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Pre-board Exam 2024-
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Solved Model Paper
Tips, Test Paper, and Guess Paper for Preparation
Mathematics SSC II
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Prepare effectively for your Mathematics SSC II Pre-board Exam 2024 with our comprehensive Solved Model Paper. This resource offers meticulously crafted solutions to a wide range of questions typically encountered in the exam, providing you with invaluable practice and reinforcement of key concepts.
In addition to the Solved Model Paper, our resource includes essential tips to enhance your exam preparation, test papers for further practice, and guess papers to assess your readiness for the exam. These resources are designed to support you in understanding the exam pattern, managing your time efficiently, and mastering the concepts necessary for success.
Utilize our Solved Model Paper and supplementary resources to boost your confidence, sharpen your problem-solving skills, and approach the Mathematics SSC II Pre-board Exam 2024 with assurance and readiness. With our comprehensive support, you can strive for excellence and achieve your academic goals.
Prepare effectively for your Mathematics SSC II Pre-board Exam 2024 with our comprehensive Solved Model Paper. This resource offers meticulously crafted solutions to a wide range of questions typically encountered in the exam, providing you with invaluable practice and reinforcement of key concepts.
In addition to the Solved Model Paper, our resource includes essential tips to enhance your exam preparation, test papers for further practice, and guess papers to assess your readiness for the exam. These resources are designed to support you in understanding the exam pattern, managing your time efficiently, and mastering the concepts necessary for success.
Utilize our Solved Model Paper and supplementary resources to boost your confidence, sharpen your problem-solving skills, and approach the Mathematics SSC II Pre-board Exam 2024 with assurance and readiness. With our comprehensive support, you can strive for excellence and achieve your academic goals.
Mathematics FBISE Class 10
Section - A
(1) Which of the following typesrepresents (x − 3)(x + 3) = 0 ?
A. Quadratic equation B. Linear equation
C. Cubic equation D. Pure quadratic equation
(1) The equation (x − 3)(x + 3) = 0 represents a quadratic equation. Therefore, the answer is A. Quadratic equation.
(2) For what value of k, 2x2 + kx + 3 = 0 has equal roots?
A. 2√6 B. 24
C. ±2√6 D. ±6√2
(2) For a quadratic equation to have equal roots, the discriminant must be zero. The given equation is 2x^2 + kx + 3 = 0. The discriminant is given by b^2 - 4ac. In this case, b = k, a = 2, and c = 3. So we have k^2 - 4(2)(3) = 0. Simplifying this equation gives k^2 - 24 = 0. Solving for k, we get k = ±2√6. Therefore, the answer is C. ±2√6.
(3) If z ∝ (w + 3) and w = 3, z = 6. What is value of 2z when w = 9 ?
A. 12 B. 24
C. 6 D. 4.5
(3) The given relationship is z ∝ (w + 3), which means z is directly proportional to (w + 3). When w = 3, z = 6. To find the value of 2z when w = 9, we can substitute w = 9 into the proportional relationship and solve for z. We get z ∝ (9 + 3), which simplifies to z ∝ 12. Therefore, z = k * 12, where k is a constant. Since z = 6 when w = 3, we can substitute these values into the equation to solve for k. We have 6 = k * 12, which gives k = 1/2. Now we can find the value of 2z when w = 9 by substituting w = 9 into the proportional relationship. We get 2z = 2 * (9 + 3) = 2 * 12 = 24. Therefore, the answer is B. 24.
(4) If α and β are the roots of 2x2 − 6x − 4 = 0. What is value of α2β3 + α3β2 ?
A. −12 B. 12
C. 6 D. −6
(4) The given quadratic equation is 2x^2 − 6x − 4 = 0. Let α and β be the roots of this equation. We need to find the value of α^2β^3 + α^3β^2. Using Vieta's formulas, we know that α + β = -b/a and αβ = c/a. In this case, a = 2, b = -6, and c = -4. So we have α + β = 6/2 = 3 and αβ = -4/2 = -2. Now we can calculate the desired value. We have α^2β^3 + α^3β^2 = α^2β^2(β + α) = (αβ)^2(β + α). Substituting the known values, we get (-2)^2(3) = 4 * 3 = 12. Therefore, the answer is B. 12.
(5) To find the partial fractions of x^3 / (x^3 + 1), we can factor the denominator as (x + 1)(x^2 - x + 1). Therefore, the partial fraction decomposition will have the form:
x^3 / (x^3 + 1) = A / (x + 1) + (Bx + C) / (x^2 - x + 1).
To find A, B, and C, we can multiply both sides of this equation by the denominator (x^3 + 1) and then equate the numerators. After simplification, we get:
x^3 = A(x^2 - x + 1) + (Bx + C)(x + 1).
Expanding and collecting like terms, we get:
x^3 = (A + B)x^2 + (-A + B + C)x + (A + C).
By comparing the coefficients of the powers of x, we can solve for A, B, and C. Equating the coefficients, we have:
A + B = 0,
-A + B + C = 0,
A + C = 1.
Solving these equations simultaneously yields A = 1/3, B = -1/3, and C = 1/3. Therefore, the answer is D. 1 + (1/3)x - (1/3)/(x^2 - x + 1).
(6) If x = 7, ∑f = 30 and ∑fx = 120 + 3k then value of k is
A. 30 B. −30
C. −11 D. 11
C. -11 (Given ∑f = 30 and ∑fx = 120 + 3k. We know ∑f represents the sum of the function's values for all elements in the set. Since x = 7 is the only element, ∑f = f(7) = 30 and ∑fx = f(7) * 7 = 120 + 3k. Solving for k, we get k = -11.)
C. 4/3 (Use the trigonometric identity tan^2 θ + 1 = sec^2 θ. We are given sinθ = 4/5 and secθ = 5/3. Substituting, we get tan^2 θ = (5/3)^2 - 1 = 16/9, so tan θ = ± 4/3. Since sinθ is positive in Quadrant I, tan θ = 4/3.)
(8) What is the elevation of Sun if a pole of 6m high casts a shadow of 2√3m ?
A. 30° B. 45°
C. 60° D. 90°
C. 60° (Use the tangent function relationship: tan θ = opposite side / adjacent side. Here, opposite side = pole height = 6m and adjacent side = shadow length = 2√3m. Solving for θ, we get tan θ = √3. The inverse tangent function (arctan) gives θ ≈ 60°.)(9)- 3π/4 Radians=
A. 115° B. 135° C. 150° D. 30°
A. 135° (-3π/4 radians is equivalent to -135° because 180° = π radians.)(10)- The system of measurement in which the angle is measured in radians is called:
A. CGS System B. Sexagesimal system C. MKS System D. Circular system
D. Circular system(Radians are used to measure angles in a circular system.)(11)- A collection of well-defined objects is called:
A. Subset B. Power set C. Set D. none
C. Set (A set is a collection of well-defined objects.)(12)- The different number of ways to describe a set are:
A. 1 B. 2 C. 3 D. 4
D. 4 (A set can be described in various ways, like listing elements, using set-builder notation, or using Venn diagrams.)(13)- A set with no elements is called:
A. subset B. Empty set C. singleton set D. Superset
B. Empty set (An empty set has no elements.)(14)- The set having only one element is called:
A. Subset B. Empty Set C. Singleton set D. Superset
C. Singleton set (A set with only one element is called a singleton set.)(15)- A histogram is a set of adjacent:
A. Square B. Rectangles C. Circles D. Triangle
B. Rectangles (A histogram is a graphical representation of data using bars or rectangles.)Boost Your Brainpower
Proven Strategies for Top Scores on FBISE Class 10 Math MCQs
SECTION – B
Q.2 Attempt ALL parts.
Section B (continued):
ii. Roots of 2θ and 2φ
Since θ and φ are the roots of y^2 − 7y + 9 = 0, we know:
- θ + φ = 7 (Sum of roots)
- θφ = 9 (Product of roots)
To find the equation whose roots are 2θ and 2φ, substitute x = 2θ: (2θ)^2 - 7(2θ) + 9 = 0 4θ^2 - 14θ + 9 = 0
Therefore, the equation whose roots are 2θ and 2φ is: 4x^2 - 14x + 9 = 0
iii. Direct Proportion (P and Q)
We know P is directly proportional to Q, which can be written as P = kQ for some constant k. We are given P = 12 when Q = 4. Substitute these values to find k: 12 = k * 4 k = 3
Therefore, the equation connecting P and Q is P = 3Q.
When Q = 8: P = 3 * 8 P = 24
iv. De Morgan's Laws (Set Theory)
We need to verify that (A ∩ B)' = A' U B'.
A = {2, 4, 6} and B = {1, 3, 5} A' = {1, 3, 5, 7, 8, 9, 10} (Complement of A) B' = {2, 4, 6, 7, 8, 9, 10} (Complement of B)
(A ∩ B)' = the complement of the intersection of A and B. Since A and B have no elements in common (their intersection is empty), its complement will contain all elements in U (universal set). In this case, U = {1, 2, 3, ..., 10}. So, (A ∩ B)' = {1, 2, 3, ..., 10}.
Now, let's check A' U B': A' U B' = {1, 3, 5, 7, 8, 9, 10} U {2, 4, 6, 7, 8, 9, 10}
Since sets contain unique elements, combining A' and B' removes duplicates, resulting in the same set as (A ∩ B)'. Therefore, De Morgan's Law (A ∩ B)' = A' U B' is verified.
v. Descriptive Statistics (Mean, Median, Mode)
The data is: 4, 1, 2, 1, 0, 0, 3, 2, 3, 3
Mean: Add all values and divide by the number of values (n = 10). Mean = (4 + 1 + 2 + 1 + 0 + 0 + 3 + 2 + 3 + 3) / 10 = 19 / 10 = 1.9
Median: Order the data: 0, 0, 1, 1, 2, 2, 3, 3, 3, 4 Since we have an even number of values, the median is the average of the two middle values: Median = (2 + 3) / 2 = 2.5
Mode: The most frequent value is 3.
vi. Trigonometric Identities (tan, sec, cosec)
We are given tan θ = 4/3 and sin θ < 0 (which means θ is in Quadrant III).
sec θ: Use the identity sec^2 θ = 1 + tan^2 θ. sec^2 θ = 1 + (4/3)^2 = 25/9 sec θ = ±√(25/9) Since θ is in Quadrant III, sec θ is negative. sec θ = -√(25/9) = -5/3
cosec θ: Use the identity cosec^2 θ = 1/sin^2 θ. Since we don't have the exact value of sin θ, we can't calculate cosec θ directly.
Proving 1 + cot^2 θ = cosec^2 θ:
cot θ = 1/tan θ = 3/4 cot^2 θ = (3/
FAQs (Frequently Asked Questions):
What does the SSC II Pre-board Exam 2024 Solved Model Paper include?
The Solved Model Paper for the Mathematics SSC II Pre-board Exam 2024 encompasses comprehensive solutions to questions typically encountered in the exam. It serves as a valuable resource for students to practice and prepare effectively.
How can the Solved Model Paper aid in preparation for the SSC II Mathematics Pre-board Exam?
The Solved Model Paper offers a structured approach to revising key concepts, practicing numerical problems, and familiarizing oneself with the exam pattern. By studying the solutions provided, students can enhance their understanding and boost their confidence before the actual exam.
Are there any additional resources provided apart from the Solved Model Paper?
Yes, apart from the Solved Model Paper, additional resources such as tips for exam preparation, test papers for further practice, and guess papers to assess one's readiness are also included. These resources aim to provide comprehensive support to students as they gear up for the examination.
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The tips provided are carefully curated to offer effective strategies and techniques for efficient exam preparation. Students are encouraged to incorporate these tips into their study routine, manage their time effectively, focus on conceptual understanding, and practice regularly to achieve optimal results in the exam.
Are the test papers included in the resource aligned with the exam syllabus?
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What is the significance of the guess paper in the preparation process?
The guess paper offers students an opportunity to gauge their preparedness and identify areas of strength and improvement before the exam. By attempting the questions in the guess paper, students can assess their knowledge, identify common question patterns, and refine their exam-taking strategies.
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