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 question two equations numbered I and II are given, you have to solve both the equation and choose the correct answer. I. x2 + 14x + 45 = 0 II. y2 + 16y + 63 = 0

  1. X > Y

  2. X < Y

  3. X ≤ Y

  4. X ≥ Y

  5. X = Y or the relationship can’t be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Solving I gives x = -5 and x = -9. Solving II gives y = -7 and y = -9. Comparison: -5 > -7 (so x > y here), but -9 = -9 (equal). Since we have both x > y and x = y cases, and no x < y cases, the relationship cannot be established as a single consistent inequality.

Multiple choice

In each question two equations numbered I and II are given, you have to solve both the equation and choose the correct answer.(I. x^2 - 8x + 15 = 0 )(II. y^2 - 12y + 36 = 0)
  1. $X > Y$
  2. $X < Y$
  3. $X ≤ Y$
  4. $X ≥ Y$

  5. X = Y or the relationship can’t be established
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Equation I: x^2-8x+15=0 factors to (x-3)(x-5)=0, so x=3 or x=5. Equation II: y^2-12y+36=0 is (y-6)^2=0, so y=6 (repeated root). Since both values of x (3 and 5) are less than y=6, the relationship is X < Y.

Multiple choice

In each question two equations numbered I and II are given, you have to solve both the equation and choose the correct answer.\newline I. (2x^2 + 7x + 5 = 0) \newline II. (3y^2 + 5y + 2 = 0)

  1. X > Y

  2. X < Y

  3. X ≤ Y

  4. X ≥ Y

  5. X = Y or the relationship can’t be established
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Solving equation I: 2x² + 7x + 5 = 0 gives x = -5/2, -1. Solving equation II: 3y² + 5y + 2 = 0 gives y = -2/3, -1. When both equal -1, x = y. In all other comparisons, x < y. Therefore x ≤ y is the correct relationship.

Multiple choice

In each question two equations numbered I and II are given, you have to solve both the equation and choose the correct answer.\nI. (x^2 + 10x + 24 = 0)\nII. (y^2 - \sqrt{625} = 0)

  1. X > Y

  2. X < Y

  3. X ≤ Y

  4. X ≥ Y

  5. X = Y or the relationship can’t be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Solving equation I: x² + 10x + 24 = 0 gives x = -4, -6. Solving equation II: y² - 625 = 0 gives y = 25, -25. Comparing values: when x = -4 and y = -25, x > y; but when x = -6 and y = 25, x < y. Since the relationship changes, it cannot be established.

Multiple choice

In each question two equations numbered I and II are given, you have to solve both the equation and choose the correct answer. ( I. 2x^2 – 13x + 18= 0 ) ( II. y^2 – 7y + 12 = 0 )

  1. X > Y

  2. X < Y

  3. X ≤ Y

  4. X ≥ Y

  5. X = Y or the relationship can’t be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Solving equation I: 2x² - 13x + 18 = 0 gives x = 2, 9/2 (or 4.5). Solving equation II: y² - 7y + 12 = 0 gives y = 3, 4. When x = 2 and y = 3, x < y; when x = 4.5 and y = 4, x > y. The relationship changes, so it cannot be established.

Multiple choice

In each question two equations numbered I and II are given, you have to solve both the equation and choose the correct answer. $I. (x^2 - 50x + 589 = 0)$ $II. (y^2 - 19y - 416 = 0)$

  1. X > Y

  2. X < Y

  3. X ≤ Y

  4. X ≥ Y

  5. X = Y or the relationship can’t be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Solve each quadratic equation by factoring or using the quadratic formula. For equation I: x² - 50x + 589 = 0 gives x = 31 or x = 19. For equation II: y² - 19y - 416 = 0 gives y = 32 or y = -13. Since x takes values between 19 and 31 while y ranges from -13 to 32, the ranges overlap significantly and no definite relationship can be established between x and y.

Multiple choice

In each question two equations numbered I and II are given, you have to solve both the equation and choose the correct answer. $I. (x^2 - 53x + 462 = 0) II. (y^2 - 88y + 1932 = 0)$

  1. X > Y

  2. X < Y

  3. X ≤ Y

  4. X ≥ Y

  5. X = Y or the relationship can’t be established
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Factor the quadratic equations. For equation I: x² - 53x + 462 = 0 factors to (x - 42)(x - 11) = 0, so x = 42 or x = 11. For equation II: y² - 88y + 1932 = 0 factors to (y - 46)(y - 42) = 0, so y = 46 or y = 42. Comparing all values: the smallest x (11) is less than the smallest y (42), and the largest x (42) is less than or equal to the largest y (46). Therefore x ≤ y for all combinations.

Multiple choice

In each question two equations numbered I and II are given, you have to solve both the equation and choose the correct answer. I. (x^2 - 10x + 18.24 = 0) II. (y^2 - 6y + 8.75 = 0)

  1. X > Y

  2. X < Y

  3. X ≤ Y

  4. X ≥ Y

  5. X = Y or the relationship can’t be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

For equation I: x² - 10x + 18.24 = 0, the discriminant is 100 - 72.96 = 27.04, giving x = (10 ± 5.2)/2 = 7.6 or 2.4. For equation II: y² - 6y + 8.75 = 0, the discriminant is 36 - 35 = 1, giving y = (6 ± 1)/2 = 3.5 or 2.5. The x values range from 2.4 to 7.6 while y values range from 2.5 to 3.5. Since the ranges overlap and x can be both greater and smaller than y, no definite relationship exists.

Multiple choice

In each question two equations numbered I and II are given, you have to solve both the equation and choose the correct answer.\nI. (x^2 + 9x + 20 = 0)\nII. (8y^2 - 15y + 7 = 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
B Correct answer
Explanation

Solve I: x²+9x+20=0 factors to (x+4)(x+5)=0, so x = -4, -5. Solve II: 8y²-15y+7=0 factors to (8y-7)(y-1)=0, so y = 7/8, 1. All x values (-4, -5) are less than all y values (7/8, 1). Therefore X < Y is always true. Option B is correct.

Multiple choice

In each question two equations numbered I and II are given, you have to solve both the equation and choose the correct answer. $I. (x^2 - x - 12 = 0) II. (y^2 + 5y + 6 = 0)$
  1. $\(X > Y\)$
  2. $\(X < Y\)$
  3. $\(X \le Y\)$
  4. $\(X \ge Y\)$

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

Solve equation I: x² - x - 12 = 0 factors to (x-4)(x+3) = 0, giving x = 4 or x = -3. Solve equation II: y² + 5y + 6 = 0 factors to (y+2)(y+3) = 0, giving y = -2 or y = -3. When we compare all possible values: x=4 vs y=-2 gives X > Y; x=4 vs y=-3 gives X > Y; x=-3 vs y=-2 gives X < Y; x=-3 vs y=-3 gives X = Y. Since the relationship varies (X can be greater than, less than, or equal to Y depending on which root values we choose), the relationship cannot be established uniquely.

Multiple choice

In the following questions two equations I and II are given. You have to solve both the equations and give answer. ( (I) ) ( 10x^2 – 5x = 2x + 3 ) ( (II) ) ( 3y^2 – 5y + 2 = 0 )

  1. If x > y

  2. If x ≤ y

  3. If x < y

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

Solving (I) gives x = 1 or -0.6. Solving (II) gives y = 2/3 or 1. Comparing all pairs: when x=1, y could be 2/3 (x>y) or 1 (x=y); when x=-0.6, y could be 2/3 (x

Multiple choice

In each question two equations numbered I and II are given, you have to solve both the equations and choose the correct answer. $I. (3x^2 + 8x + 5 = 0) II. (2y^2 + y - 1 = 0)$
  1. $\(x > y\)$
  2. $\(x < y\)$
  3. $\(x \le y\)$
  4. $\(x \ge y\)$

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

Solving equation I: 3x² + 8x + 5 = 0 gives (3x + 5)(x + 1) = 0, so x = -5/3 or x = -1. Solving equation II: 2y² + y - 1 = 0 gives (2y - 1)(y + 1) = 0, so y = 1/2 or y = -1. Comparing values: x = -5/3 < y = 1/2, and x = -5/3 < y = -1. But x = -1 = y = -1. Since x is sometimes less than y and sometimes equal to y, the relationship is x ≤ y.

Multiple choice

In each question two equations numbered I and II are given, you have to solve both the equations and choose the correct answer. $I. (5x^2 - 7x - 6 = 0) II. (5y^2 + 23y + 12 = 0)$
  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

Solving equation I: 5x² - 7x - 6 = 0 gives (5x + 3)(x - 2) = 0, so x = -3/5 or x = 2. Solving equation II: 5y² + 23y + 12 = 0 gives (5y + 3)(y + 4) = 0, so y = -3/5 or y = -4. Comparing all values: x = 2 > y = -3/5, x = 2 > y = -4, x = -3/5 = y = -3/5, x = -3/5 > y = -4. Since x is always greater than or equal to y, x ≥ y.

Multiple choice

In each question two equations numbered I and II are given, you have to solve both the equations and choose the correct answer. $I. (x^2 + x - 20 = 0)$ $II. (2y^2 + 13y + 15 = 0)$
  1. $\(x > y\)$
  2. $\(x < y\)$
  3. $\(x \le y\)$
  4. $\(x \ge y\)$

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

Solving equation I: x² + x - 20 = 0 gives (x + 5)(x - 4) = 0, so x = -5 or x = 4. Solving equation II: 2y² + 13y + 15 = 0 gives (2y + 3)(y + 5) = 0, so y = -3/2 or y = -5. Comparing values: x = -5 = y = -5, x = 4 > y = -3/2, x = 4 > y = -5. Since x is sometimes equal to y and sometimes greater than y, the exact relationship cannot be established.

Multiple choice

In each question two equations numbered I and II are given, you have to solve both the equations and choose the correct answer. \ I. (x^2 - 14x - 32 = 0) \ II. (y^2 + 4y - 60 = 0)
  1. $\(x > y\)$
  2. $\(x < y\)$
  3. $\(x \le y\)$
  4. $\(x \ge y\)$

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

Solving equation I: x² - 14x - 32 = 0 gives (x - 16)(x + 2) = 0, so x = 16 or x = -2. Solving equation II: y² + 4y - 60 = 0 gives (y + 10)(y - 6) = 0, so y = -10 or y = 6. Comparing all values: x = 16 > y = -10, x = 16 > y = 6, x = -2 > y = -10, x = -2 < y = 6. Since x is sometimes greater than and sometimes less than y, the relationship cannot be established.