RS Aggarwal Class 8 Maths Chapter 23 Ex 23.1 Solutions 2024 | Download Free PDF

RS Aggarwal Class 8 Maths Chapter 23 Ex 23.1 Solutions: In this exercise, the student will know about line graphs & linear graphs. The solutions of Chapter 23 Exercise 23.1- Line Graphs and Linear Graphs enable the students to study the way to draw graphs, and line graphs, read those graphs, read single and double line graphs, and understand the whole concept of linear graphs.

RS Aggarwal Class 8 Maths Chapter 23 Ex 23.1 Solutions are designed according to the CBSE syllabus & guidelines. These solutions enable the students to know the topics relating to line graphs & linear graphs in detail. Each answer is illustrated in detail & in an easy language to assist students to understand these topics quickly.

Access RS Aggarwal Class 8 Maths Chapter 23 Solutions PDF

Download RS Aggarwal Class 8 Maths Chapter 23 Ex 23.1 Solutions

RS Aggarwal Class 8 Maths Chapter 23 Ex 23.1 Solutions

Access The RS Aggarwal Class 8 Maths Chapter 23 Ex 23.1 Solutions

Question 1.
Solution:
Total expenditure = Rs. 4000 + 5400 + 2800 + 1800 + 400 = Rs. 14400

Construction of pie chart :
1. Draw a circle of any convenient radius.
2. Draw a horizontal radius.
3. Staring from this radius, draw sectors of central angle 100°, 135°, 70°, 45° and 10° respectively.
4. Shade these sectors with different colors or designs as shown in the figure.
This is the required pie chart.

Question 2.
Solution:
Total number of creatures 900

(i) Draw a circle with a suitable radius.
(ii) Draw a horizontal radius.
(iii) Starting from this radius, draw sectors whose central angles are 60°, 160°, 70°, 50° and 20° respectively.
(iv) Now shade each sector with different colours or designs as shown in the figure.

Question 3.
Solution:
Total number of students = 350 + 245 + 210 + 175 + 280 = 1260

(i) Draw a circle with a suitable radius.
(ii) Draw a horizontal radius.
(iii) Starting from this radius draw sectors whose central angles are 100°, 70°, 60°, 50° and 80° respectively.
(iv) Now shade each sector with different colours or designs as shown in the figure.

Question 4.
Solution:

(i) Draw a circle with a suitable radius.
(ii) Draw a horizontal radius.
(iii) Starting from this radius, draw sectors whose actual angles are 105°, 60°, 30°, 120° and 45° respectively.
(iv) Now shade each sector with different colours or design as shown in the figure.

Question 5.
Solution:
Here total number of workers = 1080

Now (i) Draw a circle with a suitable radius
(ii) Draw a horizontal radius
(iii) Starting from this radius, draw sectors whose central angle are 150°, 90°, 85°, 35° respectively.
(iv) Now shade the sectors with different colours or designs as shown in the figure.

Question 6.
Solution:
Total marks obtained by Sudhir
= 105 + 75 + 150 + 120 + 90 = 540

(i) Draw a circle with a suitable radius
(ii) Draw a horizontal radius
(iii) Starting from this radius, draw sectors whose central angles are 70°, 50°, 100°, 80° and 60° respectively
(iv) Now shade these sectors with different colours or designs as shown in the figure.

Question 7.
Solution:
Total number of fruits = 26 + 30 + 21 + 5 + 8 = 90

(i) Draw a circle with a suitable radius.
(ii) Draw a horizontal radius.
(iii) Starting from this radius, draw sectors of central angles 104°, 120°, 84°, 20° and 32° respectively.
(iv) Shade these sectors with different colours or designs as shown in the figure.

Question 8.
Solution:
Total number of million of tonnes of food grains = 57 + 76 + 38 + 19 = 190 million of tonnes

(i) Draw a circle with suitable radius.
(ii) Draw a horizontal radius.
(iii) Starting with this radius, draw sectors of central angles 108°, 144°, 72° and 366 respectively.
(iv) Shade these sectors with different colours or designs as shown in the figure.

Question 9.
Solution:
Total percentage = 25 + 45 + 20 + 10 = 100%

(i) Draw a circle with a suitable radius.
(ii) Draw a horizontal radius.
(iii) Starting from this radius, draw sectors of central angles 90°, 162°, 72° and 36° respectively.
(iv) Shade these sectors with different colours or designs as shown in the figure.

Question 10.
Solution:
Total percentage = 20 + 40 + 25 + 15 = 100%

(i) Draw a circle with a suitable radius.
(ii) Draw a horizontal radius.
(iii) Starting from this radius, draw sectors of central angles 72°, 144°, 90° and 54° respectively.
(iv) Now shade these sectors with different colours or designs as shown in the figure.

 

Important Definition for RS Aggarwal Class 8 Maths Chapter 23 Ex 23.1 Solutions

• Line Graph

It is a graph that uses lines & points to represent change over time. It is a line that shows the relation between the points or a chart that shows a line joining several points. The graph represents quantitative data between two changing variables with a curve or line that joins a series of consecutive data points. Linear graphs compare these two variables on a horizontal axis & a vertical axis.

• Types of Line Graphs

(i) Simple Line Graph: In this type of line graph, only one line is plotted on the graph.

(ii) Multiple Line Graph: In this type of line graph, more than one line is plotted on the same set of axes. A multiple-line graph can compare similar items over the same period.

(iii) Compound Line Graph: If information can be bifurcated into 2 or more types of data, it is known as a compound line graph. Lines are drawn to illustrate the basic portion of a total. The top line shows the total and the line below shows part of the total. The distance between every two lines shows the size of each part.

• Steps to Make a Line Graph

(i) Utilize the data from the data table to choose a suitable scale.

(ii) Draw & label the scale on the horizontal (x-axis) axes & vertical (y-axis).

(iii) List each item & place the points on the graph.

(iv) Join the points with line segments.

• Linear Graph Equation

The linear graph makes a straight line & represented by an equation of y = mx + c, where m is the gradient of the graph and c is the y-intercept of the graph.

The gradient between any two points (x1, y1) and (x2, y2) are any two points on the linear or straight line.

The value of gradient m is the ratio of the difference of y-coordinates to the difference of x-coordinates i.e. m = y2 – y1/x2 – x1 Or y – y1 = m (x – x1)

The linear equation can also be written as ax + by + c = 0, where a, b, and c are constants

Know more at the official website.

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