Quantitative Aptitude

Algebra

3,715 Questions
All Polynomial Roots and Coefficients 2451 Zero-Product Root Finding 667 Linear System Manipulation 162 Polynomial Theorems and Reduction 85 Algebraic Identity Simplification 82 Surds and Indices 70 Modulus and Absolute Value 54 Algebraic Expression Substitution 46 Algebraic Fraction Simplification 33 Logarithmic Equations 25 Repeated Series Sum 12 Linear Equation Word Problems 9

Algebra Questions

Multiple choice

In each of the following questions, two equations are given. You have to solve them and give answer- $I. (a(2 - a) = -8) II. (b(2 - b) = -15)$

  1. If a > b

  2. If a ≥ b

  3. If a < b

  4. If a ≤ b

  5. If a = b or relationship can not be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Solve both quadratic equations: For I, a(2-a)=-8 gives a²-2a-8=0, so a=4 or a=-2. For II, b(2-b)=-15 gives b²-2b-15=0, so b=5 or b=-3. When a=4, it could be > b=-3 but < b=5. When a=-2, it's always less than both b values. Since the relationship changes depending on which roots you pick, no single inequality holds for all possible solutions.

Multiple choice

In each of the following questions, two equations are given. You have to solve them and- ( I. x^2 = 3136 )( II. y^2 = 2916 )

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

  5. If x = y or relationship can not be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

x^2 = 3136 gives x = ±56. y^2 = 2916 gives y = ±54. When x = -56 and y = 54, we have x < y. When x = 56 and y = -54, we have x > y. When x = 56 and y = 54, x > y. When x = -56 and y = -54, x < y. Since the relationship changes based on which roots we choose, it cannot be established uniquely.

Multiple choice

In each of the following questions, two equations are given. You have to solve them and $I. (x^2 - 5x + 6 = 0) II. (y^2 - 9y + 14 = 0)$ Give answer

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

  5. If x = y or relationship can not be established
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Solve both quadratic equations. For x^2 - 5x + 6 = 0, factors are (x-2)(x-3)=0, so x = 2 or 3. For y^2 - 9y + 14 = 0, factors are (y-2)(y-7)=0, so y = 2 or 7. Since x can be 2 or 3 and y can be 2 or 7, x ≤ y is always true (when x=2, y=2 or 7; when x=3, y=7).

Multiple choice

$I. (x^2 = 6+x) (II. (y^2 - 2y = 8)$ Solve the given equations and find the relationship between x and y.

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

  5. If x = y or relationship cannot be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Solve equation I: x^2 - x - 6 = 0 gives x = (1 ± √25)/2 = (1 ± 5)/2, so x = 3 or x = -2. Solve equation II: y^2 - 2y - 8 = 0 gives y = (2 ± √36)/2 = (2 ± 6)/2, so y = 4 or y = -2. Comparing values: when x = 3 and y = 4, x < y; when x = 3 and y = -2, x > y; when x = -2 and y = 4, x < y; when x = -2 and y = -2, x = y. Since the relationship varies, option E is correct.

Multiple choice

In each of the following questions, two equations I and II are given. You have to solve them and $I. ( x = (729)^{1/3} ) II. ( y^2 = 81 )$ Give answer

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

  5. If x = y or relationship can not be established
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Solving equation I: x = (729)^(1/3) = 9 (since 9³ = 729). Solving equation II: y² = 81, so y = ±9. Therefore x = 9 and y can be 9 or -9. This means x ≥ y (x is greater than or equal to y, since 9 ≥ 9 and 9 ≥ -9).

Multiple choice

In each of the following questions, two equations numbered I and II are given. You have to solve them and $( (I) \quad 14x^2 – 5x – 1 = 0 \\ (II) \quad 2y^2 + 3y + 1 = 0 )$ Give answer

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

  5. If x = y or relationship can not be established
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Equation I: 14x² - 5x - 1 = 0. Using quadratic formula: x = [5 ± sqrt(25 - 4(14)(-1))]/28 = [5 ± sqrt(25 + 56)]/28 = [5 ± sqrt(81)]/28 = [5 ± 9]/28. So x = 14/28 = 1/2 or x = -4/28 = -1/7. Equation II: 2y² + 3y + 1 = 0. Factors: (2y + 1)(y + 1) = 0. So y = -1/2 or y = -1. Both values of x (1/2, -1/7) are greater than both values of y (-1/2, -1). Therefore x > y.

Multiple choice

In the following questions two equations numbered I and II are given. You have to solve both the equations and give answer- $I. (4x^2 + 12x + 5 = 0) II. (6y^2 + 27y + 12 = 0)$

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

  5. If x = y or the relation can't be determined
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Equation I: 4x² + 12x + 5 = 0. Discriminant = 144 - 80 = 64. Roots: x = (-12 ± 8)/8, so x = -1/2 or x = -5/2. Equation II: 6y² + 27y + 12 = 0. Discriminant = 729 - 288 = 441. Roots: y = (-27 ± 21)/12, so y = -1/2 or y = -4. Comparing: x can equal y (both -1/2) or x can be > y or < y. So no unique relation can be established.

Multiple choice

In the following questions two equations numbered I and II are given. You have to solve both the equations and give answer-\newlineI. (15x^2 + 18x + 3 = 0)\newlineII. (12.5y^2 - 24y + 16 = 0)

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

  5. If x = y or the relation can't be determined
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Equation I: 15x² + 18x + 3 = 0. Dividing by 3: 5x² + 6x + 1 = 0. Factorizing: (5x + 1)(x + 1) = 0. So x = -1/5 or x = -1. Equation II: 12.5y² - 24y + 16 = 0. Multiplying by 2: 25y² - 48y + 32 = 0. Factorizing: (5y - 8)(5y - 4) = 0. So y = 8/5 = 1.6 or y = 4/5 = 0.8. When x = -1 or -0.2 and y = 0.8 or 1.6, we have x < y in both cases. Therefore x ≤ y is correct.

Multiple choice

(\text{If } m \text{ and } n (m>n) \text{ are the roots of the equation } 7(x+2a)^2 + 3a^2 = 5a (7x+23a)\text{ where } a > 0, \text{ then what is } 3m -n \text{ equal to?})

  1. 12a

  2. 14a

  3. 15a

  4. 18a
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Expanding: 7(x² + 4ax + 4a²) + 3a² = 35ax + 115a². This gives 7x² + 28ax + 28a² + 3a² - 35ax - 115a² = 0, so 7x² - 7ax - 84a² = 0, or x² - ax - 12a² = 0. Sum of roots = a, product = -12a². If roots are m and n (m > n): m + n = a, mn = -12a². Then 3m - n = 2m + (m - n) = 2m + √((m+n)² - 4mn) = 2m + √(a² + 48a²) = 2m + 7a. Since m > n and mn < 0, one root is positive, one negative. Solving: m = 4a, n = -3a. So 3m - n = 12a - (-3a) = 15a.

Multiple choice

Solve the following equations and find the values of (x) and (y): I. (16x^2 + 20x + 6 = 0) II. (10y^2 + 38y + 24 = 0)

  1. If x < y

  2. If x > y

  3. If x ≤ y

  4. If x ≥ y

  5. x = y or relationship can not be established
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For I: 16x^2 + 20x + 6 = 0. Using quadratic formula: x = [-20 ± √(400-384)]/32 = [-20 ± 4]/32, giving x = -16/32 = -0.5 or x = -24/32 = -0.75. For II: 10y^2 + 38y + 24 = 0. y = [-38 ± √(1444-960)]/20 = [-38 ± √484]/20 = [-38 ± 22]/20, giving y = -16/20 = -0.8 or y = -60/20 = -3. Comparing: x values (-0.5, -0.75) are both greater than y values (-0.8, -3). Thus x > y.

Multiple choice

$I. (18x^2 + 18x + 4 = 0)$ $II. (12y^2 + 29y + 14 = 0)$ Solve the given equations and find the values of x and y.

  1. If x < y

  2. If x > y

  3. If x ≤ y

  4. If x ≥ y

  5. If x = y or relationship can not be established.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For I: 18x^2 + 18x + 4 = 0. Discriminant = 324 - 288 = 36. x = [-18 ± 6]/36, giving x = -12/36 = -1/3 or x = -24/36 = -2/3. For II: 12y^2 + 29y + 14 = 0. Discriminant = 841 - 672 = 169. y = [-29 ± 13]/24, giving y = -16/24 = -2/3 or y = -42/24 = -7/4 = -1.75. Since x = -1/3, -2/3 and y = -2/3, -1.75, we have x ≥ y (because -2/3 = -2/3 and -1/3 > -2/3 > -1.75).

Multiple choice

$I. (8x^2 + 6x = 5) II. (12y^2 - 22y + 8 = 0)$ Solve the given equations and find the values of x and y.

  1. If x < y

  2. If x > y

  3. If x ≤ y

  4. If x ≥ y

  5. If x = y or relationship can not be established.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For I: 8x^2 + 6x - 5 = 0. Using formula: x = [-6 ± √(36+160)]/16 = [-6 ± √196]/16 = [-6 ± 14]/16, giving x = 8/16 = 0.5 or x = -20/16 = -1.25. For II: 12y^2 - 22y + 8 = 0. y = [22 ± √(484-384)]/24 = [22 ± 10]/24, giving y = 32/24 = 4/3 ≈ 1.33 or y = 12/24 = 0.5. Comparing: x values are 0.5 and -1.25; y values are 1.33 and 0.5. Since 0.5 = 0.5 and -1.25 < both y values, x ≤ y is correct.