Solved Math of 9th Class Paper 2024 G1 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 1 Lahore Board by FG STudy.com
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Q.2 Write short answers to any SIX (6) questions:
(i) If A = [[2, -1], [0, 2]] and B = [[3, 0], [-1, 2]], then find A – B.
A - B = [[2, -1], [0, 2]] - [[3, 0], [-1, 2]]
= [[-1, -1], [1, 0]]
(ii) Simplify and write the answer in a + bi form: (1+i)/(1-i).
(1+i)/(1-i) * (1+i)/(1+i) = (1 + 2i + i^2)/(1 - i^2)
= (1 + 2i - 1)/(1 + 1)
= 2i / 2 = i
(iii) Simplify: x^3/3 + x^3.
x^3/3 + x^3 = x^3/3 + 3x^3/3 = 4x^3/3
(iv) Find the value of a if log5 670 = a.
a = log(670)/log(5) ≈ 2.8261 / 0.69897 ≈ 4.04
(v) Solve: √i.
√i = ±(1/√2 + i/√2) = ±(√2/2 + i√2/2)
(vi) Reduce the rational expression (243x5y10)/215 to its lowest terms.
(243x^5y^10)/215 (Already in simplest form)
(vii) Factorize: 5x3 – 20x.
5x^3 - 20x = 5x(x^2 - 4) = 5x(x - 2)(x + 2)
(viii) Factorize: 1 – 125x3.
1 - 125x^3 = (1)^3 - (5x)^3 = (1 - 5x)(1 + 5x + 25x^2)
Q.3 Write short answers to any SIX (6) questions:
(i) Find L.C.M. of 102xyz, 85x2yz, 187xyz2.
(ii) Define non-strict inequality.
Non-strict inequalities include ≤ (less than or equal to) and ≥ (greater than or equal to).
(iii) Solve for x: 3x – 5 = 4.
3x - 5 = 4 → 3x = 9 → x = 3
(iv) Find the value of m and c by expressing 2y – 4x = 2 in the form y = mx + c.
2y = 4x + 2 → y = 2x + 1 (m = 2, c = 1)
(v) Draw the graph of y = 3x – 5.
Graph is a straight line with slope 3 and y-intercept -5.
(vi) Define an isosceles triangle.
An isosceles triangle has at least two sides of equal length.
(vii) Find the distance between points A(3, -5) and B(2, -4).
Distance = √((2 - 3)^2 + (-4 + 5)^2) = √(1 + 1) = √2
9th class Mathematics solved Past Paper 2024 Group 1 Lahore Board by FG STudy.com
Question 3 (viii): Find the value of unknowns for the given congruent triangles
To solve for unknowns in congruent triangles, we use the ASA congruence rule.
Solution:
-
- Given: ∠B = ∠C = 55°, and ∠A = (5x + 5)°.
- The sum of angles in a triangle equals 180°.
\( (5x + 5) + 55 + 55 = 180 \)
\( 5x + 115 = 180 \)
\( 5x = 65 \quad \Rightarrow \quad x = 13 \)
The value of \( x \) is: 13.
Question 3 (ix): Find the unknowns in the given figure
In this figure, vertical opposite angles and linear pairs are used to determine the unknowns.
Solution:
-
- The angle opposite \( 75^\circ \) is also \( 75^\circ \) (Vertical Angles).
- Using the Linear Pair Property:
\( x + 75 = 180 \)
\( x = 105^\circ \)
- Since vertical opposite angles are equal, \( y \) is also \( 105^\circ \).
The values are:
- \( x = 105^\circ \)
- \( y = 105^\circ \)
9th class Mathematics solved Past Paper 2024 Group 1 Lahore Board by FG Study.com
Question 4 (vii): Find the area of the given figure
The given figure is a rectangle with length \( l = 6 \, \text{cm} \) and height \( h = 2 \, \text{cm} \).
The formula for area of a rectangle is:
\( \text{Area} = \text{length} \times \text{height} \)
Substitute the values:
\( \text{Area} = 6 \times 2 = 12 \, \text{cm}^2 \)
Answer: \( 12 \, \text{cm}^2 \)
Question 5 (a): Solve by Matrix Inversion
Given the system of equations:
- \( -4x – y = -9 \)
- \( 3x + y = 5 \)
The equations in matrix form are:
\[
A = \begin{bmatrix} -4 & -1 \\ 3 & 1 \end{bmatrix}, \, X = \begin{bmatrix} x \\ y \end{bmatrix}, \, B = \begin{bmatrix} -9 \\ 5 \end{bmatrix}
\]
Using the matrix inversion formula \( X = A^{-1} B \), find \( A^{-1} \), then calculate \( X \).
The solution is: \( x = 1, \, y = 2 \).
Answer: \( x = 1, \, y = 2 \)
Question 6 (a): Solve using log table
Find the value of:
\( \frac{(438)^3 \times \sqrt{0.056}}{(388)^4} \)
Using logarithmic properties:
- Find log of \( 438 \), \( 388 \), and \( 0.056 \) using log tables.
- Apply the formula for logs of powers and roots.
- Subtract and simplify the values.
Final Answer: Approximately 0.014
9th class Mathematics solved Past Paper 2024 Group 1 Lahore Board by FG Study.com
Question 7 (a): Factorize by factor theorem
Given expression:
\( x^3 – 2x^2 – x + 2 \)
Using factor theorem, check for roots:
Substitute \( x = 1 \):
\( 1^3 – 2(1)^2 – 1 + 2 = 0 \)
Thus, \( x – 1 \) is a factor. Perform polynomial division:
\( x^3 – 2x^2 – x + 2 = (x – 1)(x^2 – x – 2) \)
Factorize \( x^2 – x – 2 \):
\( (x – 1)(x – 2)(x + 1) \)
Answer: \( (x – 1)(x – 2)(x + 1) \)
Question 8 (a): Solve the equation
Given:
\( \frac{2}{3x + 6} = \frac{1}{6} + \frac{1}{2x + 4} \)
Combine terms and simplify to form a linear equation. Solve for \( x \).
Final Answer: \( x = 2 \).
9th class Mathematics solved Past Paper 2024 Group 1 Lahore Board by FG STudy.com
Solution for Question 9 (Option 1):
Theorem: Any point equidistant from the endpoints of a line segment lies on its perpendicular bisector.
Proof:
- Let AB be a line segment with endpoints \( A(x_1, y_1) \) and \( B(x_2, y_2) \).
- Assume a point \( P(x, y) \) is equidistant from \( A \) and \( B \): \( PA = PB \).
- Using the distance formula:
\( PA = \sqrt{(x – x_1)^2 + (y – y_1)^2} \) and \( PB = \sqrt{(x – x_2)^2 + (y – y_2)^2} \). - Given \( PA = PB \), squaring both sides:
\( (x – x_1)^2 + (y – y_1)^2 = (x – x_2)^2 + (y – y_2)^2 \). - After simplification, it shows that \( P \) lies on the perpendicular bisector of \( AB \).
Conclusion: Any point equidistant from the endpoints of a line segment lies on its perpendicular bisector.
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Solution for Question 9 (Option 2):
Theorem: Triangles on the same base and between the same altitudes are equal in area.
Proof:
-
- Let \( \triangle ABC \) and \( \triangle ABD \) share the same base \( AB \) and lie between the same height (altitude).
- The area of a triangle is given by:
\( \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} \). - For both triangles, the base \( AB \) and height are the same.
- Thus, their areas are equal:
Area of \( \triangle ABC \) = Area of \( \triangle ABD \)
Conclusion: Triangles on the same base and between the same altitudes are equal in area.
