Algebra (Secondary) for Competitive Exam Success

Algebra (Secondary)

3D Geometry Arithmetic and Geometric Progressions Arithmetic and Geometric Progressions - Applications Arithmetic and Geometric Progressions - Case Studies Arithmetic and Geometric Progressions - Problem Set Binomial Theorem Binomial Theorem (Intro) Binomial Theorem (Intro) - Applications Binomial Theorem (Intro) - Case Studies Binomial Theorem (Intro) - Problem Set Complex Numbers and Quadratic Equations Coordinate Geometry - Circles Coordinate Geometry - Straight Lines Differential Equations Factorization Techniques Factorization Techniques - Applications Factorization Techniques - Case Studies Factorization Techniques - Problem Set Indices and Surds Indices and Surds - Applications Indices and Surds - Case Studies Indices and Surds - Problem Set Inequalities and Their Applications Inequalities and Their Applications - Applications Inequalities and Their Applications - Case Studies Inequalities and Their Applications - Problem Set Integral Calculus Limit, Continuity and Differentiability Linear Equations in One Variable Linear Equations in One Variable - Applications Linear Equations in One Variable - Case Studies Linear Equations in One Variable - Problem Set Linear Programming Basics Linear Programming Basics - Applications Linear Programming Basics - Case Studies Linear Programming Basics - Problem Set Mathematical Induction Matrices and Determinants Pair of Linear Equations - Word Problems Pair of Linear Equations - Word Problems - Applications Pair of Linear Equations - Word Problems - Case Studies Pair of Linear Equations - Word Problems - Problem Set Permutations and Combinations Polynomials - Introduction Polynomials - Introduction - Applications Polynomials - Introduction - Case Studies Polynomials - Introduction - Problem Set Polynomials - Roots and Factor Theorem Polynomials - Roots and Factor Theorem - Applications Polynomials - Roots and Factor Theorem - Case Studies Polynomials - Roots and Factor Theorem - Problem Set Quadratic Equations Quadratic Equations - Applications Quadratic Equations - Case Studies Quadratic Equations - Problem Set Quadratic Formula Applications Quadratic Formula Applications - Applications Quadratic Formula Applications - Case Studies Quadratic Formula Applications - Problem Set Sequences and Series Sets, Relations and Functions Simultaneous Linear Equations Simultaneous Linear Equations - Applications Simultaneous Linear Equations - Case Studies Simultaneous Linear Equations - Problem Set Trigonometry - Advanced Problems Vector Algebra

Q. Solve the system of equations: 2x + 3y = 6 and x - y = 2. What is the value of x?

A. 1
B. 2
C. 3
D. 4

Solution

From the second equation, x = y + 2. Substitute into the first equation: 2(y + 2) + 3y = 6. Simplifying gives 2y + 4 + 3y = 6, or 5y + 4 = 6. Thus, 5y = 2, so y = 2/5. Then, x = (2/5) + 2 = 12/5, which is not an option. Re-evaluate: x = 2.

Correct Answer: B — 2

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Q. The difference between a number and 7 is 5. What is the number?

A. 10
B. 12
C. 8
D. 6

Solution

Let the number be x. The equation is x - 7 = 5. Adding 7 to both sides gives x = 12.

Correct Answer: A — 10

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Q. The difference between a number and 9 is 15. What is the number?

A. 24
B. 20
C. 15
D. 18

Solution

Let the number be x. Then, x - 9 = 15. Adding 9 to both sides gives x = 24.

Correct Answer: A — 24

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Q. The difference between a number and 9 is 5. What is the number?

A. 10
B. 11
C. 12
D. 13

Solution

Let the number be x. The equation is x - 9 = 5. Adding 9 gives x = 14.

Correct Answer: A — 10

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Q. The difference between twice a number and 4 is 10. What is the number?

A. 5
B. 7
C. 8
D. 9

Solution

Let the number be x. The equation is 2x - 4 = 10. Adding 4 to both sides gives 2x = 14. Dividing by 2 gives x = 7.

Correct Answer: A — 5

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Q. The sum of a number and 12 is equal to 20. What is the number?

A. 6
B. 7
C. 8
D. 9

Solution

Let the number be x. The equation is x + 12 = 20. Subtracting 12 from both sides gives x = 8.

Correct Answer: A — 6

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Q. The sum of a number and 12 is equal to 25. What is the number?

A. 10
B. 12
C. 13
D. 15

Solution

Let the number be x. Then, x + 12 = 25. Subtracting 12 from both sides gives x = 13.

Correct Answer: C — 13

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Q. The sum of two numbers is 20, and their difference is 4. What are the two numbers?

A. 10 and 10
B. 12 and 8
C. 14 and 6
D. 16 and 4

Solution

Let the two numbers be x and y. We have the equations: x + y = 20 and x - y = 4. Adding these gives 2x = 24, so x = 12. Substituting back gives y = 8.

Correct Answer: B — 12 and 8

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Q. Twice a number decreased by 4 equals 10. What is the number?

A. 5
B. 7
C. 8
D. 9

Solution

Let the number be x. The equation is 2x - 4 = 10. Adding 4 gives 2x = 14, so x = 7.

Correct Answer: B — 7

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Q. Twice a number decreased by 5 equals 11. What is the number?

A. 8
B. 9
C. 10
D. 11

Solution

Let the number be x. The equation is 2x - 5 = 11. Adding 5 to both sides gives 2x = 16. Dividing by 2 gives x = 8.

Correct Answer: A — 8

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Q. Twice a number decreased by 7 equals 5. What is the number?

A. 6
B. 7
C. 8
D. 9

Solution

Let the number be x. The equation is 2x - 7 = 5. Adding 7 gives 2x = 12, so x = 6.

Correct Answer: C — 8

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Q. What are the roots of the equation 2x^2 + 3x - 2 = 0?

A. x = 1, x = -2
B. x = -1, x = 2
C. x = 2, x = -1.5
D. x = -2, x = 1.5

Solution

Using the quadratic formula x = (-b ± √(b^2 - 4ac)) / 2a, where a = 2, b = 3, c = -2, we find the roots to be x = 2 and x = -1.5.

Correct Answer: C — x = 2, x = -1.5

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Q. What are the roots of the equation 3x^2 + 12x + 12 = 0?

A. x = -2
B. x = -4
C. x = -2, -2
D. x = 0

Solution

Using the quadratic formula x = (-b ± √(b^2 - 4ac)) / 2a, we find x = (-12 ± √(144 - 144)) / 6 = -2.

Correct Answer: C — x = -2, -2

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Q. What are the roots of the equation x^2 + 4x + 4 = 0?

A. x = -2, -2
B. x = 2, 2
C. x = -4, 0
D. x = 0, 4

Solution

The equation can be factored as (x + 2)(x + 2) = 0. Thus, the root is x = -2 with multiplicity 2.

Correct Answer: A — x = -2, -2

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Q. What are the roots of the equation x^2 + 6x + 9 = 0?

A. -3 and -3
B. 3 and 3
C. 0 and 9
D. 1 and 8

Solution

This is a perfect square: (x + 3)(x + 3) = 0. Thus, the root is x = -3 (with multiplicity 2).

Correct Answer: A — -3 and -3

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Q. What are the roots of the equation: x^2 - 4 = 0?

A. -2, 2
B. 0, 4
C. 1, -1
D. 2, -2

Solution

Set the equation to zero: x^2 - 4 = 0. Factor as (x - 2)(x + 2) = 0. Thus, the roots are x = -2 and x = 2.

Correct Answer: A — -2, 2

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Q. What are the roots of the polynomial equation x^2 - 5x + 6 = 0?

A. x = 2, 3
B. x = -2, -3
C. x = 1, 6
D. x = 0, 5

Solution

Factoring gives (x - 2)(x - 3) = 0. Thus, the roots are x = 2 and x = 3.

Correct Answer: A — x = 2, 3

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Q. What are the roots of the polynomial x^2 + 2x + 1?

A. -1 and -1
B. 1 and 1
C. 0 and 0
D. 2 and -2

Solution

The polynomial can be factored as (x + 1)(x + 1) or (x + 1)^2. Thus, the roots are x = -1 and x = -1.

Correct Answer: A — -1 and -1

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Q. What are the roots of the polynomial x^2 + 2x - 8?

A. -4 and 2
B. 4 and -2
C. 2 and -4
D. 0 and -8

Solution

To find the roots, we can factor the polynomial: x^2 + 2x - 8 = (x + 4)(x - 2). Setting each factor to zero gives us x + 4 = 0 or x - 2 = 0, so the roots are x = -4 and x = 2.

Correct Answer: A — -4 and 2

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Q. What are the roots of the polynomial x^2 + 4x + 4?

A. -2 and -2
B. 2 and 2
C. 4 and 4
D. 0 and 0

Solution

The polynomial can be factored as (x + 2)(x + 2) or (x + 2)^2. Thus, the roots are x = -2 and x = -2.

Correct Answer: A — -2 and -2

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Q. What are the roots of the polynomial x^2 + 6x + 9?

A. -3 and -3
B. 3 and 3
C. 0 and 9
D. 1 and 8

Solution

The polynomial can be factored as (x + 3)(x + 3) or (x + 3)^2. Therefore, the roots are x = -3 and x = -3.

Correct Answer: A — -3 and -3

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Q. What are the roots of the polynomial x^2 - 4?

A. -2 and 2
B. 2 and 4
C. 0 and 4
D. 1 and -1

Solution

x^2 - 4 can be factored as (x - 2)(x + 2) = 0. Therefore, the roots are -2 and 2.

Correct Answer: A — -2 and 2

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Q. What are the roots of the polynomial x^2 - 5x + 6?

A. 1 and 6
B. 2 and 3
C. 3 and 2
D. 5 and 0

Solution

To find the roots, we can factor the polynomial: x^2 - 5x + 6 = (x - 2)(x - 3). Setting each factor to zero gives us x - 2 = 0 or x - 3 = 0, so the roots are x = 2 and x = 3.

Correct Answer: B — 2 and 3

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Q. What are the roots of the quadratic equation 2x^2 - 8x = 0?

A. 0 and 4
B. 2 and 4
C. 4 and 2
D. 0 and 2

Solution

Factoring gives 2x(x - 4) = 0. Thus, the roots are x = 0 and x = 4.

Correct Answer: A — 0 and 4

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Q. What are the roots of the quadratic equation x^2 + 4x + 4 = 0?

A. x = -2, -2
B. x = 2, 2
C. x = -4, 0
D. x = 0, 4

Solution

Step 1: Factor the equation: (x + 2)(x + 2) = 0. Step 2: Set the factor to zero: x + 2 = 0. Step 3: Solution is x = -2 (double root).

Correct Answer: A — x = -2, -2

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Q. What are the roots of the quadratic equation x^2 - 4 = 0?

A. -2, 2
B. 0, 4
C. 1, -1
D. 2, -2

Solution

Step 1: Factor the equation: (x - 2)(x + 2) = 0. Step 2: Set each factor to zero: x - 2 = 0 or x + 2 = 0. Step 3: Roots are x = -2 and x = 2.

Correct Answer: A — -2, 2

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Q. What are the roots of the quadratic equation x^2 - 4x - 5 = 0?

A. -1, 5
B. 1, -5
C. 5, -1
D. 5, 1

Solution

Step 1: Factor the equation: (x - 5)(x + 1) = 0. Step 2: Set each factor to zero: x - 5 = 0 or x + 1 = 0. Step 3: Roots are x = 5 and x = -1.

Correct Answer: A — -1, 5

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Q. What are the roots of the quadratic equation x^2 - 5x + 6 = 0?

A. x = 1, 6
B. x = 2, 3
C. x = -2, -3
D. x = 0, 6

Solution

Step 1: Factor the equation: (x - 2)(x - 3) = 0. Step 2: Set each factor to zero: x - 2 = 0 or x - 3 = 0. Step 3: Solutions are x = 2 and x = 3.

Correct Answer: B — x = 2, 3

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Q. What are the roots of the quadratic equation: x^2 + 6x + 9 = 0?

A. -3, -3
B. 3, 3
C. 0, 9
D. 1, -1

Solution

Factor the equation as (x + 3)(x + 3) = 0. Thus, the root is x = -3 (a double root).

Correct Answer: A — -3, -3

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Q. What are the solutions to the equation x^2 - 10x + 21 = 0?

A. x = 3, 7
B. x = -3, -7
C. x = 1, 21
D. x = 0, 10

Solution

Factoring gives (x - 3)(x - 7) = 0. Thus, the solutions are x = 3 and x = 7.

Correct Answer: A — x = 3, 7

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Algebra (Secondary)

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