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 $( (I) \quad 4x + 2y = 8.5 ) ( (II) \quad 2x + 4y = 9.5 )$

  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

Multiply equation (II) by 2: 4x + 8y = 19. Subtract from equation (I) multiplied by 2 (8x + 4y = 17): (8x + 4y) - (4x + 8y) = 17 - 19, giving 4x - 4y = -2, so x - y = -0.5, meaning x = y - 0.5. Therefore x < y always. Alternatively, solve directly: from (I), 2x = 4.25 - y, substitute into (II) to get y = 3.167, x = 1.417, confirming x < y.

Multiple choice

In each of these questions two equations are given. You have to solve these equations and find out the value of x and y and give the answer.(\text{(I)}) (x^2 + 7x + 12 = 0) (\text{(II)}) (y^2 + 5y + 6 = 0)

  1. If x < y

  2. If x > y

  3. If x ≤ y

  4. If x ≥ y

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

Equation (I): x² + 7x + 12 = 0 factors to (x+3)(x+4)=0, so x = -3 or -4. Equation (II): y² + 5y + 6 = 0 factors to (y+2)(y+3)=0, so y = -2 or -3. Comparing all pairs: (-3,-2) gives x

Multiple choice

In each of these questions two equations are given. You have to solve these equations and find out the value of x and y and give the answer. $( (I) \quad 4x + 3y – 16 = 0 ) ( (II) \quad 2x + 2y – 9 = 0 )$

  1. If x < y

  2. If x > y

  3. If x ≤ y

  4. If x ≥ y

  5. If x = y or if the relation cannot be established.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Solve the system: 4x + 3y = 16 and 2x + 2y = 9. Multiply the second equation by 2 to get 4x + 4y = 18. Subtract the first equation: (4x + 4y) - (4x + 3y) = 18 - 16, giving y = 2. Substitute back: 2x + 2(2) = 9, so 2x = 5 and x = 2.5. Since x = 2.5 and y = 2, we have 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. $(I) (3x - 2y = 10) (II) (5x - 6y = 6)$

  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

From equation I: 3x - 2y = 10. From equation II: 5x - 6y = 6. Multiply equation I by 3: 9x - 6y = 30. Subtract equation II: (9x - 6y) - (5x - 6y) = 30 - 6, giving 4x = 24, so x = 6. Substitute back: 3(6) - 2y = 10, so 18 - 2y = 10, giving y = 4. Since x=6 and y=4, we have 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) ) ( x^2 + x - 12 = 0 ) ( (II) ) ( y^2 - 5y + 6 = 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
E Correct answer
Explanation

Equation I: x² + x - 12 = 0 gives (x+4)(x-3) = 0, so x = -4 or 3. Equation II: y² - 5y + 6 = 0 gives (y-2)(y-3) = 0, so y = 2 or 3. Comparing all possible pairs: when x=-4, xy when y=2. Since the relationship varies depending on which root values we choose, it cannot be established.

Multiple choice

(\text{Solve the following equations and determine the relationship between x and y:})\n(\text{(I)}\quad x^2 + 9x + 18 = 0 )\n(\text{(II)}\quad y^2 - 13y + 40 = 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
C Correct answer
Explanation

Equation I: x² + 9x + 18 = 0 gives (x+6)(x+3) = 0, so x = -6 or -3. Equation II: y² - 13y + 40 = 0 gives (y-8)(y-5) = 0, so y = 8 or 5. All values of x are negative (-6, -3) while all values of y are positive (8, 5). Therefore x < y is always true regardless of which specific root values we compare.

Multiple choice

In each of the following questions, two equations (I) and (II) are given. You have to solve them and- $( (I) x^2 + 17x + 30 = 0 )$ $( (II) y^2 - 13y = -30 )$ 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
C Correct answer
Explanation

Solving equation (I): x² + 17x + 30 = 0. Factor: (x + 15)(x + 2) = 0, so x = -15 or x = -2. Solving equation (II): y² - 13y = -30, so y² - 13y + 30 = 0. Factor: (y - 10)(y - 3) = 0, so y = 10 or y = 3. Comparing values: x values are -15 and -2 (both negative), y values are 10 and 3 (both positive). Therefore x < y for all possible combinations, making answer C correct.

Multiple choice

In each of the following questions, two equations (I) and (II) are given. You have to solve them and $( (I) \ 3x + 2y = 14 \\ (II) \ x + 4y = 13 )$ 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

Solving the system: Multiply equation (II) by 3 to get 3x + 12y = 39. Subtract (I) from this: (3x+12y) - (3x+2y) = 39-14, giving 10y = 25, so y = 2.5. Substituting in (II): x + 4(2.5) = 13, so x + 10 = 13, giving x = 3. Since x = 3 and y = 2.5, we have x > y. Option B (x ≥ y) is incorrect because x is strictly greater than y, not equal to it. Option C (x < y) and D (x ≤ y) are both false since 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. $I. (x^2 + 3x + 2 = 0) II. (y^2 + 5y + 6 = 0)$

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

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

Solve x²+3x+2=0 gives x=-1,-2. Solve y²+5y+6=0 gives y=-2,-3. So x can be -1 or -2, y can be -2 or -3. When x=-1, x>-2 and x>-3. When x=-2, x=-2=y is possible, x>-3. Thus x ≥ y always holds.

Multiple choice

In the following questions two equations I and II are given. You have to solve both the equations and give answer. $I. (2x + 4y = 13)$ $II. (4x + 2y = 14)$

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

  5. If x = y or the relation cannot be determined.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Solving the system: 2x+4y=13 and 4x+2y=14. Multiply first equation by 2: 4x+8y=26. Subtract second equation: (4x+8y)-(4x+2y)=26-14 gives 6y=12, so y=2. Substitute: 2x+8=13, x=2.5. Thus x=2.5 > y=2.

Multiple choice

In the following questions two equations I and II are given. You have to solve both the equations and give answer. $I. (x^2 + 8x + 15 = 0)$ $II. (y^2 + 11y + 30 = 0)$

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

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

Solve x²+8x+15=0: (x+3)(x+5)=0, so x=-3,-5. Solve y²+11y+30=0: (y+5)(y+6)=0, so y=-5,-6. Compare: x=-3 > y=-5,-6 always. x=-5 = y=-5 possible, x=-5 > y=-6. Thus x ≥ y holds in all cases.

Multiple choice

In the following question two equations numbered I and II are given. You have to solve both the question and give answer if - $(I. \quad x^2 - 9x + 20 = 0 \ II. \quad y^2 - 13y + 42 = 0)$
  1. $\(x > y\)$
  2. $\(x \ge y\)$
  3. $\(x < y\)$
  4. $\(x \le y\)$

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

Equation I: x²-9x+20=0 factors to (x-4)(x-5)=0, so x=4 or 5. Equation II: y²-13y+42=0 factors to (y-6)(y-7)=0, so y=6 or 7. All values of y (6,7) are greater than all values of x (4,5). Therefore x < y is always true. Option C is correctly marked.

Multiple choice

In each of the following questions, two equations are given. You have to solve them and $( (I) \ 12x^2 - 7x + 1 = 0 ) ( (II) \ 15y^2 - 16y + 4 = 0 )$

  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
C Correct answer
Explanation

Solve equation (I): 12x² - 7x + 1 = 0. Using quadratic formula: x = (7 ± √(49 - 48)) / 24 = (7 ± 1) / 24. So x = 8/24 = 1/3 or x = 6/24 = 1/4. Values are x = 0.33 or x = 0.25. For equation (II): 15y² - 16y + 4 = 0. Using formula: y = (16 ± √(256 - 240)) / 30 = (16 ± 4) / 30. So y = 20/30 = 2/3 ≈ 0.667 or y = 12/30 = 2/5 = 0.4. Comparing: x (0.33 or 0.25) < y (0.667 or 0.4). Therefore x < y is correct.

Multiple choice

In each of the following questions, two equations are given. You have to solve them and $I. (x^2 + 3x - 18 = 0) II. (y^2 + 9y + 18 = 0)$

  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

I: x² + 3x - 18 = 0 gives (x + 6)(x - 3) = 0, so x = -6 or 3. II: y² + 9y + 18 = 0 gives (y + 6)(y + 3) = 0, so y = -6 or -3. For x = 3: x > y (both -6, -3). For x = -6: x = y when y = -6, and x > y when y = -3. Therefore x ≥ y is always true.