WBBSE Solutions Class 8 Chapter 18 Graphs

Graphs

Let’s find out the points in the graph paper.

We can see many points in the graph paper. Let’s try to write the co-ordinates of these points.

We see, A point is 4 units away from the origin on the x-axis.

Question 1. What are the coordinates of point A ?
Solution:

The coordinates of point A

The distance of point A from the y-axis is 4 units and from the x-axis is 0 unit.

∴ The coordinates of point A are (4, 0).

Understand, the coordinates of the point B are (9,0); the coordinates of point C are (14,0)

The coordinates of the point (D) are (18, 0).

So we got, the y- coordinate of any point on x-axis is 0 (zero).

We see that the point E is situated at 3 units away from the origin and it is on the y-axis.

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Question 2. What are the coordinates of point E ?
Solution:

The coordinates of point E

Point E lies on y-axis and 3 units away from x-axis but it is at 0 unit away from y-axis.

∴ The coordinates of point E are (0, 3).

Now we understand, the coordinates, of point F are (0.7).

The coordinates of point G are (0,10).

The coordinates of point H are (0, 14).

So we got, the x-coordinate of any point on the y-axis is 0

Let’s write the coordinates of the points I and J from the graph paper.

We see, the point I is 5 units away from the y-axis and 3 units away from the x-axis.

∴ The coordinates of point I are (5, 3).

The point J is 10 units away from the y-axis and 3 units away from the x-axis.

∴ The coordinates of J are (10,3).

The coordinates of K are (13,3)

The coordinates of L are (17,3)

Similarly let’s write the coordinates of the points M, N, P, Q, R, S and T.

The coordinates of M are (3,4)

The coordinates of N are (3,6)

The coordinates of P are (3,9)

The coordinates of Q are (3,13)

The coordinates of R are (8,9)

The coordinates of S are (10,11)

The coordinates of T are (14,12)

Now let’s join the points of the graph paper with a pencil and see which three or more than three points are collinear.

We see that the points A, B, C, and D lie on the X axis and they are collinear (collinear/noncollinear).

The Points E,F,G and H are on the y-axis and they are collinear.

The points I, J, K and L are collinear and M, N,P and Q are collinear. However, the points R, S and T are noncollinear.

Graphs Exercise 18.1

Question 1. Let’s plot the points A (4, 0), B (0, 6), C (2, 5), D (7, 1), E(6, 5) and F (10, 5) on a graph sheet.

Given

A (4, 0), B (0, 6), C (2, 5), D (7, 1), E(6, 5) and F (10, 5)

Coordinates of point A are (4, 0)

Coordinates of point B are (0, 6)

Coordinates of point C are (2, 5)

Coordinates of point D are (7, 1)

Coordinates of point E are (6, 5)

Coordinates of point F are (10, 5)

Question 2. Let’s plot the points (1, 1), (3, 7), (9, 1) and (12, 1) on a graph sheet and verify whether they lie on a line.
Solution:

Given

(1, 1), (3, 7), (9, 1) and (12, 1)

Question 3. On a graph sheet let’s write 4 collinear points and write the coordinates of the points. Similarly, plotting the points on the graph sheet and joining them we get some pictures. What will we call that?
Solution:

I took on the graph paper 4 collinear points A, B, C and D.

Plotting the points on the graph sheet from the coordinates of the points and joining the points we get a picture which is called a (Graph).

If joining the points we get a straight line segment, then this graph is called a linear graph.

1. Let’s plot the points A(2,0), B(6, 0) and C(4, .3) on the graph paper and join the points A, B ; B, C and C, A. Let’s find out what we have got.

Joining the points we have got a [triangle (triangle/quadrilateral).

2. Joining the points P(2,2) Q(4,4), R(6,6) we got a (collinear/non collinear) graph.

3. Let’s join the points P(1,2), A (2,3), T(3,4), and H(4,5) and see whether we have got a linear graph.
Solution:

Given

P(1,2), A (2,3), T(3,4), and H(4,5)

Path is a linear graph

Coordinates of P are (1, 2)

Coordinates of A are (2, 3)

Coordinates of T are (3, 4)

Coordinates of H are (4, 5)

Graphs Exercise 18.2

Question 1. I took 4 copies for Rs. 20. Let’s write the number and price of exercise books on the following table and draw a graph from the data.

Let’s find out from the graph the price of 6 exercise books and the number of exercise books for Rs. 45.
Solution:

Given

I took 4 copies for Rs. 20

1. First draw x- and y-axes.

2. Let us consider the length of one smallest square along x-axis = copy and the length of one smallest square along the y-axis = 1 copy and the length of one smallest square along the y-axis = Re. 1

3. From the data of graph plot points. (4, 20), (8, 40), (10, 50) and (12, 60)

4. Joining the points got AB straight line. So it is a linear graph.

5. On x-axis took the 6 exercise books and drew a ray parallel to the y-axis which cuts straight line at P point. Coordinates of P are (6, 30).

∴ Coordinates of point P are (6,30).

6. On the y-axis from the point of Rs. 45 drew a ray parallel to x- axis which cuts AB at Q. Coordinates of Q are (9, 45)

∴ Got from the graph that 9 exercise books are got in 45 rupees.

Graphs Exercise 18.3

Question 1. Let’s draw a graph of time and distance on the graph paper based on the information given below and find the distance covered in 4 hours and find how long it will take to cover 150 km.

Solution:

1. First of all, draw on the graph paper x- and y-axes.

2. Let distance is measured on x-axis and time along y-axis. Let the length of one smallest square along x-axis = 5 km and the length of one smallest square along y-axis = 1 hour

3. Plot the points on the graph paper (50, 2), (75, 3) and (125, 5) from the data given above.

4. Joining the above points got AB straight line. So it is a linear graph.

5. On the y-axis from the point of 4 hours drew a ray parallel to the x-axis which cuts AB at P.

Coordinates of point P are (100, 4)

∴ Got from the graph that in 4 hours, 100 km distance is covered.

6. On x-axis from the point of 150 km drew a ray parallel to y-axis which cuts AB at point Q. Coordinates of point Q are (150, 6).

∴ To cover 150 distance 6 hours are required.

Graphs Exercise 18.3

Question 1. For the graph paper beside we consider: 1 unit = length of 2 sides of the smallest squares along both the axes. Let’s write the coordinates of the points.
Solution:

Coordinates of A are (2, 0)

Coordinates of B are (5, 0)

Coordinates of C are (0, 4)

Coordinates of D are (0, 6)

Coordinates of E are (0, 8)

Coordinates of F are (4, 2)

Coordinates of G are (7, 2)

Coordinates of-H are (10, 2)

Coordinates of I are (2, 3)

Coordinates of J are (3, 4)

Coordinates of K are (4, 5)

2. Let’s find which three points are collinear.
Solution:

CDE, UK, FGH, CIF, and JFB are collinear.

3. Let’s find which three points are non-collinear.
Solution:

A, B and C are non-collinear.

Question 2. Let’s plot the points (1,0), (0, 5), (2, 1), (3,3), (1,3), (2,5) and (0,0) on the graph paper.
Solution:

Given

(1,0), (0, 5), (2, 1), (3,3), (1,3), (2,5) and (0,0)

O = (0, 0)

A = (1,0)

B = (0; 5)

C = (2, 1)

D = (3, 3)

E = (1, 3)

F = (2, 5)

Question 3.1. Let’s put the points (1,1), (2,2) and (6,6) on the graph paper and see whether they are collinear.

2. Let’s put three non-collinear points on the graph paper.Let’s put three more

3. collinear points other than the above three collinear points on the graph paper and write their coordinates.

A=(1,1)

B=(2,2)

C=(6,6)

A,B and C are collinear.

2. Lets plot any three non-collinear are points on the graph paper.
Solution:

D = (13, 5)

E = (15, 10)

F = (20, 10)

D, E and F are three non-collinear points.

3. After the three points above, we plot three collinear points and write their coordinates.
Solution:

P = (5, 10)

Q = (10, 15)

R = (15, 20)

P,Q and R are three collinear points.

Let 1 guava = length of 2 sides of the smallest square along x- axis and 1 rupee = length of a side of one smallest square along y-axis.

1. Let’s write the relation between the number of guava and the price of guava from graph.
Solution: There is direct relation between the number of guavas and price.

2. Let’s write the price of 4 guavas.
Solution:

4 guavas cost Rs. 12.

3. Let’s write the number of guavas in Rs. 30 from the graph.
Solution:

10 guavas are bought in Rs. 30.

4. Let’s write the number of guavas in Rs. 9.
Solution:

3 guavas are bought in Rs.9.

Question 5. Let’s see the graph beside and find the answers to the questions given below

Let 1 hdur = length of the 2 sides of the smallest squares along x- axis and 5 km = length of a side of the smallest square along y-axis.

1. Let’s write the relation between time and distance.
Solution:

Distance is in direct relation with time.

2. Let’s write the distance covered in 3. hours.
Solution:

90 km distance is covered in 3 hours.

3. Let’s write the time required to cover 120 km.
Solution:

To cover 120 km distance 4 hours are required.

4. Let’s find the velocity in km per hour.
Solution:

Velocity’is 30 kmph.

5. Let’s find and write the distance covered in 2 hours 30 minutes from the graph.
Solution:

In 2 hours 30 minutes, 75 km distance is covered.

6. Let’s find and write the time required to cover 45 km.
Solution:

To cover 45 km distance 1 hour 30 minutes are-required.

Question 6. Let’s draw the graph on the graph paper of the following data and see whether it is a linear graph.

Solution:

1. Let on the x-axis the length of one smallest square = 1. Pencil. Let along the y-axis the length of one smallest square = Rs. 1
2. Plot the points (3, 6), (5, 10), (7, 14) and (8, 16) on graph from the above data.
3. Joining the points got AB straight line. So it is a linear graph.

Question 7. Let’s draw the graph on the graph paper of the foilwing data and see whether it is a linear graph.

Solution:

1. Let the length of the smallest square on the axis = 1 hour; on the y-axis the length of one smallest square = 5 km.
2. Let on the plot the points (2, 40), (4, 80), (6, 120) and (8, 160)on the graph from the above data.
3. Joining the points got AB straight line. So it is a linear graph.

Question 8. Let’s draw the graph of the following data on the graph paper and see whether it is a linear graph.

Solution:

1. Let on the the x-axis the length of 5 sides of the smallest square =1 bag. Let on the y-axis the length of one smallest square = Rs. 5
2. Plot the points (1, 50), (2,100), (3, 150) and (4, 200) on the graph paper from the above data.
3. Joining the points got AB straight line. So the graph is linear.

Question 9. Let’s, draw the graph of the following data on the graph paper and see whether it is a linear graph.

Solution:

1. Let on the x-axis the length of 5 sides of the smallest square = 1 book and on the y-axis the length of one side of the smallest square = Rs. 5.
2. Plot the points (2, 50), (3, 75), (5, 125) and (8, 200) on the
   graph paper from the above data.
3. Joining the points got AB straight line. So the graph is linear.

Question 10. Let’s draw the graph on the graph paper of the following data and see whether it is a linear graph.

Solution:

1. Let on the x-axis the length of 5 sides of the smallest square = 1 over. On the y-axis the length of one side of the smallest square = 1 run.
2. Plot the points (1, 4), (3, 12), (5, 20) and (7, 24) on the graph paper from the above data.
3. Joining the points got AB graph. So the graph is not linear.