Graphs
Class 8th Mathematics Term 2 Tamilnadu Board Solution
- A (2, 3) Plot the following points in the graph paper and find out where they…
- B (- 3, 2) Plot the following points in the graph paper and find out where they…
- C (-5, -5) Plot the following points in the graph paper and find out where they…
- D (5, -8) Plot the following points in the graph paper and find out where they…
- E (6, 0) Plot the following points in the graph paper and find out where they…
- F (-4, 0) Plot the following points in the graph paper and find out where they…
- G (0, 9) Plot the following points in the graph paper and find out where they…
- H (0, -3) Plot the following points in the graph paper and find out where they…
- J (7, 8) Plot the following points in the graph paper and find out where they…
- O (0, 0) Plot the following points in the graph paper and find out where they…
- State in which quadrant each of the following points lie without actually…
- Determine the quadrants and the coordinates of the points A, B, C, D, E, F, G, H…
- (2, 7), (-2, - 3) Plot the following points and draw a line through the points.…
- (5, 4), (8, -5) Plot the following points and draw a line through the points.…
- (-3, 4), (-7, -2) Plot the following points and draw a line through the points.…
- (-5, 3), (5, -1) Plot the following points and draw a line through the points.…
- (2, 0), (6, 0) Plot the following points and draw a line through the points.…
- (0, 7), (4, -4) Plot the following points and draw a line through the points.…
- y = 0 Draw the graph of the following equations:
- x = 5 Draw the graph of the following equations:
- x = -7 Draw the graph of the following equations:
- y = 4 Draw the graph of the following equations:
- y = -3 Draw the graph of the following equations:
- A (3, 1), B(3, 6), C(-5, 6), D(-5, 1) Plot the following points and find out…
- A(-2, -4), B(5, -4), C(5, 4), D(-2, 4) Plot the following points and find out…
- A(3, 3), B(-3, 3), C(-3, -3), D(3, -3) Plot the following points and find out…
- O(0, 0), A(0, 7), B(-7, 7), C(-7, 0) Plot the following points and find out the…
- A(0, -2), B(-4, -6), C(4 ,-6) Plot the following points and find out the area…
- A(1, 2), B(9, 2), C(7, 4), D(3, 4) Plot the following points and find out the…
- A(-4, 1), B(-4, 7), C(-7, 10), D(-7, 4) Plot the following points and find out…
- Find the perimeter of the rectangle and squares of the previous problems 6 (i).…
- Find the perimeter of the rectangle and squares of the previous problems 6…
- Find the perimeter of the rectangle and squares of the previous problems 6…
- Find the perimeter of the rectangle and squares of the previous problems 6…
- |c|c|c|c|c|c|c| x&5&5&5&5&5&5 y&1&2&3&4&5&6 Draw a linear graph for the…
- |c|c|c|c|c|c| x&1&2&3&4&5 y&1&2&3&4&5 Draw a linear graph for the following…
- Draw the linear graph and find the missing entries. |c|c|c|c|c|c| x&1&2&3&4&-…
- Draw the following graph of side-area relationship of a square.
- Draw the graph of y = 7x.
- If Akbar is driving a car at a uniform speed of 40 km/hr. Draw the distance time…
- Eliza deposited ₹ 20,000 in a bank at the rate of 10% per annum. Draw a linear…
Exercise 3.1
Question 1.Plot the following points in the graph paper and find out where they lie?
A (2, 3)
Answer:
x co-ordinate = 2
y co-ordinate = 3
point A is 2 units to the right of Y-axis and 3 units above X-axis
Question 2.
Plot the following points in the graph paper and find out where they lie?
B (– 3, 2)
Answer:
x co-ordinate = -3
y co-ordinate = 2
point B is 3 units to the left of Y-axis and 2 units above X-axis
Question 3.
Plot the following points in the graph paper and find out where they lie?
C (–5, –5)
Answer:
x co-ordinate = -5
y co-ordinate = -5
point C is 5 units below X-axis and 5 units to the left of Y-axis
Question 4.
Plot the following points in the graph paper and find out where they lie?
D (5, –8)
Answer:
x co-ordinate = 5
y co-ordinate = -8
point D is 5 units to the right of Y-axis and 8 units below X-axis
Question 5.
Plot the following points in the graph paper and find out where they lie?
E (6, 0)
Answer:
x co-ordinate = 6
y co-ordinate = 0
point E is 6 units to the right of Y-axis and 0 units from X-axis i.e. y = 0
Question 6.
Plot the following points in the graph paper and find out where they lie?
F (–4, 0)
Answer:
x co-ordinate = -4
y co-ordinate = 0
point F is 4 units to the left of Y-axis and 0 units from X-axis i.e. y = 0
Question 7.
Plot the following points in the graph paper and find out where they lie?
G (0, 9)
Answer:
x co-ordinate = 0
y co-ordinate = 9
point G is 9 units above X-axis and 0 units from Y-axis i.e. x = 0
Question 8.
Plot the following points in the graph paper and find out where they lie?
H (0, –3)
Answer:
x co-ordinate = 0
y co-ordinate = -3
point H is 3 units below X-axis and 0 units from Y-axis i.e. x = 0
Question 9.
Plot the following points in the graph paper and find out where they lie?
J (7, 8)
Answer:
x co-ordinate = 7
y co-ordinate = 8
point J is 7 units above X-axis and 8 nits to the right of Y-axis
Question 10.
Plot the following points in the graph paper and find out where they lie?
O (0, 0)
Answer:
x co-ordinate = 0
y co-ordinate = 0
point O is origin intersection of X and Y axis is origin
Question 11.
State in which quadrant each of the following points lie without actually plotting the points.
i. (8, 15)
ii. (–15, 2)
iii. (–20, –10)
iv. (6, –9)
v. (0, 18)
vi. (–17, 0)
vii. (9, 0)
viii. (–100, –200)
ix. (200, 500)
x. (–50, 7500)
Answer:
Refer this table
i. x co-ordinate = 8 and y co-ordinate = 15
both x and y are positive therefore point (8, 15) lies in the 1st quadrant
ii. x co-ordinate = -15 and y co-ordinate = 2
x co-ordinate is negative and y co-ordinate is positive therefore point (–15, 2) lies in 2nd quadrant
iii. x co-ordinate = -20 and y co-ordinate = -10
both x and y are negative therefore point (-20, -10) lies in the 3rd quadrant
iv. x co-ordinate = 6 and y co-ordinate = -9
x co-ordinate is positive and y co-ordinate is negative therefore point (6, -9) lies in 4th quadrant
v. x co-ordinate = 0 and y co-ordinate = 18
as x = 0 the point lies on Y-axis because equation of Y-axis is x = 0
therefore point (0, 18) lies on Y-axis
vi. x co-ordinate = -17 and y co-ordinate = 0
as y = 0 the point lies on X-axis because equation of X-axis is y = 0
therefore point (-17, 0) lies on X-axis
vii. x co-ordinate = 9 and y co-ordinate = 0
as y = 0 the point lies on X-axis because equation of X-axis is y = 0
therefore point (9, 0) lies on X-axis
viii. x co-ordinate = -100 and y co-ordinate = -200
both x and y are negative therefore point (-100, -200) lies in the 3rd quadrant
ix. x co-ordinate = 200 and y co-ordinate = 500
both x and y are positive therefore point (200, 500) lies in the 1st quadrant
x. x co-ordinate = -50 and y co-ordinate = 7500
x co-ordinate is negative and y co-ordinate is positive therefore point (–50, 7500) lies in 2nd quadrant
Question 12.
Determine the quadrants and the coordinates of the points A, B, C, D, E, F, G, H and O in the given figure.
Answer:
If a point is left to the Y-axis then the x co-ordinate will be negative and similarly if a point if below X-axis its y co-ordinate will be negative
Point A
A is on the Y-axis implies x co-ordinate = 0
A is 4 units above X-axis implies y co=ordinate = 4
Therefore, co-ordinates of A are (0, 4)
Point B
B is 2 units above X-axis implies y = 2
B is 3 units left to the Y-axis implies x = -3
Therefore, co-ordinates of B are (-3, 2)
As from figure given B lies in the second quadrant
Point C
C lies on X-axis implies y = 0
C is 5 units to the left of Y-axis implies x = -5
Therefore, co-ordinates of C are (-5, 0)
Point D
D is 6 units below X-axis implies y = -6
D is 4 units to the left of Y-axis implies x = -4
Therefore, co-ordinates of D are (-6, -4)
As seen in figure D lies in 3rd quadrant
Point E
E is on the Y-axis implies x = 0
E is 3 units below X-axis implies y = -3
Therefore, co-ordinates of E are (0, -3)
Point F
F is 7 units to the right of Y-axis implies x = 7
F is 1 unit below X-axis implies y = -1
Therefore, co-ordinates of F are (7, -1)
As seen in figure F lies in 4th quadrant
Point G
G is on X-axis implies y = 0
G is 4 units to the right of Y-axis implies x = 4
Therefore, co-ordinates of G are (4, 0)
Point H
H is 6 units to the right of Y-axis implies x = 6
H is 3 units above X-axis implies y = 3
Therefore, co-ordinates of H are (6, 3)
As seen in figure H lies in 1st quadrant
Point O
O is at intersection of X-axis and Y-axis
Intersection of X and Y axis is (0, 0)
Therefore, co-ordinates of O are (0, 0)
Question 13.
Plot the following points and draw a line through the points.
(2, 7), (–2, – 3)
Answer:
Let the points be P (2, 7) and Q (-2, -3)
Consider point P
x co-ordinate = 2
y co-ordinate = 7
point P is 2 units to the right of Y-axis and 7 units above X-axis
Consider point Q
x co-ordinate = -2
y co-ordinate = -3
point Q is 3 units below X-axis and 2 units to the left of Y-axis
Question 14.
Plot the following points and draw a line through the points.
(5, 4), (8, –5)
Answer:
Let the points be P (5, 4) and Q (8, -5)
Consider point P
x co-ordinate = 5
y co-ordinate = 4
point P is 5 units to the right of Y-axis and 4 units above X-axis
Consider point Q
x co-ordinate = 8
y co-ordinate = -5
point Q is 5 units below X-axis and 8 units to the right of Y-axis
Question 15.
Plot the following points and draw a line through the points.
(–3, 4), (–7, –2)
Answer:
Let the points be P (-3, 4) and Q (-7, -2)
Consider point P
x co-ordinate = -3
y co-ordinate = 4
point P is 3 units to the left of Y-axis and 4 units above X-axis
Consider point Q
x co-ordinate = -7
y co-ordinate = -2
point Q is 2 units below X-axis and 7 units to the left of Y-axis
Question 16.
Plot the following points and draw a line through the points.
(–5, 3), (5, –1)
Answer:
Let the points be P (-5, 3) and Q (5, -1)
Consider point P
x co-ordinate = -5
y co-ordinate = 3
point P is 5 units to the left of Y-axis and 3 units above X-axis
Consider point Q
x co-ordinate = 5
y co-ordinate = -1
point Q is 1 units below X-axis and 5 units to the right of Y-axis
Question 17.
Plot the following points and draw a line through the points.
(2, 0), (6, 0)
Answer:
Let the points be P (2, 0) and Q (6, 0)
Consider P
x co-ordinate = 2
y co-ordinate = 0
since y = 0 the point P lies on the X-axis
Consider Q
x co-ordinate = 6
y co-ordinate = 0
since y = 0 the point Q lies on the X-axis
since both the points P and Q lie on the X-axis the line passing through both the points would be X-axis
Question 18.
Plot the following points and draw a line through the points.
(0, 7), (4, –4)
Answer:
Let the points be P (0, 7) and Q (4, -4)
Consider P
x co-ordinate = 0
y co-ordinate = 7
as x = 0 the point P lies on the Y-axis because it is 7 units above the X-axis and 0 units to the left or right from the Y-axis
Consider Q
x co-ordinate = 4
y co-ordinate = -4
point Q is 4 units to the right of Y-axis and 4 units below Y-axis
Question 19.
Draw the graph of the following equations:
y = 0
Answer:
y = 0 means the y co-ordinate is fixed and we can take any value of x
to draw graph we will require two points
let us consider point P whose x co-ordinate be 2 and y co-ordinate is already given as 0
⇒ P (2, 0)
Now consider point Q whose x co-ordinate be 4 and y co-ordinate is already given as 0
⇒ Q (4, 0)
Plot points P and Q and join them to get the graph of y = 0
Question 20.
Draw the graph of the following equations:
x = 5
Answer:
x = 5 means the x co-ordinate is fixed and we can take any value of y
to draw graph we will require two points
let us consider point P whose y co-ordinate be 2 and x co-ordinate is already given as 5
⇒ P (5, 2)
Now consider point Q whose y co-ordinate be 4 and x co-ordinate is already given as 5
⇒ Q (5, 4)
Plot points P and Q and join them to get the graph of x = 5
Question 21.
Draw the graph of the following equations:
x = –7
Answer:
x = -7 means the x co-ordinate is fixed and we can take any value of y
to draw graph we will require two points
let us consider point P whose y co-ordinate be 2 and x co-ordinate is already given as -7
⇒ P (-7, 2)
Now consider point Q whose y co-ordinate be 4 and x co-ordinate is already given as -7
⇒ Q (-7, 4)
Plot points P and Q and join them to get the graph of x = -7
Question 22.
Draw the graph of the following equations:
y = 4
Answer:
y = 4 means the y co-ordinate is fixed and we can take any value of x
to draw graph we will require two points
let us consider point P whose x co-ordinate be 2 and y co-ordinate is already given as 4
⇒ P (2, 4)
Now consider point Q whose x co-ordinate be 4 and y co-ordinate is already given as 4
⇒ Q (4, 4)
Plot points P and Q and join them to get the graph of y = 4
Question 23.
Draw the graph of the following equations:
y = –3
Answer:
y = -3 means the y co-ordinate is fixed and we can take any value of x
to draw graph we will require two points
let us consider point P whose x co-ordinate be 2 and y co-ordinate is already given as -3
⇒ P (2, -3)
Now consider point Q whose x co-ordinate be 4 and y co-ordinate is already given as -3
⇒ Q (4, -3)
Plot points P and Q and join them to get the graph of y = -3
Hint: x = a is always a straight line parallel to Y-axis and passing through (a, 0) whereas y = a is always a straight line parallel to X-axis and passing through (0, a)
Question 24.
Plot the following points and find out the area of enclosed figures.
A (3, 1), B(3, 6), C(–5, 6), D(–5, 1)
Answer:
plot points A, B, C and D and join them as shown
As seen in the figure ABCD is a rectangle
Area of rectangle = length × breadth
Length = AD = AE + ED
= 3 + 5
= 8 cm
ED = 5 and not -5 because distance of point D from Y-axis is 5 and distance cannot be negative
⇒ AD = 8 cm
B is at height of 6 units from X-axis and A is at height of 1 units from X-axis therefore distance between points A and B would be 6 – 1
Breadth = AB = 6 – 1 = 5 cm
Area of rectangle = 8 × 5 = 40 sq. cm
Question 25.
Plot the following points and find out the area of enclosed figures.
A(–2, –4), B(5, –4), C(5, 4), D(–2, 4)
Answer:
plot points A, B, C and D and join them as shown
As seen in the figure ABCD is a rectangle
Area of rectangle = length × breadth
Length = CD = CF + FD
= 5 + 2
= 7 cm
⇒ CD = 7 cm
Breadth = AD = AE + ED
= 4 + 4
= 8 cm
⇒ AD = 8 cm
Area of rectangle = AD × CD
= 7 × 8 = 56 sq. cm
Question 26.
Plot the following points and find out the area of enclosed figures.
A(3, 3), B(–3, 3), C(–3, –3), D(3, –3)
Answer:
plot points A, B, C and D and join them as shown
As seen in the figure ABCD is a rectangle
Area of rectangle = length × breadth
Length = CD = CF + FD
= 3 + 3
= 6 cm
⇒ CD = 6 cm
Breadth = AD = AE + ED
= 3 + 3
= 6 cm
⇒ AD = 6 cm
Area of rectangle = CD × AD
= 6 × 6 = 36 sq. cm
Question 27.
Plot the following points and find out the area of enclosed figures.
O(0, 0), A(0, 7), B(–7, 7), C(–7, 0)
Answer:
plot points O, A, B and C and join them as shown
As seen in the figure OABC is a rectangle
Area of rectangle = length × breadth
Length = CO = 7 cm
⇒ CO = 7 cm
Breadth = OA = 7 cm
⇒ OA = 7 cm
Area of rectangle = CO × AO
= 7 × 7 = 49 sq. cm
Question 28.
Plot the following points and find out the area of enclosed figures.
A(0, –2), B(–4, –6), C(4 ,–6)
Answer:
Plot points A, B and C to form triangle as shown
For the ΔABC BC is the base and AD is the height
From figure
BC = BD + DC
= 4 + 4
= 8 cm
⇒ BC = 8 cm
AD = 6 – 2
= 4 cm
⇒ AD = 4cm
Area of triangle = 1/2 × base × height
= 1/2 × BC × AD
= 1/2 × 8 × 4
= 4 × 4
= 16 sq. cm
Area of ΔABC = 16 sq. cm
Question 29.
Plot the following points and find out the area of enclosed figures.
A(1, 2), B(9, 2), C(7, 4), D(3, 4)
Answer:
plot points A, B, C and D and join them as shown
Here quadrilateral ABCD so formed is a trapezium
Area of trapezium = 1/2 × (sum of parallel sides) × height
Here AB || DC
AB = 9 – 1 = 8 cm
CD = 7 – 3 = 4 cm
DG is the height of trapezium
DG = DF – GF
= 4 – 2
= 2 cm
DG = 2 cm
∴ Area of trapezium = 1/2 × (AB + CD) × DG
= 1/2 × (8 + 4) × 2
= 12 sq. cm
Area of trapezium ABCD = 12 sq. cm
Question 30.
Plot the following points and find out the area of enclosed figures.
A(–4, 1), B(–4, 7), C(–7, 10), D(–7, 4)
Answer:
plot points A, B, C and D and join them as shown
Here quadrilateral ABCD so formed is a trapezium
Area of trapezium = 1/2 × (sum of parallel sides) × height
Here AB || DC
AB = 7 – 1 = 6 cm
CD = 10 – 4 = 6 cm
DG is the height of trapezium
DG = DF – GF
= 7 – 4
= 3 cm
DG = 3 cm
∴ Area of trapezium = 1/2 × (AB + CD) × DG
= 1/2 × (6 + 6) × 3
= 1/2 × 12 × 3
= 6 × 3
= 18 sq. cm
Area of trapezium ABCD = 18 sq. cm
Question 31.
Find the perimeter of the rectangle and squares of the previous problems 6 (i).
Answer:
6 (i) A (3, 1), B(3, 6), C(–5, 6), D(–5, 1)
plot points A, B, C and D and join them as shown
As seen in the figure ABCD is a rectangle
Length = AD = AE + ED
= 3 + 5
= 8 cm
ED = 5 and not -5 because distance of point D from Y-axis is 5 and distance cannot be negative
⇒ AD = 8 cm
B is at height of 6 units from X-axis and A is at height of 1 units from X-axis therefore distance between points A and B would be 6 – 1
Breadth = AB = 6 – 1 = 5 cm
Perimeter of rectangle ABCD = 2 × (length + breadth)
= 2 × (AD + AB)
= 2 × (8 + 5)
= 26 cm
Perimeter of rectangle ABCD = 26 cm
Question 32.
Find the perimeter of the rectangle and squares of the previous problems 6 (ii).
Answer:
6 (ii) A(–2, –4), B(5, –4), C(5, 4), D(–2, 4)
plot points A, B, C and D and join them as shown
As seen in the figure ABCD is a rectangle
Length = CD = CF + FD
= 5 + 2
= 7 cm
⇒ CD = 7 cm
Breadth = AD = AE + ED
= 4 + 4
= 8 cm
⇒ AD = 8 cm
Perimeter of rectangle ABCD = 2 × (length + breadth)
= 2 × (CD + AD)
= 2 × (7 + 8)
= 2 × 15
= 30 cm
Perimeter of rectangle ABCD = 30 cm
Question 33.
Find the perimeter of the rectangle and squares of the previous problems 6 (iii).
Answer:
6 (iii) A(3, 3), B(–3, 3), C(–3, –3), D(3, –3)
plot points A, B, C and D and join them as shown
As seen in the figure ABCD is a rectangle
Length = CD = CF + FD
= 3 + 3
= 6 cm
⇒ CD = 6 cm
Breadth = AD = AE + ED
= 3 + 3
= 6 cm
⇒ AD = 6 cm
Perimeter of rectangle ABCD = 2 × (length + breadth)
= 2 × (CD + AD)
= 2 × (6 + 6)
= 2 × 12
= 24 cm
Perimeter of rectangle ABCD = 24 cm
Question 34.
Find the perimeter of the rectangle and squares of the previous problems 6 (iv).
Answer:
6 (iv) O(0, 0), A(0, 7), B(–7, 7), C(–7, 0)
plot points O, A, B and C and join them as shown
As seen in the figure OABC is a rectangle
Length = CO = 7 cm
⇒ CO = 7 cm
Breadth = OA = 7 cm
⇒ OA = 7 cm
Perimeter of rectangle OABC = 2 × (length + breadth)
= 2 × (CO + OA)
= 2 × (7 + 7)
= 2 × 14
= 28 cm
Perimeter of rectangle ABCD = 28 cm
Exercise 3.2
Question 1.Draw a linear graph for the following data.
Answer:
When x = 5, y = 1 consider it as point A with co-ordinates (5, 1)
When x = 5, y = 2 consider it as point B with co-ordinates (5, 2)
When x = 5, y = 3 consider it as point C with co-ordinates (5, 3)
When x = 5, y = 4 consider it as point D with co-ordinates (5,4)
When x = 5, y = 5 consider it as point E with co-ordinates (5, 5)
When x = 5, y = 6 consider it as point F with co-ordinates (5, 6)
Plot the points A, B, C, D, E and F and join them to get the required linear graph
Question 2.
Draw a linear graph for the following data.
Answer:
When x = 1, y = 1 consider it as point A with co-ordinates (1, 1)
When x = 2, y = 2 consider it as point B with co-ordinates (2, 2)
When x = 3, y = 3 consider it as point C with co-ordinates (3, 3)
When x = 4, y = 4 consider it as point D with co-ordinates (4,4)
When x = 5, y = 5 consider it as point E with co-ordinates (5, 5)
Plot the points A, B, C, D, E and F and join them to get the required linear graph
Question 3.
Draw the linear graph and find the missing entries.
Answer:
Consider the missing entries as a, b and c as follows
As given the graph is linear i.e. it would be straight line
When x = 1, y = 6 consider it as point A with co-ordinates (1, 6)
When x = 2, y = 12 consider it as point B with co-ordinates (2,12)
Plot points A and B and draw a straight line passing through it
When x = 3, y = a consider it as point C with co-ordinates (3, a)
But from graph point C has co-ordinates (3,18)
⇒ a = 18
When x = 4, y = b consider it as point D with co-ordinates (4, d)
But from graph point D has co-ordinates (4, 24)
⇒ b = 24
When x = c, y = 30 consider it as point E with co-ordinates (c, 30)
But from graph point C has co-ordinates (5,30)
⇒ c = 30
Question 4.
Draw the following graph of side–area relationship of a square.
Answer:
Consider size as x co-ordinate and area as y co-ordinate
When x = 2, y = 4 consider it as point A with co-ordinates (2, 4)
When x = 3, y = 9 consider it as point B with co-ordinates (3, 9)
When x = 4, y = 16 consider it as point C with co-ordinates (4, 16)
When x = 5, y = 25 consider it as point D with co-ordinates (5,25)
When x = 6, y = 36 consider it as point E with co-ordinates (6,36)
Plot points A, B, C, D and E and join them to get the required graph
Question 5.
Draw the graph of y = 7x.
Answer:
y = 7x is a linear equation therefore its graph would be a straight line
now to draw a straight line we require atleast two points
consider any value of x and put it in y = 7x to get corresponding value of y
let x = 0 after putting x = 0 in y = 7x we get y = 0 consider this point as A with co-ordinates (0, 0)
let x = 1 after putting x = 1 in y = 7x we get y = 7 consider this point as B with co-ordinates (1, 7)
plot points A and B and draw a line passing through them, we get graph of y = 7x
Question 6.
If Akbar is driving a car at a uniform speed of 40 km/hr. Draw the distance time graph. Also find the time taken by Akbar to cover a distance of 200 km.
Answer:
Seed is uniform i.e. 40 km/hr
Which means Akbar covers 40 km distance in 1 hour
Using speed = 
So, in two hours Akbar will travel 40 × 2 = 80 km distance
Let ‘a’ be the time required to cover 200 km
⇒ 40 = 
⇒ a = 
⇒ a = 5 hours
Representing in tabular form
Consider time on X-axis and distance on Y-axis
Consider point A with co-ordinates (1, 40) and point B with co-ordinates (2, 80)
Plot points A and B and draw a line passing through them is the required graph
Scale:
On X-axis 1 cm = 1 hour
On Y-axis 1 cm = 10 km
Question 7.
Eliza deposited ₹ 20,000 in a bank at the rate of 10% per annum. Draw a linear graph showing the time and simple interest relationship. Also, find the simple interest for 4 years.
Answer:
Principal amount = P = 20,000 rs
Rate = r = 10%
Time = t
Simple interest = I = 
⇒ I = 
⇒ I = 2000t …(i)
Simple interest for 4 years implies t = 4
Put t = 4 in equation (i)
⇒ I = 2000 × 4
⇒ I = 8000 rs
Consider equation (i)
Linear graph means a straight line
Now to draw a straight line we need atleast two points
Consider t time on X-axis and I simple interest on Y-axis
Take any two values of t and find corresponding values of I
If t = 1 implies I = 2000 × 1 = 2000
Consider this as a point A with co-ordinates (1, 2000)
If t = 2 implies I = 2000 × 2 = 4000
Consider this as a point B with co-ordinates (1, 4000)
Plot points A and B and draw a line passing through them
Scale for graph
On X-axis 1cm = 1 year
On Y-axis 1 cm = 1000rs
