RS Aggarwal Solutions Class 10 Ex 9A

Question 1:

The following table gives the literacy rate (in percentage) in 40 cities. Find the mean literacy rate, choosing a suitable method/p>

Literacy rate (%) 45 – 55 55 – 65 65 – 75 75 – 85 85 – 95
Number of cities 4 11 12 9 4

Solution:

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

Literacy rate (%) Number of cities (fi) Class mark (xi) (fixi)
45 – 55 4 50 200
55 – 65 11 60 660
65 – 75 12 70 840
75 – 85 9 80 720
85 – 95 4 90 360
Total \(\sum f_{i}=40\) \(\sum (f_{i}\times x_{i})=2780\)

The mean of given data is given by

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

= \(\frac{2780}{40}\)

= 69.5

Thus, the mean literacy rate is 69.5%

Question 2:

Find the mean of the following frequency distribution using step-deviation method.

Class 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50
Frequency 7 10 15 8 10

Solution:

Let us choose a = 25, h = 10, then di = xi – 25 and ui = \(\frac{x_{i}-25}{10}\).

Using Step-deviation method, the given data is shown as follows:

Class Frequency (fi) Class mark (xi) di = xi – 25 ui = \(\frac{x_{i}-25}{10}\) fiui
0 – 10 7 5 -20 -2 -14
10 – 20 10 15 -10 -1 -10
20 – 30 15 25 0 0 0
30 – 40 8 35 10 1 8
40 – 50 10 45 20 2 20
Total \(\sum f_{i}=50\) \(\sum (f_{i}\times u_{i})=4\)

The mean of given data is given by

\(\bar{x}=a+\left ( \frac{\sum f_{i}u_{i}}{\sum f_{i}} \right )\times h\)

= 25 + \(\frac{4}{50}\) x 10

= 25 + \(\frac{4}{5}\)

= \(\frac{125+4}{5}\)

= \(\frac{129}{5}\)

= 25.8

Thus, the mean is 25.8.

Question 3:

Find the mean of the following data, using step-deviation method:

Class 5 – 15 15 – 25 25 – 35 35 – 45 45 – 55 55 – 65 65 – 75
Frequency 6 10 16 15 24 8 7

Solution:

Let us choose a = 40, h = 10, then di = xi – 40 and ui = \(\frac{x_{i}-40}{10}\).

Using Step-deviation method, the given data is shown as follows:

Class Frequency (fi) Class mark (xi) di = xi – 40 ui = \(\frac{x_{i}-40}{10}\) fiui
5 – 15 6 10 -30 -3 -18
15 – 25 10 20 -20 -2 -20
25 – 35 16 30 -10 -1 -16
35 – 45 15 40 0 0 0
45 – 55 24 50 10 1 24
55 – 65 8 60 20 2 16
65 – 75 7 70 30 3 21
Total \(\sum f_{i}=86\) \(\sum (f_{i}\times u_{i})=7\)

The mean of given data is given by/p>

\(\bar{x}=a+\left ( \frac{\sum f_{i}u_{i}}{\sum f_{i}} \right )\times h\)

= 40 + \(\frac{7}{86}\) x 10

= 40 + \(\frac{70}{86}\)

= 40 + 0.81

= 40.81

Thus, the mean is 40.81.

Question 4:

The weights of tea in 70 packets are shown in the following table

Weights (in grams) 200 – 201 201 – 202 202 – 203 203 – 204 204 – 205 205 – 206
Number of packets 13 27 18 10 1 1

Find the mean weight of packets using step-deviation method.

Solution:

Let us choose a = 202.5, h = 1, then di = xi – 202.5 and ui = \(\frac{x_{i}-202.5}{1}\).

Using Step-deviation method, the given data is shown as follows:

Weight (in grams) Number of packets (fi) Class mark (xi) di = xi – 202.5 ui = \(\frac{x_{i}-202.5}{1}\) fiui
200 – 201 13 200.5 -2 -2 -26
201 – 202 27 201.5 -1 -1 -27
202 – 203 18 202.5 0 0 0
203 – 204 10 203.5 1 1 10
204 – 205 1 204.5 2 2 2
205 – 206 1 205.5 3 3 3
Total \(\sum f_{i}=70\) \(\sum (f_{i}\times u_{i})=-38\)

The mean of given data is given by

\(\bar{x}=a+\left ( \frac{\sum f_{i}u_{i}}{\sum f_{i}} \right )\times h\)

= 202.5 + \(\frac{-38}{70}\) x 1

= 202.5 – 0.542

= 201.96

Hence, the mean is 201.96 g.

Question 5:

Find the mean of the following frequency distribution using a suitable method.

Class 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70
Frequency 25 40 42 33 10

Solution:

Let us choose a = 45, h = 10, then di = xi – 45 and ui = \(\frac{x_{i}-45}{10}\).

Using Step-deviation method, the given data is shown as follows:

Class Frequency (fi) Class mark (xi) di = xi – 45 ui = \(\frac{x_{i}-45}{10}\) fiui
20 – 30 25 25 -20 -2 -50
30 – 40 40 35 -10 -1 -40
40 – 50 42 45 0 0 0
50 – 60 33 55 10 1 33
60 – 70 10 65 20 2 20
Total \(\sum f_{i}=150\) \(\sum (f_{i}\times u_{i})=-37\)

The mean of given data is given by

\(\bar{x}=a+\left ( \frac{\sum f_{i}u_{i}}{\sum f_{i}} \right )\times h\)

= 45 – \(\frac{37}{150}\) x 10

= 40 – \(\frac{37}{15}\)

= 45 – 2.466

= 42.534

Thus, the mean is 42.534

Question 6:

In an annual examination, marks (out of 90) obtained by students of Class X in mathematics are given below:

Marks obtained 0 – 15 15 – 30 30 – 45 45 – 60 60 – 75 75 – 90
Number of students 2 4 5 20 9 10

Find the mean marks.

Solution:

Let us choose a = 52.5, h = 15, then di = xi – 52.5 and ui = \(\frac{x_{i}-52.5}{15}\).

Using Step-deviation method, the given data is shown as follows:

Marks obtained Number of students (fi) Class mark (xi) di = xi – 52.5 ui = \(\frac{x_{i}-52.5}{15}\) fiui
0 – 15 2 7.5 -45 -3 -6
15 – 30 4 22.5 -30 -2 -8
30 – 45 5 37.5 -15 -1 -5
45 – 60 20 52.5 0 0 0
60 – 75 9 67.5 15 1 9
75 – 90 10 82.5 30 2 20
Total \(\sum f_{i}=50\) \(\sum (f_{i}\times u_{i})=10\)

The mean of given data is given by

\(\bar{x}=a+\left ( \frac{\sum f_{i}u_{i}}{\sum f_{i}} \right )\times h\)

= 52.5 + \(\frac{10}{50}\) x 15

= 52.5 + 3

= 55.5

Thus, the mean is 55.5.

Question 7:

Find the arithmetic mean of the following frequency distribution using step-deviation method:

Age (in years) 18 – 24 24 – 30 30 – 36 36 – 42 42 – 48 48 – 54
Number of workers 6 8 12 8 4 2

Solution:

Class Frequency (fi) Mid values (xi) ui = \(\frac{x_{i}-A}{h}=\frac{x_{i}-33}{6}\) fiui
18 – 24 6 21 -2 -12
24 – 30 8 27 -1 -8
30 – 36 12 33 = A 0 0
36 – 42 8 39 1 8
42 – 48 4 45 2 8
48 – 54 2 51 3 6
Total \(\sum f_{i}=40\) \(\sum (f_{i}\times u_{i})=2\)

Now, A = 33, h = 6, \(\sum f_{i}=40\) and \(\sum (f_{i}\times u_{i})=2\)

Therefore, Mean, \(\bar{x}=A+\left \{ h\times \frac{\sum (f_{i}u_{i})}{\sum f_{i}} \right \}\)

= 33 + \(\left \{ 6\times \frac{2}{40} \right \}\)

= 33 + 0.3

= 33.3

Therefore, \(\bar{x}\) = 33.3 years.

Question 8:

Find the mean of the following data using step-deviation method:

Class 500 – 520 520 – 540 540 – 560 560 – 580 580 – 600 600 – 620
Frequency 14 9 5 4 3 5

Solution:

Class Frequency (fi) Mid values (xi) ui = \(\frac{x_{i}-A}{h}=\frac{x_{i}-550}{20}\) fiui
500 – 520 14 510 -2 -28
520 – 540 9 530 -1 -9
540 – 560 5 550 = A 0 0
560 – 580 4 570 1 4
580 – 600 3 590 2 6
600 – 620 5 610 3 15
Total \(\sum f_{i}=40\) \(\sum (f_{i}\times u_{i})=-12\)

Now, A = 550, h = 20, \(\sum f_{i}=40\) and \(\sum (f_{i}\times u_{i})=-12\)

Therefore, Mean, \(\bar{x}=A+\left \{ h\times \frac{\sum (f_{i}u_{i})}{\sum f_{i}} \right \}\)

= 550 + \(\left \{ 20\times \frac{(-12)}{40} \right \}\)

= 550 – 6

= 544

Therefore, \(\bar{x}\) = 544.

Question 9:

Find the mean age from the following frequency distribution:

Age (in years) 25 – 29 30 – 34 35 – 39 40 – 44 45 – 49 50 – 54 55 – 59
Number of persons 4 14 22 16 6 5 3

Solution:

Converting the series into exclusive form, we get:

Class Frequency (fi) Mid values (xi) ui = \(\frac{x_{i}-A}{h}=\frac{x_{i}-42}{5}\) fiui
24.5 – 29.5 4 27 -3 -12
29.5 – 34.5 14 32 -2 -28
34.5 – 39.5 22 37 -1 -22
39.5 – 44.5 16 42 = A 0 0
44.5 – 49.5 6 47 1 6
49.5 – 54.5 5 52 2 10
54.5 – 59.5 3 57 3 9
Total \(\sum f_{i}=70\) \(\sum (f_{i}\times u_{i})=-37\)

Now, A = 42, h = 5, \(\sum f_{i}=70\) and \(\sum (f_{i}\times u_{i})=-37\)

Therefore, Mean, \(\bar{x}=A+\left \{ h\times \frac{\sum (f_{i}u_{i})}{\sum f_{i}} \right \}\)

= 42 + \(\left \{ 5\times \frac{(-37)}{70} \right \}\)

= 42 – 2.64

= 39.36

Therefore, \(\bar{x}\) = 39.36.

Therefore, Mean age = 39.36 years.

Question 10:

The following table shows the age distribution of patients of malaria in a village during a particular month:

Age (in years) 5 – 14 15 – 24 25 – 34 35 – 44 45 – 54 55 – 64
Number of cases 6 11 21 23 14 5

Find the average age of the patients.

Solution:

Converting the series into exclusive form, we get:

Class Frequency (fi) Mid values (xi) ui = \(\frac{x_{i}-A}{h}=\frac{x_{i}-29.5}{10}\) fiui
4.5 – 14.5 6 9.5 -2 -12
14.5 – 24.5 11 19.5 -1 -11
24.5 – 34.5 21 29.5 = A 0 0
34.5 – 44.5 23 39.5 1 23
44.5 – 54.5 14 49.5 2 28
54.5 – 64.5 5 59.5 3 15
Total \(\sum f_{i}=80\) \(\sum (f_{i}\times u_{i})=43\)

Now, A = 29.5, h = 10, \(\sum f_{i}=80\) and \(\sum (f_{i}\times u_{i})=43\)

Therefore, Mean, \(\bar{x}=A+\left \{ h\times \frac{\sum (f_{i}u_{i})}{\sum f_{i}} \right \}\)

= 29.5 + \(\left \{ 10\times \frac{(43)}{80} \right \}\)

= 29.5 + 5.375

= 34.875

Therefore, \(\bar{x}\) = 34.875.

Therefore, The average age of the patients is 34.87 years.

Question 11:

Weights of 60 eggs were recorded as given below:

Weight (in grams) 75 – 79 80 – 84 85 – 89 90 – 94 95 – 99 100 – 104 105 – 109
Number of eggs 4 9 13 17 12 3 2

Calculate their mean weight to the nearest gram.

Solution:

Let us choose a = 92, h = 5, then di = xi – 92 and ui = \(\frac{x_{i}-92}{5}\).

Using Step-deviation method, the given data is shown as follows:

Weight (in grams) Number of eggs (fi) Class mark (xi) di = xi – 92 ui = \(\frac{x_{i}-92}{5}\) fiui
74.5 – 79.5 4 77 -15 -3 -12
79.5 – 84.5 9 82 -10 -2 -18
84.5 – 89.5 13 87 -5 -1 -13
89.5 – 94.5 17 92 0 0 0
94.5 – 99.5 12 97 5 1 12
99.5 – 104.5 3 102 10 2 6
104.5 – 109.5 2 107 15 3 6
Total \(\sum f_{i}=60\) \(\sum (f_{i}\times u_{i})=-19\)

The mean of given data is given by

\(\bar{x}=a+\left ( \frac{\sum f_{i}u_{i}}{\sum f_{i}} \right )\times h\)

= 92 + \(\frac{-19}{60}\) x 5

= 92 – 1.58

= 90.42

\(\approx\) 90

Thus, the mean weight to the nearest gram is 90g.

Question 12:

The following table shows the marks scored by 80 students in an examination:

Marks Less than 5 Less than 10 Less than 15 Less than 20 Less than 25 Less than 30 Less than 35 Less than 40
Number of students 3 10 25 49 65 73 78 80

Calculate the mean mark correct to 2 decimal places.

Solution:

Let us choose a = 17.5, h = 5, then di = xi – 17.5 and ui = \(\frac{x_{i}-17.5}{5}\).

Using Step-deviation method, the given data is shown as follows:

Marks Number of students (cf) Frequency (fi) Class mark (xi) di = xi – 17.5 ui = \(\frac{x_{i}-17.5}{5}\) fiui
0 – 5 3 3 2.5 -15 -3 -9
5 – 10 10 7 7.5 -10 -2 -14
10 – 15 25 15 12.5 -5 -1 -15
15 – 20 49 24 17.5 0 0 0
20 – 25 65 16 22.5 5 1 16
25 – 30 73 8 27.5 10 2 16
30 – 35 78 5 32.5 15 3 15
35 – 40 80 2 37.5 20 4 8
Total \(\sum f_{i}=80\) \(\sum (f_{i}\times u_{i})=17\)

The mean of given data is given by

\(\bar{x}=a+\left ( \frac{\sum f_{i}u_{i}}{\sum f_{i}} \right )\times h\)

= 17.5 + \(\frac{17}{80}\) x 5

= 17.5 + 1.06

= 18.56

Thus, the mean mark correct to 2 decimal places is 18.56.


Practise This Question

Write the following in the form of a word equation:
Sodium hydroxide reacts with calcium bromide to form calcium hydroxide and sodium bromide.In the formula CaBr2​ what does Ca stand for and what is its valency?