Class 9 Graphical Solution Mock Test (ICSE)

Topics in Graphical Solution

• Graph of a linear equation
• Solving simultaneous equations graphically
• Consistent and inconsistent systems
• Area of the region bounded by lines and axes

Graphical Solution practice questions with answers

1. Q1Multiple choice·1 mark
   If the graphical representations of two simultaneous linear equations intersect at the point $(-2, 3)$, then the solution to the system is:

   1. (a)$x = -2, y = 3$
   2. (b)$x = -3, y = 2$
   3. (c)$x = 3, y = -2$
   4. (d)$x = 2, y = -3$
   Show answer and step marks

   1. Correct option: (a) $x = -2, y = 3$. The coordinates of the point of intersection of two graphical lines represent the unique solution of the simultaneous linear equations, so $x = -2$ and $y = 3$.+1

   Answer: (a)
2. Q2Multiple choice·1 mark
   Which of the following points lies on the graph of the linear equation $5x - 2y = 10$?

   1. (a)$(1, 3)$
   2. (b)$(4, 3)$
   3. (c)$(2, 0)$
   4. (d)$(0, 5)$
   Show answer and step marks

   1. Correct option: (c) $(2, 0)$. Substitute the coordinates $(2, 0)$ into the equation: $5(2) - 2(0) = 10 - 0 = 10$, which satisfies the equation.+1

   Answer: (c)
3. Q3Multiple choice·1 mark
   The area of the triangular region formed by the line $3x + 4y = 12$ with the coordinate axes is:

   1. (a)$6$ sq. units
   2. (b)$12$ sq. units
   3. (c)$3$ sq. units
   4. (d)$24$ sq. units
   Show answer and step marks

   1. Correct option: (a) $6$ sq. units. Find the intercepts: when $x = 0$, $y = 3$; when $y = 0$, $x = 4$. The base and height of the right-angled triangle are $4$ and $3$ units. Area = $\frac{1}{2} \times 4 \times 3 = 6$ sq. units.+1

   Answer: (a)
4. Q4Multiple choice·1 mark
   The area of the region bounded by the line $x + y = 5$, the $x$-axis, and the ordinates $x = 1$ and $x = 4$ is:

   1. (a)$7.5$ sq units
   2. (b)$12.5$ sq units
   3. (c)$5$ sq units
   4. (d)$10$ sq units
   Show answer and step marks

   1. Correct option: (a) $7.5$ sq units+1

   Answer: (a)
5. Q5Long answer·4 marks
   Find the solutions for the linear equations $2x - y = 2$ and $4x + 3y = 24$ by constructing their tables of values using at least three points for each. Draw both lines on a graph sheet and find the coordinates of the vertices of the triangular region formed by these two lines and the $x$-axis. Hence, calculate the area of this region.
   Show answer and step marks

   1. Table for $2x - y = 2$ with any three correct points, e.g. $(0, -2)$, $(1, 0)$, $(2, 2)$.+0.5
   2. Table for $4x + 3y = 24$ with any three correct points, e.g. $(0, 8)$, $(3, 4)$, $(6, 0)$.+0.5
   3. Plotting both lines accurately on the graph sheet.+0.5
   4. Reading the solution from the point of intersection: $x = 3$, $y = 4$.+0.5
   5. Identifying the coordinates of the vertices of the triangular region as $(1, 0)$, $(6, 0)$, and $(3, 4)$.+1
   6. Calculating the area: Base along the $x$-axis = $6 - 1 = 5$ units, Height = $4$ units. Area = $\frac{1}{2} \times 5 \times 4 = 10$ sq. units.+1

   Answer: Solution: $x = 3$, $y = 4$; vertices: $(1, 0)$, $(6, 0)$, and $(3, 4)$; Area = $10$ sq. units
6. Q6Long answer·4 marks
   Consider the two linear equations $4x - 3y = 12$ and $4x + 3y = 12$. (a) Find the coordinates of the points where the line $4x - 3y = 12$ intersects the $x$-axis and the $y$-axis. (b) Find the coordinates of the points where the line $4x + 3y = 12$ intersects the $x$-axis and the $y$-axis. (c) Calculate the area of the triangular region enclosed by these two lines and the $y$-axis.
   Show answer and step marks

   1. For $4x - 3y = 12$, setting $y = 0$ gives $x = 3$ (point $(3, 0)$); setting $x = 0$ gives $y = -4$ (point $(0, -4)$).+1
   2. For $4x + 3y = 12$, setting $y = 0$ gives $x = 3$ (point $(3, 0)$); setting $x = 0$ gives $y = 4$ (point $(0, 4)$).+1
   3. Base of the triangular region along the $y$-axis from $(0, -4)$ to $(0, 4)$ has length $4 - (-4) = 8$ units.+1
   4. Height of the triangle corresponding to this base is $3$ units, so Area = $\frac{1}{2} \times 8 \times 3 = 12$ sq. units.+1

   Answer: (a) $(3, 0)$ and $(0, -4)$; (b) $(3, 0)$ and $(0, 4)$; (c) $12$ sq. units

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