# RS Aggarwal Class 10 Solutions Chapter 9 - Mean Median Mode Ex 9A(9.1)

## RS Aggarwal Class 10 Chapter 9 - Mean Median Mode Ex 9A(9.1) Solutions Free PDF

The RS Aggarwal Class 10 solutions provide students an easy and effective approach to solve difficult mathematics question. These solutions are one of the best and most preferred study materials while preparing for the final exam. The Class 10 solutions of RS Aggarwal we provide are specifically designed to aid students to easy their exam preparation and score good marks in the exam.

By practicing these solutions students can excel in their Class 10 board exam and will get an idea to solve difficult and tricky questions using different methods. These solutions can help you to solve complex RS Aggarwal Maths textbook questions in a stepwise manner. By practicing these Class 10 RS Aggarwal Maths Solutions you can tackle difficult questions and solve them correctly during the exam to score desired marks.

## Download PDF of RS Aggarwal Class 10 Solutions Chapter 9– Mean Median Mode Ex 9A (9.1)

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.

### Key Features of RS Aggarwal Class 10 Solutions Chapter 9 – Mean Median Mode Ex 9A (9.1)

• It describes multiple ways to tackle different problems.
• It provides detailed and complete explanations of the questions.
• Students should practice the RS Aggarwal Class 10 Solutions to improve their marks significantly.
• It helps in boosting the confidence level of the students.

#### Practise This Question

A girl child is born when she inherits ______ chromosome from father and ________ chromosome from mother.