 RS Aggarwal Class 10 Solutions Chapter 9 Mean Median Mode

RS Aggarwal Class 10 Chapter 9 Mean Median Mode Solutions Free PDF

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

Access CBSE Sample Paper for class 10 Maths here.

Access NCERT Book for class 10 Maths here.

Practise This Question

Identify the correct sequence of regeneration process in Planaria.