Solved Math of 9th Class Paper 2024 G2 Lahore Board Mathematics (Science) – Essay Type Questions Session: 2020-2022 to 2023-2025
Total Marks: 60
Time Allowed: 2 Hours 10 Minutes
9th class Mathematics solved Past Paper 2024 Group 2 Lahore Board by FG STudy.com
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Q-2(i) Matrix Multiplication
Given:
\( A = \begin{bmatrix} -4 & 2 \\ 2 & 1 \end{bmatrix}, \quad B = \begin{bmatrix} -3 \\ -2 \end{bmatrix} \)
To find \( AB \):
\( AB = \begin{bmatrix} -4 & 2 \\ 2 & 1 \end{bmatrix} \times \begin{bmatrix} -3 \\ -2 \end{bmatrix} \)
Step-by-step calculation:
- First row: \((-4)(-3) + (2)(-2) = 12 – 4 = 8\)
- Second row: \((2)(-3) + (1)(-2) = -6 – 2 = -8\)
Final result:
\( AB = \begin{bmatrix} 8 \\ -8 \end{bmatrix} \)
Q-2(ii) Simplify \( x^{5^2} \div (x^5)^2 \)
Step-by-step solution:
- \( 5^2 = 25 \), so \( x^{5^2} = x^{25} \).
- \( (x^5)^2 = x^{5 \times 2} = x^{10} \).
- Using the rule of exponents for division: \( x^{25} \div x^{10} = x^{25-10} = x^{15} \).
Final result:
\( x^{15} \)
Q-2(ii) Find the value of \(x\) and \(y\) if \(x + iy + 1 = 4 – 3i\):
Separating real and imaginary parts:
- Real part: \(x + 1 = 4 \Rightarrow x = 3\)
- Imaginary part: \(iy = -3i \Rightarrow y = -3\)
Answer: \(x = 3, y = -3\)
Q-2(iv) Find the value of \(x\):
Given:
\(\log_{64} 8 = \frac{x}{2}\)
\(\log_{64} 8 = \frac{1}{2}\) (since \(64^{1/2} = 8\))
\(\frac{1}{2} = \frac{x}{2} \Rightarrow x = 1\)
Answer: \(x = 1\)
Q-2(v) Find the value of \(x\):
Given:
\(\log x = 0.1821\)
Using the property of logarithms:
\(x = 10^{0.1821}\)
Approximation:
\(x \approx 1.52\)
Answer: \(x \approx 1.52\)
Q-2(vi) Simplify:
\(\left(\sqrt{2} + \frac{1}{\sqrt{3}}\right)\left(\sqrt{2} – \frac{1}{\sqrt{3}}\right)\)
Using the difference of squares formula:
\((a + b)(a – b) = a^2 – b^2\)
\(\left(\sqrt{2}\right)^2 – \left(\frac{1}{\sqrt{3}}\right)^2 = 2 – \frac{1}{3}\)
Final result:
\(2 – \frac{1}{3} = \frac{6}{3} – \frac{1}{3} = \frac{5}{3}\)
Answer: \(\frac{5}{3}\)
Q-2(vii) If \(x = \sqrt{3} + 2\), find \(x + \frac{1}{x}\):
Given:
\(x = \sqrt{3} + 2\)
\(\frac{1}{x} = \frac{1}{\sqrt{3} + 2}\)
Rationalizing the denominator:
\(\frac{1}{x} = \frac{\sqrt{3} – 2}{(\sqrt{3} + 2)(\sqrt{3} – 2)} = \frac{\sqrt{3} – 2}{-1} = -\sqrt{3} + 2\)
Adding \(x + \frac{1}{x}\):
\((\sqrt{3} + 2) + (-\sqrt{3} + 2) = 4\)
Answer: \(x + \frac{1}{x} = 4\)
Q-2(viii) Factorize:
Given:
\(4x^2 + 12x + 5\)
Splitting the middle term:
\(4x^2 + 10x + 2x + 5\)
Grouping terms:
\((4x^2 + 10x) + (2x + 5)\)
\(2x(2x + 5) + 1(2x + 5)\)
Factoring:
\((2x + 5)(2x + 1)\)
Answer: \((2x + 5)(2x + 1)\)
Q-2(ix) Factorize:
Given:
\(1 – 27y^3\)
Using the difference of cubes formula:
\(a^3 – b^3 = (a – b)(a^2 + ab + b^2)\)
Here, \(a = 1\) and \(b = 3y\):
\(1 – 27y^3 = (1 – 3y)(1^2 + 1(3y) + (3y)^2)\)
\(= (1 – 3y)(1 + 3y + 9y^2)\)
Answer: \((1 – 3y)(1 + 3y + 9y^2)\)
Q-3 (i) Find the LCM of the expressions x² – 25x + 100 and x² – x – 20
To find the LCM, we factorize both expressions:
x² – 25x + 100 = (x – 10)(x – 10)
x² – x – 20 = (x – 5)(x + 4)
The LCM will be the product of all unique factors:
LCM = (x – 10)(x – 5)(x + 4)
Q-3 (ii) Solve the inequality:
\[ \frac{4 – \frac{1}{2}x}{2} \geq -7 + \frac{1}{4}x \]
Solution:
\[ \frac{4 – \frac{1}{2}x}{2} = 2 – \frac{1}{4}x \]
\[ 2 – \frac{1}{4}x \geq -7 + \frac{1}{4}x \]
\[ 2 + 7 \geq \frac{1}{2}x \]
\[ 9 \geq \frac{1}{2}x \]
\[ x \leq 18 \]
Q-3 (iii) Solve the equation:
\[ \frac{x – 3}{2} – \frac{x – 2}{2} = -1 \]
Solution:
\[ \frac{(x – 3) – (x – 2)}{2} = \frac{-1}{2} \]
\[ \frac{-1}{2} = -1 \]
Q-3 (iv) Find the values of \(m\) and \(c\) by expressing the line \(2x – 3y = -5\) in the form \(y = mx + c\):
Solution:
\[ 2x – 3y = -5 \]
\[ -3y = -2x – 5 \]
\[ y = \frac{2}{3}x + \frac{5}{3} \]
Here, \(m = \frac{2}{3}\) and \(c = \frac{5}{3}\).
Q-3 (v) Draw the graph of \(x = -3\):
This equation represents a vertical line passing through \(x = -3\) on the Cartesian plane.
Q-3 (vi) Find the distance between the points \(A(6, -2)\) and \(B(6, -3)\):
Solution:
Distance formula: \[ d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2} \]
\[ d = \sqrt{(6 – 6)^2 + (-3 – (-2))^2} \]
\[ d = \sqrt{0^2 + (-1)^2} = \sqrt{1} = 1 \]
The distance is \(1\) unit.
Q-3 (vii) Find the midpoint of the line segment joining the points \(A(2, -6)\) and \(B(3, -6)\):
Solution:
Midpoint formula: \[ \text{Midpoint} = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \]
\[ \text{Midpoint} = \left(\frac{2 + 3}{2}, \frac{-6 + (-6)}{2}\right) \]
\[ \text{Midpoint} = \left(2.5, -6\right) \]
Q-3 (viii) Define congruent triangles:
Definition:
Congruent triangles are triangles that have the same size and shape. This means:
- Their corresponding sides are equal in length.
- Their corresponding angles are equal in measure.
Notation: If \(\triangle ABC\) is congruent to \(\triangle DEF\), we write:
\[ \triangle ABC \cong \triangle DEF \]
Q-3 (ix) Define a parallelogram:
Definition:
A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel and equal in length.
Properties:
- Opposite angles are equal.
- Adjacent angles are supplementary (add up to \(180^\circ\)).
- Diagonals bisect each other.