RS Aggarwal Solutions Class 10 Mean Median Mode

The mean of some grouped data can be defined as the sum of all the numbers in a group divided by how many numbers are present in a group. Median can simply be defined as the middle number which can be found out through a set of numbers, from the order value. The mode can be defined as a number which occurs the most often in a group of numbers. It is important for students to know how to calculate the mean, median and mode as this will help students make calculations from a grouped set of data.

Learn more about RS Aggarwal Class 10 Solutions Chapter 9 Mean, Median, Mode of Grouped Data can be found below:

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 is 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 the 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 the 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/p>

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


Practise This Question

Light travels from one point to another in a straight line which is known as ___ propogation of light.