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 (I) and (II) are given. You have to solve them and answer the question.(\text{(I)}) (4x^2 + 19x + 21 = 0)(\text{(II)}) (2y^2 - 23y + 63 = 0)

  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
C Correct answer
Explanation

Solving equation (I): 4x²+19x+21=0 factors to (4x+7)(x+3)=0, giving x=-7/4=-1.75 or x=-3. Solving equation (II): 2y²-23y+63=0 factors to (2y-7)(y-9)=0, giving y=3.5 or y=9. Since x=-1.75 or -3 (both negative) and y=3.5 or 9 (both positive), x

Multiple choice

In each of the following questions, two equations (I) and (II) are given. You have to solve them and give answer. $(I) (x^2 - 1 = 0) (II) (y^2 + 4y + 3 = 0)$

  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

Equation I: x² - 1 = 0 gives x² = 1, so x = ±1 (x can be 1 or -1). Equation II: y² + 4y + 3 = 0 factors as (y + 3)(y + 1) = 0, so y = -3 or y = -1. Comparing all values: x = 1 > y = -3, x = 1 > y = -1, x = -1 > y = -3, x = -1 = y = -1. In all cases, x ≥ y is true (x is never less than y).

Multiple choice

(\text{Solve the following equations and compare the values of x and y:}\ \text{(I) } x^2 - 7x + 12 = 0 \ \text{(II) } y^2 - 12y + 32 = 0)

  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

Equation I: x² - 7x + 12 = 0 factors as (x - 3)(x - 4) = 0, so x = 3 or x = 4. Equation II: y² - 12y + 32 = 0 factors as (y - 4)(y - 8) = 0, so y = 4 or y = 8. Comparing: When x = 3, y can be 4 or 8 (x < y in both cases). When x = 4, y can be 4 (x = y) or 8 (x < y). In all cases, x ≤ y is true. x is never greater than y.

Multiple choice

In each of the following questions, two equations are given. You have to solve them and determine the relationship between x and y. $( (I) \quad x^2 - 9x + 18 = 0 )$ $( (II) \quad y^2 - 11y + 30 = 0 )$

  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

Solving equation I: x² - 9x + 18 = 0 factors to (x-3)(x-6) = 0, so x = 3 or x = 6. Solving equation II: y² - 11y + 30 = 0 factors to (y-5)(y-6) = 0, so y = 5 or y = 6. When x = 6 and y = 6, we have x = y. When x = 3 and y = 5, we have x < y. When x = 3 and y = 6, we have x < y. When x = 6 and y = 5, we have x > y. Since different value combinations give different relationships, the relationship between x and y cannot be uniquely established.

Multiple choice

In the following questions two equation (I) and (II) are given. You have to solve both the equation and give answer. $I. (x^2 - 14x + 48 = 0) II. (y^2 + 7y + 12 = 0)$

  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
A Correct answer
Explanation

Solving equation I: x² - 14x + 48 = 0 factors to (x-8)(x-6) = 0, so x = 8 or 6. Solving equation II: y² + 7y + 12 = 0 factors to (y+4)(y+3) = 0, so y = -4 or -3. All values of x (8, 6) are greater than all values of y (-4, -3). Therefore x > y is always true, making option A correct.

Multiple choice

In the following questions two equation (I) and (II) are given. You have to solve both the equation and give answer. $I. (x^2 + 9x + 20 = 0) II. (y^2 + 16y + 63 = 0)$

  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
A Correct answer
Explanation

Solving equation I: x² + 9x + 20 = 0 factors to (x+5)(x+4) = 0, so x = -5 or -4. Solving equation II: y² + 16y + 63 = 0 factors to (y+9)(y+7) = 0, so y = -9 or -7. For all combinations: when x=-5, x > y (since -5 > -7, -5 > -9); when x=-4, x > y (since -4 > -7, -4 > -9). Therefore x > y always, confirming option A.

Multiple choice

In the following questions two equation (I) and (II) are given. You have to solve both the equation and give answer. I. (x^2 + 6 = 5x) II. (y^2 + 13y = -42)

  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
A Correct answer
Explanation

Solving equation I: x² - 5x + 6 = 0 factors to (x-2)(x-3) = 0, so x = 2 or 3. Solving equation II: y² + 13y + 42 = 0 factors to (y+6)(y+7) = 0, so y = -6 or -7. All positive values of x (2, 3) are greater than all negative values of y (-6, -7). Therefore x > y, making option A correct.

Multiple choice

(\text{Solve the following equations and choose the correct option:}\\text{(I) } a^2 + 3a + 2 = 0\\text{(II) } b^2 + 5b + 6 = 0)

  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
B Correct answer
Explanation

a²+3a+2=0 factors to (a+1)(a+2)=0, so a=-1 or a=-2. b²+5b+6=0 factors to (b+2)(b+3)=0, so b=-2 or b=-3. In all cases: a=-1>-2, a=-1>-3, a=-2>-3. Therefore a is always greater than or equal to b.

Multiple choice

In each of the following question, two equation (I) and (II) is given. You have to solve them and answer the question. ( (I) ) ( 6x^2 + 23x + 20 = 0 ) ( (II) ) ( 6y^2 + 41y + 63 = 0 )

  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

Equation I: 6x²+23x+20=0 factors to (2x+5)(3x+4)=0, giving x = −2.5 or x = −1.33. Equation II: 6y²+41y+63=0 factors to (2y+9)(3y+7)=0, giving y = −4.5 or y = −2.33. Since x values (−2.5, −1.33) and y values (−4.5, −2.33) overlap in range (x can be greater or less than y), the relationship cannot be established.

Multiple choice

(\text{In each of the following question, two equation (I) and (II) is given. You have to solve them and answer the question})\n(\text{(I) } 8x^2 + 3x = 38 \text{ (II) } 6y^2 + 34 = 29y)

  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
C Correct answer
Explanation

Solve equation (I): 8x² + 3x - 38 = 0 factors to (8x - 19)(x + 2) = 0, giving x = 19/8 = 2.375 or x = -2. Solve equation (II): 6y² - 29y + 34 = 0 factors to (2y - 17)(3y - 2) = 0, giving y = 17/2 = 8.5 or y = 2/3. Taking positive values: x = 2.375, y = 8.5, so x < y. Since x < y is true, x ≤ y is also true.

Multiple choice

In each of the following questions, (I) and (II) equations are given. You have to solve them and Give answer. $I. (x^2 + 5x + 6 = 0)$ $II. (y^2 + 7y + 12 = 0)$ 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 determined
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Solving x² + 5x + 6 = 0 gives (x+2)(x+3) = 0, so x = -2 or x = -3. Solving y² + 7y + 12 = 0 gives (y+3)(y+4) = 0, so y = -3 or y = -4. For all possible combinations: (-2,-3): -2 ≥ -3; (-2,-4): -2 ≥ -4; (-3,-3): -3 ≥ -3; (-3,-4): -3 ≥ -4. Every combination satisfies x ≥ y.

Multiple choice

$In each of the following questions, (I) and (II) equations are given. You have to solve them and \ I. (x^2 + 2x + 1 = 0)\ II. (y^2 = 4)$ 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 determined
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

x² + 2x + 1 = 0 is (x+1)² = 0, giving x = -1 (repeated root). y² = 4 gives y = 2 or y = -2. When y = 2, we have x < y since -1 < 2. When y = -2, we have x > y since -1 > -2. Since the relationship changes depending on which value of y we choose, it cannot be determined.

Multiple choice

In each of the following questions (I) and (II) equations are given. You have to solve them and \ I. (x^2 - 8x + 15 = 0) \ II. (y^2 = 25) \ 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 determined
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

x² - 8x + 15 = 0 factors to (x-3)(x-5) = 0, so x = 3 or x = 5. y² = 25 gives y = 5 or y = -5. Testing combinations: when x=3,y=5: xy; when x=5,y=5: x=y; when x=5,y=-5: x>y. Since we get different relationships (xy), the relationship cannot be uniquely determined.

Multiple choice

In each of the following question, two equation (I) and (II) is given. You have to solve them and answer the question.( (I) 4x^2 + 51x + 135 = 0 ) ( (II) 2y^2 + 11y + 15 = 0 )

  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
A Correct answer
Explanation

Solve equation (I): 4x² + 51x + 135 = 0 factors to (x + 12)(4x + 11.25) = 0 (checking: 4 × 11.25 = 45, not 135). Let's use quadratic formula or re-factor: (4x + 15)(x + 9) = 4x² + 51x + 135. So x = -15/4 = -3.75 or x = -9. Solve equation (II): 2y² + 11y + 15 = 0 factors to (2y + 5)(y + 3) = 0. So y = -5/2 = -2.5 or y = -3. Compare: when x = -9 and y = -2.5, we have x < y. When x = -3.75 and y = -3, we have x < y. Thus x < y, confirming option A.