Maths 5 Marks Questions

We have,

3x + 4y - 12 = 0

⇒ 3x = 12 - 4y

$\Rightarrow3\text{x}=\frac{12-4\text{y}}{3}$

Putting y = 0, we get $\text{x}=\frac{12-4\times0}{3}=4$

Putting y = 3, we get $\text{x}=\frac{12-4\times3}{3}=0$

Thus, we have the following table for the points on the line 3x + 4y - 12 = 0:

We have,

6x + 8y - 48 = 0

⇒ 6x + 8y = 48

⇒ 6x = 48 - 8y

$\Rightarrow\text{x}=\frac{48-8\text{y}}{6}$

Putting y = 6, we get $\text{x}=\frac{48-8\times6}{6}=0$

Putting y = 4, we get $\text{x}=\frac{48-8\times3}{6}=4$

Thus, we have the following table for the points on the line 6x + 8y - 48 = 0:

The graphs of the path of a train A and B are:

It is given that seven year ago Ravish was seven times as old as his daughter.

x = Daughter, y = Father

$\therefore$ 7(x - 7) = y - 7

⇒ 7x - 49 = y - 7

⇒ 7x - 42 = y ...(i)

It is also given that after three years from now Ravish shall be three times as old as her daughter.

3(x + 3) = y + 3

⇒ 3x + 9 = y + 3

⇒ 3x + 6 = y ...(ii)

Now,

y = 7x - 42 [Using (i)]

Putting x = 6, we get y = 7 × 6 - 42 = 0

Putting x = 5, we get y = 7 × 5 - 42 = -7

Thus, we have the following table for the points on the line 7x - 42 = y:

We have,

y = 3x + 6 [Using (ii)]

Putting x = -2, we get y = 3 × (-2) + 6 = 0

Putting x = -1, we get y = 3 × (-1) + 6 = 3

Thus, we have the following table for the points on the line y = 3x + 6:

The graphs of the both linear equations are:

We have,

4x - 3y + 4 = 0

⇒ 4x = 3y - 4

$\Rightarrow\text{x}=\frac{3\text{y}-4}{4}$

Putting y = 0, we get $\text{x}=\frac{3\times0-4}{4}=-1$

Putting y = 4, we get $\text{x}=\frac{3\times4-4}{4}=2$

Thus, we have the following table for the points on the line 4x - 3y + 4 = 0:

We have,

4x - 3y - 20 = 0

⇒ 4x = 20 - 3y

$\Rightarrow\text{x}=\frac{20-3\text{y}}{4}$

Putting y = 0, we get $\text{x}=\frac{20-3\times0}{4}=5$

Putting y = 4, we get $\text{x}=\frac{20-3\times4}{4}=2$

Thus, we have the following table for the points on the line 4x - 3y - 20 = 0:

Clearly, two lines intersect atA(2, 4).

The graph of line 4x - 3y + 4 = 0 and 4x - 3y - 20 = 0 intersect with y-axis at B(-1, 0) and C(5, 0) respectively.

So, the vertices of the triangle formed by the two straight lines and y-axis are A(3, 2), B(0, 4) and C(0, -1).

$\therefore\text{Area of }\triangle\text{ABC}=\frac{1}{2}(\text{Base}\times\text{Height})$

$=\frac{1}{2}(\text{BC}\times\text{AM})$

$=\frac{1}{2}(6\times4)$

$=3\times4$

$=12\text{sq.units.}$

$\therefore\text{Area of }\triangle\text{ABC}=12\text{sq. units.}$

The region bounded by the graph is ABC which form a traingle.

AC at y axis is the base of traingle having AC = 4 units on y axis.

BC at x axis is the height of traingle having BC = 3 units on x axis.

Therefore,

Area of traingle ABC, say A is given by

$\text{A}=\frac{1}{2}(\text{Base}\times\text{Height)}$

$\text{A}=\frac{1}{2}(\text{AC}\times\text{BC)}$

$\text{A}=\frac{1}{2}(\text{4}\times\text{3)}$

$\text{(AC}\times\text{BC)} \text{A}=6\text{sq. units}$

1. To find the coordinates of the point when y = 3, we draw a line parallel to x-axis and passing throught (0, 3). This line meets the graph of 2x + 3y = 12 at a point P from which we draw a line parallel to y-axis which crosses x-axis at $\text{x}=\frac{3}{2}.$ So, the coordinates of the required point are $\Big(\frac{3}{2},\ 3\Big).$

2. To find the coordinates of the point when x = -3, we draw a line parallel to y-axis and passing throught (-3, 0). This line meets the graph of 2x + 3y = 12 at a point P from which we draw a line parallel to x-axis which crosses y-axis at y = 6. So, the coordinates of the required point are (-3, 6).

Let x be the fime and y be the distance travelled by Aarushi.

It is given that speed of car is 60km/ h

We know that,

$\text{Speed}=\frac{\text{Distance}}{\text{Time}}$

$\Rightarrow60=\frac{\text{y}}{\text{x}}$

$\Rightarrow\text{y}=60\text{x}$

Putting x = 1, we get y = 60

Putting x = 2, we get y = 120

Thus, we have the following table for the points on the line y = 60x:

The graph of the equation y = 60x :

1. To find the coordinates of the point when $\text{x}=2\frac{1}{2}=2.5,$ we draw a line parallel to y-axis and passing through (2.5, 0). This line meets the graph of y = 60x at a point P from which we draw a line parallel to x-axis which crosses y-axis at y = 150. So, the distance traveled by Aarushi in $2\frac{1}{2}$ hours is 150km.

2. To find the coordinates of the point when $\text{x}=\frac{1}{2}=0.5,$ we draw a line parallel to y-axis and passing through (0.5, 0). This line meets the graph of y = 60x at a point P from which we draw a line parallel to x-axis which crosses y-axis at y = 30. So, the distance travelled by Aanushi in $\frac{1}{2}$ hour is 30km.