GSEB Class 8 Maths Solutions Chapter 15 Introduction to Graphs Exercise 15.3

 

Question 1. Draw the graphs for the following tables of values, with suitable scales on the axes?
(a) Cost of apples

(b) Distance travelled by a car(i) How much distance did the car cover during the period 7.30 a.m. to 8 a.m.?
(ii) What was the time when the car had covered a distance of 100 km since its start?
(c) Interest on deposits for a year.(i) Does the graph pass through the origin?
(ii) Use the graph to find the interest on Rs 2500 for a year.
(iii) To get an interest of Rs 280 per year, how much money should be deposited?
Answer:
(a) For the cost of apples:
I. First, draw the x-axis and y-axis so they are perpendicular to each other.
II. Choose an appropriate scale for both axes.
III. Put the number of apples along the x-axis and mark the cost (in Rs) along the y-axis.
IV. Plot the points \( (1, 5), (2, 10), (3, 15), (4, 20) \) and \( (5, 25) \).
V. Then, connect all these points. The resulting graph is a straight line.
(b) For distance travelled by a car:
I. Start by drawing the axes for your graph.
II. Choose an appropriate scale for both the x-axis and the y-axis.
III. Mark the time (in hours) along the x-axis and the distance (in km) along the y-axis.
IV. Plot the points \( (6, 40), (7, 80), (8, 120) \) and \( (9, 160) \).
V. Connect these points to get the required graph.
(i) On the graph, draw a perpendicular line from 7:30 a.m. on the x-axis until it meets the graph at point A. From point A, draw a line parallel to the x-axis to meet the y-axis at 100 km.
Therefore, the distance travelled between 7:30 a.m. and 8:00 a.m. is \( 120 \, \text{km} - 100 \, \text{km} = 20 \, \text{km} \).
(ii) The time when the car had covered a distance of 100 km since its start was 7:30 a.m.
(c) For interest on deposits for a year:
I. Begin by drawing your x and y axes.
II. Select a suitable scale for both axes.
III. Mark the deposits (in Rs) along the x-axis and the simple interest (in Rs) along the y-axis.
IV. Plot the given points: \( (1000, 80), (2000, 160), (3000, 240), (4000, 320) \) and \( (5000, 400) \).
V. Connect the points to obtain the graph.
(i) Yes, the graph does pass through the origin \( (0,0) \).
(ii) Based on the graph, the interest on Rs 2500 is Rs 200.
(iii) From the graph, an interest of Rs 280 can be obtained by depositing Rs 3500.
In simple words: First, plot the apple cost on a graph. Then, plot the car's distance over time. Finally, plot the interest earned on deposits. Look at the graphs to find specific values like car distance between times or deposit amount for a certain interest.

Exam Tip: Always label your axes clearly and choose a scale that allows all data points to fit well within the graph area for easy reading.

 

Question 2. Draw a graph for the following.
(i)

It is linear graph?
(ii)
Answer:
(i) To graph the perimeter, use the side of the square along the x-axis and the perimeter along the y-axis. When you plot the points \( (2, 8), (3, 12), (3.5, 14), (5, 20) \) and \( (6, 24) \), the resulting graph is a straight line.
This graph is a linear graph.
(ii) To graph the area, use the side of the square along the x-axis and the area (in cm²) along the y-axis. You can draw the required graph by plotting the points \( (2, 4), (3, 9), (4, 16), (5, 25) \) and \( (6, 36) \).
The graph is not a straight line, which means it is not a linear graph.
In simple words: For the perimeter, plot the side length against the perimeter, and you will see a straight line. For the area, plot the side length against the area, and you will get a curve, not a straight line.

Exam Tip: Remember that a linear relationship results in a straight line graph, while a non-linear relationship (like area with side length) will produce a curved graph.