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

(\text{In each of the following questions two equations are given. You have to solve them and give answer -}) (\text{(I)}\quad a^2 - 8a + 15 = 0 )(\text{(II)}\quad b^2 - 5b = -6)
  1. $\(a < b\)$
  2. $\(a > b\)$
  3. $\(a \le b\)$
  4. $\(a \ge b\)$
  5. $\(a = b \text{ or Relationship cannot be determined}\)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Solving a² - 8a + 15 = 0 gives (a-3)(a-5) = 0, so a = 3 or a = 5. Solving b² - 5b = -6 gives b² - 5b + 6 = 0, which factors to (b-2)(b-3) = 0, so b = 2 or b = 3. Comparing: when a=3, b can equal 3; when a=5, b can be 2 or 3. Since a ≥ b in all cases (a equals or exceeds b), option D is correct.

Multiple choice

(\text{In each of the following questions two equations are given. You have to solve them and give answer -})\n(\text{(I)}) (2a^2 + 20a + 50 = 0)\n(\text{(II)}) (b^2 = 25)
  1. $\(a < b\)$
  2. $\(a > b\)$
  3. $\(a \le b\)$
  4. $\(a \ge b\)$

  5. If a=b or Relationship cannot be determine
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Solving 2a² + 20a + 50 = 0, divide by 2: a² + 10a + 25 = 0, giving (a+5)² = 0, so a = -5 (repeated root). Solving b² = 25 gives b = 5 or b = -5. Comparing all combinations: -5 ≤ -5 (equal), -5 ≤ 5. Thus a ≤ b always holds.

Multiple choice

In each of the following questions, two equations (\bf{(I)}) and (\bf{(II)}) are given. You have to solve them and Give answer: (\bf{(I)}) (4x^2 + 9 - 12x = 0) (\bf{(II)}) (64y^2 - 80y + 25 = 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 (I): 4x²+9-12x=0 rearranges to 4x²-12x+9=0, which is (2x-3)²=0, so x=1.5 (repeated root). Solve (II): 64y²-80y+25=0 is (8y-5)²=0, so y=0.625 (repeated root). Since 1.5 > 0.625, x>y.

Multiple choice

In each of the following questions, two equations (I) and (II) are given. You have to solve them and Give answer: (I) 34x2 + 11x = 3 (II) 51y2 - 77y + 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
D Correct answer
Explanation

Solve (I): 34x²+11x-3=0 gives x=2/17≈0.118 or x=-3/2=-1.5. Solve (II): 51y²-77y+12=0 gives y=4/3≈1.33 or y=3/17≈0.176. Comparing: 0.118<0.176, 0.118<1.33, -1.5<0.176, -1.5<1.33. In all cases, x≤y.

Multiple choice

In the following question two equations are given. You have to solve both the equations and find out the values of x and y and Give answer - ( (I) 4x^2 – 7x + 3 = 0 )( (II) y^2 – 2y + 1 = 0 )
  1. $x > y$
  2. $x < y$
  3. $x ≤ y$
  4. $x ≥ y$

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

Equation (I): 4x² - 7x + 3 = 0. Factorizing: (4x-3)(x-1) = 0, so x = 3/4 or x = 1. Equation (II): y² - 2y + 1 = 0. This is (y-1)² = 0, so y = 1. Comparing: when x = 1, x = y. When x = 3/4, x < y. Since one possible x value is less than y and the other equals y, the relationship that covers all cases is x ≤ y, which is option C. Note that x can be equal to y (when both are 1), so x < y alone is incorrect.

Multiple choice

In the following question two equations are given. You have to solve both the equations and find out the values of x and y and Give answer - ( (I) x^2 – x - 12 = 0 ) ( (II) 2y^2 + 18y + 36 = 0 )
  1. $x < y$
  2. $x > y$
  3. $x ≤ y$
  4. $x ≥ y$

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

Equation (I): x² - x - 12 = 0. Factorizing: (x-4)(x+3) = 0, so x = 4 or x = -3. Equation (II): 2y² + 18y + 36 = 0. Divide by 2: y² + 9y + 18 = 0. Factorizing: (y+3)(y+6) = 0, so y = -3 or y = -6. Comparing: x values (4, -3) and y values (-3, -6). The smallest x (-3) equals the largest y (-3). All other comparisons give x > y. Therefore x ≥ y is always true, which is option D.

Multiple choice

In the following question two equations are given. You have to solve both the equations and find out the values of x and y and Give answer - ( (I) (x + 2)^{2} – 25 = 0 ) ( (II) 3y^{2} + y - 4 = 0 )

  1. x<y

  2. x>y

  3. x≤y

  4. x≥y

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

Solve equation (I): (x+2)²-25=0 gives x = 3 or -7. Solve equation (II): 3y²+y-4=0 factors to (y-1)(3y+4)=0, giving y = 1 or -4/3. When x = 3, x > y for both values. When x = -7, x < y for both values. Since the relationship changes based on which roots are compared, it cannot be established. Option E is correct.

Multiple choice

In each of the following questions two equations are given. You have to solve them and give answer - (I) (a^{2} + 5a + 6 = 0) (II) (b^{2} + 7b + 12 = 0)
  1. $If \(a < b\)$
  2. $If \(a > b\)$
  3. $If \(a \le b\)$
  4. $If \(a \ge b\)$
  5. $If \(a = b\) or Relationship cannot be determine$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For equation I: a² + 5a + 6 = 0 factors as (a+2)(a+3) = 0, giving a = -2 or a = -3. For equation II: b² + 7b + 12 = 0 factors as (b+3)(b+4) = 0, giving b = -3 or b = -4. Comparing values: a can be -2 or -3; b can be -3 or -4. When a = -2, a > b (since -2 > -3 and -2 > -4). When a = -3, a ≥ b (since -3 = -3 and -3 > -4). Therefore, in all cases a ≥ b is true, making option D correct.

Multiple choice

In the following questions two equations numbered I and II are given. You have to solve both the equations and- $I. (3x^2 + 7x + 2 = 0)$ $II. (y^2 = 4)$ Give answer

  1. If x < y

  2. If x > y

  3. If x ≤ y

  4. If x ≥ y

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

Solve equation I: 3x² + 7x + 2 = 0 gives x = -1/3 or x = -2. Solve equation II: y² = 4 gives y = ±2. When x = -1/3, compare with y = 2 (x < y) and y = -2 (x > y). When x = -2, compare with y = 2 (x < y) and y = -2 (x = y). Since x can be less than, greater than, or equal to y depending on which values we consider, no single relationship holds consistently.

Multiple choice

In the following questions two equations numbered I and II are given. You have to solve both the equations and Give answer $I. (4x^2 - 20x + 21 = 0)$ $II. (2y^2 - 13y + 20 = 0)$

  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

Solve equation I: 4x² - 20x + 21 = 0 factors as (2x-3)(2x-7), giving x = 3/2 or x = 7/2. Solve equation II: 2y² - 13y + 20 = 0 factors as (2y-5)(y-4), giving y = 5/2 or y = 4. When x = 1.5, it's less than y = 2.5 but greater than y = 4. When x = 3.5, it's greater than y = 2.5 but less than y = 4. Since the relationship varies, it cannot be established.

Multiple choice

In the following questions two equations (I) and (II) are given. You have to solve both the equations and give answer.( (I) \quad 2x^2 - 11x + 15 = 0 \quad (II) \quad 2y^2 - 15y + 28 = 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): 2x² - 11x + 15 = 0 gives x = 3 or x = 2.5. Solving equation (II): 2y² - 15y + 28 = 0 gives y = 4 or y = 3.5. All values of x are less than all values of y, so x < y.

Multiple choice

In the following questions two equations (I) and (II) are given. You have to solve both the equations and give answer. ( (I) \quad 2x^2 + 27 + 15x = 0 \quad (II) \quad 3y^2 + 14 = 13y )

  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): 2x² + 15x + 27 = 0. Discriminant = 225 - 216 = 9, so x = (-15 ± 3)/4, giving x = -3 or x = -4.5. Solving equation (II): 3y² - 13y + 14 = 0 gives y = 2 or y = 7/3. All x values (-3, -4.5) are less than all y values (2, 2.33), so x < y.

Multiple choice

In the following questions two equations (I) and (II) are given. You have to solve both the equations and give answer.( (I) \quad 2x^2 - 9x + 10 = 0 \quad (II) \quad 2y^2 - 13y + 20 = 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
D Correct answer
Explanation

Solving equation (I): 2x² - 9x + 10 = 0 gives x = 2 or x = 5/2. Solving equation (II): 2y² - 13y + 20 = 0 gives y = 5/2 or y = 4. For x = 2, both y values (2.5, 4) are greater. For x = 2.5, y can be 2.5 (equal) or 4 (greater). Thus, x is always less than or equal to y.

Multiple choice

In the following questions two equations (I) and (II) are given. You have to solve both the equation and give answer- (I) ( x^{2} -1 = 0 ) (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 the relationship can not be established.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Solving equation (I): x² - 1 = 0 gives x = ±1. Solving equation (II): y² - 12y + 32 = 0. Using quadratic formula: y = (12 ± √(144-128))/2 = (12 ± √16)/2 = (12 ± 4)/2, so y = 8 or y = 4. For all values of x (±1) and y (4, 8), we have x < y. Therefore x < y is always true. Option B is correct.

Multiple choice

In the following questions two equations (I) and (II) are given. You have to solve both the equation and give answer- $( (I) \quad 7x + 6y = 110 \ (II) \quad 4x + 3y = 59 )$

  1. If x > y

  2. If x < y

  3. If x ≥ y

  4. If x ≤ y

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

From (II): 3y = 59 - 4x, so y = (59-4x)/3. Substitute into (I): 7x + 6(59-4x)/3 = 110 gives 7x + 2(59-4x) = 110, so 7x + 118 - 8x = 110, giving x = 8. Then y = (59-32)/3 = 9. Since 8 < 9, x < y.