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. A bakery sells cakes and cookies. If the total number of items is 150 and the number of cakes is twice the number of cookies, how many cakes are there?

A. 100
B. 50
C. 75
D. 25

Solution

Let the number of cookies be x. Then the number of cakes is 2x. The equation is x + 2x = 150. Solving gives 3x = 150, so x = 50. Therefore, cakes = 2x = 100.

Correct Answer: A — 100

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Q. A bookstore sells novels and magazines. If the total number of books is 200 and the number of novels is 3 times the number of magazines, how many novels are there?

A. 150
B. 120
C. 75
D. 100

Solution

Let the number of magazines be x. Then the number of novels is 3x. The equation is x + 3x = 200. Solving gives 4x = 200, so x = 50. Therefore, novels = 3x = 150.

Correct Answer: A — 150

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Q. A car rental company charges a flat fee of $50 plus $0.20 per mile driven. If a customer paid $70, how many miles did they drive?

A. 100 miles
B. 150 miles
C. 200 miles
D. 250 miles

Solution

Let the number of miles driven be m. The total cost is given by: 50 + 0.20m = 70. Solving for m gives 0.20m = 20, so m = 100 miles.

Correct Answer: A — 100 miles

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Q. A company produces pens and pencils. If the total production is 500 items and the number of pens is 4 times the number of pencils, how many pens are produced?

A. 400
B. 100
C. 200
D. 300

Solution

Let the number of pencils be x. Then the number of pens is 4x. The equation is x + 4x = 500. Solving gives 5x = 500, so x = 100. Therefore, pens = 4x = 400.

Correct Answer: A — 400

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Q. A concert hall has 300 seats. If the number of reserved seats is 50 more than the number of general admission seats, how many reserved seats are there?

A. 125
B. 175
C. 100
D. 150

Solution

Let the number of general admission seats be x. Then the number of reserved seats is x + 50. The equation is x + (x + 50) = 300. Solving gives 2x + 50 = 300, so 2x = 250, x = 125. Therefore, reserved seats = x + 50 = 175.

Correct Answer: B — 175

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Q. A farmer has chickens and cows. If there are 20 heads and 56 legs in total, how many cows are there?

A. 8
B. 10
C. 12
D. 14

Solution

Let the number of chickens be c and the number of cows be k. We have the equations: c + k = 20 and 2c + 4k = 56. Solving these gives k = 10.

Correct Answer: B — 10

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Q. A farmer has chickens and cows. If there are 50 animals in total and the number of cows is 10 more than the number of chickens, how many cows are there?

A. 20
B. 30
C. 25
D. 15

Solution

Let the number of chickens be x. Then the number of cows is x + 10. The equation is x + (x + 10) = 50. Solving gives 2x + 10 = 50, so 2x = 40, x = 20. Therefore, cows = x + 10 = 30.

Correct Answer: A — 20

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Q. A fruit seller has apples and oranges. If the total number of fruits is 80 and the number of apples is 3 times the number of oranges, how many apples are there?

A. 60
B. 30
C. 40
D. 50

Solution

Let the number of oranges be x. Then the number of apples is 3x. The equation is x + 3x = 80. Solving gives 4x = 80, so x = 20. Therefore, apples = 3x = 60.

Correct Answer: A — 60

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Q. A number decreased by 7 equals 12. What is the number?

A. 5
B. 19
C. 7
D. 12

Solution

Let the number be x. Then x - 7 = 12. Add 7 to both sides: x = 19.

Correct Answer: B — 19

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Q. A number decreased by 7 is equal to 3. What is the number?

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

Solution

Let the number be x. The equation is x - 7 = 3. Adding 7 to both sides gives x = 10.

Correct Answer: A — 8

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Q. A number is decreased by 4 and then multiplied by 3. The result is 18. What is the number?

A. 8
B. 10
C. 6
D. 7

Solution

Let the number be x. Then, 3(x - 4) = 18. Dividing by 3 gives x - 4 = 6. Adding 4 gives x = 10.

Correct Answer: A — 8

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Q. A number is decreased by 9 and then multiplied by 2 to give 10. What is the number?

A. 14
B. 15
C. 16
D. 17

Solution

Let the number be x. The equation is 2(x - 9) = 10. Dividing by 2 gives x - 9 = 5, so x = 14.

Correct Answer: A — 14

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Q. A number is decreased by 9 and then multiplied by 4 to give 28. What is the number?

A. 5
B. 7
C. 9
D. 11

Solution

Let the number be x. The equation is 4(x - 9) = 28. Dividing by 4 gives x - 9 = 7, so x = 16.

Correct Answer: C — 9

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Q. A number is divided by 2 and then increased by 3. The result is 10. What is the number?

A. 10
B. 14
C. 12
D. 8

Solution

Let the number be x. Then, x/2 + 3 = 10. Subtracting 3 from both sides gives x/2 = 7. Multiplying by 2 gives x = 14.

Correct Answer: B — 14

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Q. A number is divided by 2 and then increased by 6 to equal 10. What is the number?

A. 4
B. 6
C. 8
D. 10

Solution

Let the number be x. The equation is (x / 2) + 6 = 10. Subtracting 6 from both sides gives x / 2 = 4. Multiplying by 2 gives x = 8.

Correct Answer: A — 4

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Q. A number is divided by 2 and then increased by 6 to get 10. What is the number?

A. 4
B. 6
C. 8
D. 10

Solution

Let the number be x. The equation is (x / 2) + 6 = 10. Subtracting 6 from both sides gives x / 2 = 4. Multiplying by 2 gives x = 8.

Correct Answer: C — 8

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Q. A number is divided by 2 and then increased by 7 to equal 15. What is the number?

A. 10
B. 12
C. 14
D. 16

Solution

Let the number be x. The equation is x/2 + 7 = 15. Subtracting 7 gives x/2 = 8, so x = 16.

Correct Answer: A — 10

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Q. A number is divided by 4 and then increased by 6 to equal 10. What is the number?

A. 8
B. 12
C. 16
D. 20

Solution

Let the number be x. The equation is x/4 + 6 = 10. Subtracting 6 from both sides gives x/4 = 4. Multiplying by 4 gives x = 16.

Correct Answer: A — 8

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Q. A number is increased by 15 and the result is 45. What is the number?

A. 30
B. 15
C. 45
D. 60

Solution

Let the number be x. Then x + 15 = 45. Subtract 15 from both sides: x = 30.

Correct Answer: A — 30

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Q. A number is increased by 25% and then decreased by 20%. If the final result is 96, what was the original number?

A. 80
B. 90
C. 100
D. 110

Solution

Let the original number be x. The equation is: x * 1.25 * 0.8 = 96. Simplifying gives x = 96 / 1 = 120, so the original number is 100.

Correct Answer: C — 100

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Q. A number is increased by 5 and then multiplied by 3 to give 24. What is the number?

A. 1
B. 3
C. 5
D. 7

Solution

Let the number be x. The equation is 3(x + 5) = 24. Simplifying gives 3x + 15 = 24. Subtract 15 from both sides: 3x = 9. Dividing by 3 gives x = 3.

Correct Answer: B — 3

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Q. A number is increased by 5 and then multiplied by 3. The result is 24. What is the number?

A. 1
B. 3
C. 5
D. 7

Solution

Let the number be x. The equation is 3(x + 5) = 24. Simplifying gives 3x + 15 = 24. Subtracting 15 from both sides gives 3x = 9. Dividing by 3 gives x = 3.

Correct Answer: B — 3

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Q. A number is multiplied by 3 and then decreased by 5. The result is 16. What is the number?

A. 7
B. 5
C. 10
D. 8

Solution

Let the number be x. Then, 3x - 5 = 16. Adding 5 to both sides gives 3x = 21. Dividing by 3 gives x = 7.

Correct Answer: A — 7

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Q. A number is multiplied by 4 and then decreased by 8 to equal 16. What is the number?

A. 4
B. 6
C. 8
D. 10

Solution

Let the number be x. The equation is 4x - 8 = 16. Adding 8 to both sides gives 4x = 24. Dividing by 4 gives x = 6.

Correct Answer: C — 8

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Q. A number is subtracted from 15 and the result is 9. What is the number?

A. 4
B. 5
C. 6
D. 7

Solution

Let the number be x. The equation is 15 - x = 9. Subtracting 15 from both sides gives -x = -6. Multiplying by -1 gives x = 6.

Correct Answer: B — 5

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Q. A number is tripled and then increased by 2 to equal 20. What is the number?

A. 5
B. 6
C. 7
D. 8

Solution

Let the number be x. The equation is 3x + 2 = 20. Subtracting 2 gives 3x = 18, so x = 6.

Correct Answer: B — 6

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Q. A rectangle has a length that is 3 times its width. If the perimeter of the rectangle is 48 cm, what is the width?

A. 4 cm
B. 6 cm
C. 8 cm
D. 10 cm

Solution

Let the width be w cm. Then the length is 3w cm. The perimeter is given by: 2(length + width) = 48. Thus, 2(3w + w) = 48. Simplifying gives 8w = 48, so w = 6 cm.

Correct Answer: B — 6 cm

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Q. A rectangle has a length that is twice its width. If the width is w, what is the area of the rectangle?

A. 2w^2
B. w^2
C. 2w
D. w

Solution

Area = Length × Width. Length = 2w, so Area = 2w × w = 2w^2.

Correct Answer: A — 2w^2

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Q. A rectangle's length is 3 times its width. If the perimeter is 48 cm, what is the width?

A. 4 cm
B. 6 cm
C. 8 cm
D. 10 cm

Solution

Let the width be x. Then, length = 3x. The perimeter is 2(length + width) = 48. So, 2(3x + x) = 48. Simplifying gives 8x = 48, thus x = 6 cm.

Correct Answer: C — 8 cm

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Q. A rectangle's length is 3 times its width. If the perimeter is 48, what is the width?

A. 4
B. 6
C. 8
D. 10

Solution

Let the width be w. Then the length is 3w. The perimeter is 2(w + 3w) = 48. Simplifying gives 8w = 48. Dividing by 8 gives w = 6.

Correct Answer: B — 6

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Algebra (Secondary)

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