(\sum f_{1}=80\) \(\sum f_{1}x_{1}=426\)

The mean of the given data is given by

\(\bar{x}=\frac{\sum f_{1}x_{1}}{\sum f_{1}}\)

= \(\frac{426}{80}\) = 5.325

Thus, the mean of the following data is 5.325

Question 4:

Find the mean, using direct method:

Class 0 ??? 10 10 ??? 20 20 ??? 30 30 ??? 40 40 ??? 50 50 ??? 60
Frequency 7 5 6 12 8 2

Solution:

Class Frequency (fi) Mid Values (xi) (fixi)
0 ??? 10 7 5 35
10 ??? 20 5 15 75
20 ??? 30 6 25 150
30 ??? 40 12 35 420
40 ??? 50 8 45 360
50 ??? 60 2 55 110
Total \(\sum f_{i}=40\) \(\sum (f_{i}\times x_{i})=1150\)

Therefore, Mean, \(\bar{x}\) = \(\frac{\sum (f_{i}\times x_{i})}{\sum f_{i}}\)

= \(\frac{1150}{40}\) = 28.75

Therefore, \(\bar{x}\) = 28.75

 

Question 5:

Calculate the mean of the following data, using direct method:

Class 25 ??? 35 35 ??? 45 45 ??? 55 55 ??? 65 65 ??? 75
Frequency 6 10 8 12 4

Solution:

Class Frequency (fi) Mid Values (xi) (fixi)
25 ??? 35 6 30 180
35 ??? 45 10 40 400
45 ??? 55 8 50 400
55 ??? 65 12 60 720
65 ??? 75 4 70 280
Total \(\sum f_{i}=40\) \(\sum (f_{i}\times x_{i})=1980\)

Therefore, Mean, \(\bar{x}\) = \(\frac{\sum (f_{i}\times x_{i})}{\sum f_{i}}\)

= \(\frac{1980}{40}\) = 49.5

Therefore, \(\bar{x}\) = 49.5

Question 6:

Compute the mean of the following data, using direct method:

Class 0 ??? 100 100 ??? 200 200 ??? 300 300 ??? 400 400 ??? 500
Frequency 6 9 15 12 8

Solution:

Class Frequency (fi) Mid Values (xi) (fixi)
0 ??? 100 6 50 300
100 ??? 200 9 150 1350
200 ??? 300 15 250 3750
300 ??? 400 12 350 4200
400 ??? 500 8 450 3600
Total \(\sum f_{i}=50\) \(\sum (f_{i}\times x_{i})=13200\)

Therefore, Mean, \(\bar{x}\) = \(\frac{\sum (f_{i}\times x_{i})}{\sum f_{i}}\)

= \(\frac{13200}{50}\) = 264

Therefore, \(\bar{x}\) = 264

Question 7:Using an appropriate method, find the mean of the following frequency distribution:

Class Interval 84 ??? 90 90 ??? 96 96 ??? 102 102 ??? 108 108 ??? 114 114 ??? 120
Frequency 8 10 16 23 12 11

Which method did you use, and why?

Solution:

The given data is shown as follows:

Class interval Frequency (fi) Class mark (xi) (fixi)
84 ??? 90 8 87 696
90 ??? 96 10 93 930
96 ??? 102 16 99 1584
102 ??? 108 23 105 2415
108 ??? 114 12 111 1332
114 ??? 120 11 117 1287
Total \(\sum f_{i}=80\) \(\sum (f_{i}\times x_{i})=8244\)

The mean of given data is given by

\(\bar{x}\) = \(\frac{\sum (f_{i}\times x_{i})}{\sum f_{i}}\)

= \(\frac{8244}{80}\) = 103.05

Thus, the mean of the following data is 103.05

Question 8:

If the mean of the following frequency distribution is 24, find the value of p.

Class 0 ??? 10 10 ??? 20 20 ??? 30 30 ??? 40 40 ??? 50
Frequency 3 4 p 3 2

Solution:

The given data is shown as follows:

Class Frequency (fi) Class mark (xi) (fixi)
0 ??? 10 3 5 15
10 ??? 20 4 15 60
20 ??? 30 p 25 25p
30 ??? 40 3 35 105
40 ??? 50 2 45 90
Total \(\sum f_{i}=12+p\) \(\sum (f_{i}\times x_{i})=270+25p\)

The mean of given data is given by

\(\bar{x}\) = \(\frac{\sum (f_{i}\times x_{i})}{\sum f_{i}}\)

\(\Rightarrow\) 24 = \(\frac{270+25p}{12+p}\)

\(\Rightarrow\) 24(12 + p) = 270 + 25p

\(\Rightarrow\) 288 + 24p = 270 + 25p

\(\Rightarrow\) 25p ??? 24p = 288 ??? 270

\(\Rightarrow\) p = 18

Hence, the value of p is 18.

Question 9:

The following distribution shows the daily pocket allowance of children of a locality. If the mean pocket allowance is Rs. 18, find the missing frequency f.

Daily pocket allowance (in Rs.) 11 ??? 13 13 ??? 15 15 ??? 17 17 ??? 19 19 ??? 21 21 ??? 23 23 ??? 25
Frequency 7 6 9 13 f 5 4

Solution:

The given data is shown as follows:

Daily pocket allowance (in Rs.) Number of children (fi) Class mark (xi) (fixi)
11 ??? 13 7 12 84
13 ??? 15 6 14 84
15 ??? 17 9 16 144
17 ??? 19 13 18 234
19 ??? 21 f 20 20f
21 ??? 23 5 22 110
23 ??? 25 4 24 96
Total \(\sum f_{i}=44+f\) \(\sum (f_{i}\times x_{i})=752+20f\)

The mean of given data is given by

\(\bar{x}\) = \(\frac{\sum (f_{i}\times x_{i})}{\sum f_{i}}\)

\(\Rightarrow\) 18 = \(\frac{752+20f}{44+f}\)

\(\Rightarrow\) 18(44 + f) = 752 + 20f

\(\Rightarrow\) 792 + 18f = 752 + 20f

\(\Rightarrow\) 20f ??? 18f = 792 ??? 752

\(\Rightarrow\) 2f = 40

\(\Rightarrow\) f = 20

Hence, the value of f is 20.

Question 10:

If the mean of the following frequency distribution is 54, find the value of p.

Class 0 ??? 20 20 ??? 40 40 ??? 60 60 ??? 80 80 ??? 100
Frequency 7 p 10 9 13

Solution:

The given data is shown as follows:

Class Frequency (fi) Class mark (xi) (fixi)
0 ??? 20 7 10 70
20 ??? 40 p 30 30p
40 ??? 60 10 50 500
60 ??? 80 9 70 630
80 ??? 100 13 90 1170
Total \(\sum f_{i}=39+p\) \(\sum (f_{i}\times x_{i})=2370+30p\)

The mean of given data is given by

\(\bar{x}\) = \(\frac{\sum (f_{i}\times x_{i})}{\sum f_{i}}\)

\(\Rightarrow\) 54 = \(\frac{2370+30p}{39+p}\)

\(\Rightarrow\) 54(39 + p) = 2370 + 30p

\(\Rightarrow\) 2106 + 54p = 2370 + 30p

\(\Rightarrow\) 54p ??? 30p = 2370 ??? 2106

\(\Rightarrow\) 24p = 264

\(\Rightarrow\) p = 11

Hence, the value of p is 11.

Question 11:

The mean of the following data is 42. Find the missing frequencies of x and y if the sum of frequencies is 100.

Class interval 0 ??? 10 10 ??? 20 20 ??? 30 30 ??? 40 40 ??? 50 50 ??? 60 60 ??? 70 70 ??? 80
Frequency 7 10 x 13 y 10 14 9

Solution:

The given data is shown as follows:

Class interval Frequency (fi) Class mark (xi) (fixi)
0 ??? 10 7 5 35
10 ??? 20 10 15 150
20 ??? 30 x 25 25x
30 ??? 40 13 35 455
40 ??? 50 y 45 45y
50 ??? 60 10 55 550
60 ??? 70 14 65 910
70 ??? 80 9 75 675
Total \(\sum f_{i}=63+x+y\) \(\sum (f_{i}\times x_{i})=2775+25x+45y\)

Sum of the frequencies = 100

\(\Rightarrow\) \(\sum f_{i}\) = 100

\(\Rightarrow\) 63 + x + y = 100

\(\Rightarrow\) x + y = 100 ??? 63

\(\Rightarrow\) x + y = 37

\(\Rightarrow\) y = 37 ??? x – – – – – – (1)

Now, the mean of the given data is given by

\(\bar{x}\) = \(\frac{\sum (f_{i}\times x_{i})}{\sum f_{i}}\)

\(\Rightarrow\) 42 = \(\frac{2775+25x+45y}{100}\)

\(\Rightarrow\) 4200 = 2775 + 25 x + 45y

\(\Rightarrow\) 4200 ??? 2775 = 25x + 45y

\(\Rightarrow\) 1425 = 25x + 45(37 ??? x) [from (1)]

\(\Rightarrow\) 1425 = 25x + 1665 ??? 45x

\(\Rightarrow\) 20x = 1665 ??? 1425

\(\Rightarrow\) 20x = 240

\(\Rightarrow\) x = 12

If x = 12, then y = 37 ??? 12 = 25

Thus, the value of x is 12 and y is 25.

Question 12:

The daily expenditure of 100 families are given below. Calculate f1 and f2 if the mean daily expenditure is Rs. 188.

Expenditure (in Rs.) 140 ??? 160 160 ??? 180 180 ??? 200 200 ??? 220 220 ??? 240
Number of families 5 25 f1 f2 5

Solution:

The given data is shown as follows:

Expenditure (in Rs.) Number of families (fi) Class mark (xi) (fixi)
140 ??? 160 5 150 750
160 ??? 180 25 170 4250
180 ??? 200 f1 190 190f1
200 ??? 220 f2 210 210f2
220 ??? 240 5 230 1150
Total \(\sum f_{i}= 35+f_{1}+f_{2} \) \(\sum (f_{i}\times x_{i})=6150+190f_{1}+210f_{2}\)

Sum of the frequencies = 100

\(\Rightarrow\) \(\sum f_{i}\) = 100

\(\Rightarrow\) 35 + f1 + f2 = 100

\(\Rightarrow\) f1 + f2 = 100 ??? 35

\(\Rightarrow\) f1 + f2 = 65

\(\Rightarrow\) f2 = 65 ??? f1 – – – – – – (1)

Now, the mean of the given data is given by

\(\bar{x}\) = \(\frac{\sum (f_{i}\times x_{i})}{\sum f_{i}}\)

\(\Rightarrow\) 188 = \(\frac{6150+190f_{1}+210f_{2}}{100}\)

\(\Rightarrow\) 18800 = 6150 + 190f1 + 210f2

\(\Rightarrow\) 18800 ??? 6150 = 190f1 + 210f2

\(\Rightarrow\) 12650 = 190f1 + 210(65 ??? f1) [from (1)]

\(\Rightarrow\) 12650 = 190f1 ??? 210f1 + 13650

\(\Rightarrow\) 20f1 = 13650 ??? 12650

\(\Rightarrow\) 20f1 = 1000

\(\Rightarrow\) f1 = 50

If f1 = 50, then f2 = 65 ??? 50 = 15

Thus, the value of f1 is 50 and f2 is 15.

Question 13:

The mean of the following frequency distribution is 57.6 and the total number of observations is 50.

Class 0 ??? 20 20 ??? 40 40 ??? 60 60 ??? 80 80 ??? 100 100 ??? 120
Frequency 7 f1 12 f2 8 5

Find f1 and f2.

Solution:

Class Frequency (fi) Mid values (xi) (fixi)
0 ??? 20 7 10 70
20 ??? 40 f1 30 30f1
40 ??? 60 12 50 600
60 ??? 80 18 ??? f1 70 1260 ??? 70f1
80 ??? 100 8 90 720
100 ??? 120 5 110 550
Total \(\sum f_{i}=50\) \(\sum (f_{i}\times x_{i})=3200-40f_{1}\)

We have:

7 + f1 + 12 + f2 + 8 + 5 = 50

\(\Rightarrow\) f1 + f2 = 18

\(\Rightarrow\) f2 = 18 ??? f1

Therefore, Mean, \(\bar{x}\) = \(\frac{\sum (f_{i}\times x_{i})}{\sum f_{i}}\)

\(\Rightarrow\) 57.6 = \(\frac{3200-40f_{1}}{50}\) [Because Mean = 57.6]

\(\Rightarrow\) 40f1 = 320

Therefore, f1 = 8

And f2 = 18 ??? 8

\(\Rightarrow\) f2 = 10

Therefore, The missing frequencies are f1 = 8 and f2 = 10.

Question 14:

During a medical check-up, the numbers of heartbeats per minutes of 30 patients were recorded and summarised as follows:

Number of heart-beats per minute 65 ??? 68 68 ??? 71 71 ??? 74 74 ??? 77 77 ??? 80 80 ??? 83 83 ??? 86
Number of patients 2 4 3 8 7 4 2

Find the mean heartbeats per minute for these patients, choosing a suitable method.

Solution:

Using Direct method, the given data is shown as follows:

Number of heartbeats per minute Number of patients (fi) Class mark (xi) (fixi)
65 ??? 68 2 66.5 133
68 ??? 71 4 69.5 278
71 ??? 74 3 72.5 217.5
74 ??? 77 8 75.5 604
77 ??? 80 7 78.5 549.5
80 ??? 83 4 81.5 326
83 ??? 86 2 84.5 169
Total \(\sum f_{i}=30\) \(\sum (f_{i}\times x_{i})=2277\)

The mean of given data is given by

\(\bar{x}\) = \(\frac{\sum (f_{i}\times x_{i})}{\sum f_{i}}\)

= \(\frac{2277}{30}\) = 75.9

Thus, the mean heartbeats per minute for these patients are 75.9.

Question 15:

Find the mean marks per student, using assumed-mean method:

Marks 0 ??? 10 10 ??? 20 20 ??? 30 30 ??? 40 40 ??? 50 50 ??? 60
Number of students 12 18 27 20 17 6

 

Solution:

Class Frequency (fi) Mid values (xi) Deviation (di)

di = (xi ??? 25)

(fidi)
0 ??? 10 12 5 -20 -240
10 ??? 20 18 15 -10 -180
20 ??? 30 27 25 = A 0 0
30 ??? 40 20 35 10 200
40 ??? 50 17 45 20 340
50 ??? 60 6 55 30 180
Total \(\sum f_{i}=100\) \(\sum (f_{i}\times d_{i})=300\)

Let A = 25 be the assumed mean. Then we have:

Mean, \(\bar{x}=A+\frac{\sum (f_{i}\times d_{i})}{\sum f_{i}}\)

= 25 + \(\frac{300}{100}\)

= 28

Therefore, \(\bar{x}\) = 28

Question 16:

Find the mean of the following frequency distribution, using the assumed-mean method:

Class 100 ??? 120 120 ??? 140 140 ??? 160 160 ??? 180 180 ??? 200
Frequency 10 20 30 15 5

Solution:

Class Frequency (fi) Mid values (xi) Deviation (di)

di = (xi ??? 150)

(fidi)
100 ??? 120 10 110 -40 -400
120 ??? 140 12 130 -20 -400
140 ??? 160 30 150 = A 0 0
160 ??? 180 15 170 20 300
180 ??? 200 5 190 40 200
Total \(\sum f_{i}=80\) \(\sum (f_{i}\times d_{i})=-300\)

Let A = 150 be the assumed mean. Then we have:

Mean, \(\bar{x}=A+\frac{\sum (f_{i}\times d_{i})}{\sum f_{i}}\)

= 150 ??? \(\frac{300}{80}\)

= 150 ??? 3.75

Therefore, \(\bar{x}\) = 146.25

Question 17:

Find the mean of the following data, using the assumed-mean method:

Class 0 ??? 20 20 ??? 40 40 ??? 60 60 ??? 80 80 ??? 100 100 ??? 120
Frequency 20 35 52 44 38 31

Solution:

Class Frequency (fi) Mid values (xi) Deviation (di)

di = (xi ??? 50)

(fidi)
0 ??? 20 20 10 -40 -800
20 ??? 40 35 30 -20 -700
40 ??? 60 52 50 = A 0 0
60 ??? 80 44 70 20 880
80 ??? 100 38 90 40 1520
100 ??? 120 31 110 60 1860
Total \(\sum f_{i}=220\) \(\sum (f_{i}\times d_{i})=2760\)

Let A = 50 be the assumed mean. Then we have:

Mean, \(\bar{x}=A+\frac{\sum (f_{i}\times d_{i})}{\sum f_{i}}\)

= 50 + \(\frac{2760}{220}\)

= 50 + 12.55

Therefore, \(\bar{x}\) = 62.55

Related Links
NCERT Books NCERT Solutions RS Aggarwal
Lakhmir Singh RD Sharma Solutions NCERT Solutions Class 6 to 12
More RS Aggarwal Solutions
RS Aggarwal Solutions Class 10 Solutions Real NumbeRSRS Aggarwal Solutions Class 10 Solutions Trigonometric Ratios Of Complementary Angles
RS Aggarwal Solutions Class 6 SolutionsRS Aggarwal Solutions Class 6 Solutions Chapter 10 Ratio Proportion And Unitary Method Exercise 10 1
RS Aggarwal Solutions Class 6 Solutions Chapter 1 Number System Exercise 1 2RS Aggarwal Solutions Class 6 Solutions Chapter 21 Concept Of Perimeter And Area Exercise 21 3
RS Aggarwal Solutions Class 6 Solutions Chapter 4 IntegeRS Exercise 4 2RS Aggarwal Solutions Class 6 Solutions Chapter 5 Fractions Exercise 5 3