5 Mark Question

Question 15 Marks

Draw the graph of the equation 2x + y = 6. Shade the region bounded by the graph and the coordinate axes. Also, find the area of the shaded region.

Answer

We have,
2x + y = 6
⇒ y = 6 - 2x ...(i)
Putting x = 3 in (i), we get y = 6 - 2 × 3 = 0
Putting x = 0 in (i), we get y = 6 - 2 × 0 = 6
Thus, we obtain the following table giving coordinates of two points on the line represented by the equation 2x + y = 6.

X	3	0
y	0	6

The graph of line 2x + y = 6:

The area enclosed by the graph of line and the coordinate axes is shaded in the graph
Now,
Required area = Area of the shaded region
⇒ Required area = Area of $\triangle\text{ABC}$
⇒ Required area $=\frac{1}{2}(\text{Base}\times\text{Height})$
⇒ Required area $=\frac{1}{2}(3\times6)$
=9sq. units.

View full question & answer→

Question 25 Marks

Draw the graph of the equation $\frac{\text{x}}{3}+\frac{\text{y}}{4}=1.$ Also, find the area of the triangle formed by the line and the coordinates axes.

Answer

We are given,$\frac{\text{x}}{3}+\frac{\text{y}}{4}=1$
4x + 3y = 12 We get,$\text{y}=\frac{12-4\text{x}}{3}$
Now, Substituting x = 0 in $\text{y}=\frac{12-4\text{x}}{3},$ we get y = 4 Substituting x = 3 in $\text{y}=\frac{12-4\text{x}}{3},$ we get Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

x	0	3
y	4	0

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
$\text{A}=\frac{1}{2}(\text{Base}\times\text{Height)}$
$\text{A}=\frac{1}{2}(\text{AC}\times\text{BC)}$
$\text{A}=\frac{1}{2}(\text{4}\times\text{3)}$
$\text{(AC}\times\text{BC)} \text{A}=6\text{sq. units}$

View full question & answer→

Question 35 Marks

Draw the graph of the equation 2x + 3y = 12. From the graph, find the coordinates of the point:

Whose y-coordinates is 3.
whose x-coordinates is -3.

Answer

Graph of the equations 2x + 3y = 12: We have, 2x + 3y = 12 ⇒ 2x = 12 - 3y$\Rightarrow\text{x}=\frac{12-3\text{y}}{2}$
Putting y = 2, we get $\text{x}=\frac{12-3\times2}{2}=3$ Putting y = 4, we get $\text{x}=\frac{12-3\times4}{2}=0$ Thus, (3, 0) and (0, 4) are two points on the line 2x + 3y =12. The graph of the line represented by the equation 2x + 3y 12:

x	0	3
y	4	2

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 $\Big(\frac{3}{2},\ 3\Big).$
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).

View full question & answer→

Question 45 Marks

Draw the graphs of the linear equations 4x - 3y + 4 = 0 and 4x + 3y - 20 = 0. Find the area bounded by these lines and x-axis.

Answer

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:

x	-1	2
y	0	4

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:

x	5	2
y	0	4

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).

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

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

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

$=3\times4$

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

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

View full question & answer→

Question 55 Marks

Aarushi was driving a car with uniform speed of 60km/ h. Draw distance-time graph. From the graph, find the distance travelled by Aarushi in:

$2\frac{1}{2}\text{ Hours}$
$\frac{1}{2}\text{ Hour}$

Answer

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:

x	1	2
y	60	120

The graph of the equation y = 60x :

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.
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.

View full question & answer→

Question 65 Marks

Ravish tells his doughter Aarushi, "Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be".. lf present ages of Aarushi and Ravish are x and y years respectively, represent this situation algebraically as well as graphically.

Answer

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:

x	6	5
y	0	-7

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:

x	-1	-2
y	3	0

The graphs of the both linear equations are:

View full question & answer→

Question 75 Marks

The path of a train A is given by th equation 3x + 4y - 12 = 0 and the path or another train B is given by the equation 6x + 8y - 48 = 0. Represent this situation graphically.

Answer

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:

x	4	0
y	0	3

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:

x	0	4
y	6	3

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

View full question & answer→

Maths 5 Mark Question

Take a timed test