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 both equation and give the answer- $I. (3x^2 + 17x + 10 = 0)$ $II. (10y^2 + 9y + 2 = 0)$

  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

I: 3x² + 17x + 10 = 0 gives (3x + 2)(x + 5) = 0, so x = -2/3 or -5. II: 10y² + 9y + 2 = 0 gives (5y + 2)(2y + 1) = 0, so y = -2/5 or -1/2. Converting to decimals: x = -0.67 or -5, y = -0.4 or -0.5. In all cases x < y.

Multiple choice

In each of the following questions, two equations are given. You have to solve them and $(I. \quad x^2 – x – 6 = 0 \ II. \quad 2y^2 + 13y + 21 = 0)$

  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
A Correct answer
Explanation

I: x² - x - 6 = 0 gives (x - 3)(x + 2) = 0, so x = 3 or -2. II: 2y² + 13y + 21 = 0 gives (2y + 7)(y + 3) = 0, so y = -7/2 = -3.5 or -3. Comparing: when x = 3, x > y (both -3.5 and -3). When x = -2, x > y (both -3.5 and -3). Therefore x > y.

Multiple choice

In each of the following questions, two equations are given. You have to solve them and\ I. (x^2 + 2x = 24)\ II. (y^2 - 10y + 21 = 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

I: x² + 2x = 24 gives x² + 2x - 24 = 0, so (x + 6)(x - 4) = 0 and x = -6 or 4. II: y² - 10y + 21 = 0 gives (y - 7)(y - 3) = 0, so y = 7 or 3. Comparing: when x = -6, x < y (both 7 and 3). When x = 4, x > 3 but x < 7. The relationship cannot be established consistently.

Multiple choice

In each of the following questions, two equations are given. You have to solve them and $(I. \quad x^2 + 13x + 40 = 0\II. \quad y^2 + 7y + 12 = 0)$

  1. If x>y

  2. If x≥y

  3. If x<y

  4. If x≤y

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

I: x² + 13x + 40 = 0 gives (x + 8)(x + 5) = 0, so x = -8 or -5. II: y² + 7y + 12 = 0 gives (y + 4)(y + 3) = 0, so y = -4 or -3. Comparing: when x = -8, x < y (both -4 and -3). When x = -5, x < y (both -4 and -3). Therefore x < y.

Multiple choice

In each of the following questions you have to solve the given equations to find m and n and- I. (m^2 + 16 = 212) II. (n^2 - 6n - 16 = 0)

  1. if m > n

  2. if m ≥ n

  3. if m < n

  4. if m ≤ n

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

Solve m² + 16 = 212, giving m = ±14. Solve n² - 6n - 16 = 0 using quadratic formula or factoring, giving n = 8 or n = -2. Since m can be 14 or -14, we have m > n when m = 14, but m < n when m = -14. The relationship cannot be uniquely established.

Multiple choice

In the following questions two equations I and II are given. You have to solve both the equations and give answer. $I. (x^2 - 12x + 35 = 0) II. (y^2 - 9y + 20 = 0)$

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

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

Solve both equations: x² - 12x + 35 = 0 gives (x-5)(x-7) = 0, so x = 5 or 7. y² - 9y + 20 = 0 gives (y-4)(y-5) = 0, so y = 4 or 5. Comparing all combinations: when x=5, y can be 4 or 5 (5≥4, 5≥5); when x=7, y can be 4 or 5 (7≥4, 7≥5). In all cases, x ≥ y is true.

Multiple choice

In the following questions two equations numbered I and II are given. You have to solve both the equations and $(I. \quad 3x + 4y = 16 )$ $(II. \quad 2x + 2y = 9)$ Given answer-

  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
C Correct answer
Explanation

I: 3x + 4y = 16. II: 2x + 2y = 9. From II: divide by 2: x + y = 4.5, so x = 4.5 - y. Substitute in I: 3(4.5-y) + 4y = 16 → 13.5 - 3y + 4y = 16 → y = 2.5. Then x = 4.5 - 2.5 = 2. So x = 2, y = 2.5, meaning x < y.

Multiple choice

In the following question two equation number I and II are given you have to solve both equation and (I. x^2 + 29x + 210 = 0 )(II. 2y^2 + 12y + 16 = 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
A Correct answer
Explanation

Solving equation I: x² + 29x + 210 = 0 gives (x+14)(x+15) = 0, so x = -14 or x = -15. Solving equation II: 2y² + 12y + 16 = 0. Divide by 2: y² + 6y + 8 = 0 gives (y+2)(y+4) = 0, so y = -2 or y = -4. Comparing: Both values of x (-14, -15) are less than both values of y (-2, -4). So x < y is false, x > y is true for all combinations.

Multiple choice

In each of the following questions, two equations are given. You have to solve them and $( I. \quad 2x + 9y = 18 \\ II. \quad x + 3y = 12 )$

  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

Multiply equation II by 3: 3x + 9y = 36. Subtract from equation I: (2x + 9y) - (3x + 9y) = 18 - 36, giving -x = -18, so x = 18. Substituting into equation II: 18 + 3y = 12, giving 3y = -6, so y = -2. Since x = 18 > y = -2, option A is correct.

Multiple choice

In each of the following questions, two equations are given. You have to solve them and $I. (x^2 - 11x = 0) II. (y^2 - 24y + 144 = 0)$

  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

For equation I: x² - 11x = 0 gives x(x - 11) = 0, so x = 0 or x = 11. For equation II: y² - 24y + 144 = 0 is (y - 12)² = 0, so y = 12 (repeated root). Comparing: x can be 0 or 11, while y = 12. In both cases x < y (0 < 12 and 11 < 12), making option C correct.

Multiple choice

$I. (x^3 - 5260 = 5388) II. (y^3 - 4867 - 4394 = 0)$ Solve the 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 can not be established
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Solve for x: x³ = 5260 + 5388 = 10648. The cube root of 10648 is x = 22 (since 22³ = 10648). Solve for y: y³ = 4867 + 4394 = 9261. The cube root of 9261 is y = 21 (since 21³ = 9261). Comparing: x = 22 > y = 21, so x > y.

Multiple choice

In each of the following questions, two equations are given. You have to solve them and I. (x^3 - 528 = 2216) II. (y^2 - 1825 + 1629 = 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: x³ - 528 = 2216, so x³ = 2744, x = ∛2744 = 14. Equation II: y² - 1825 + 1629 = 0, so y² - 196 = 0, y² = 196, y = ±14. Since x = 14 and y = 14 or -14, we have x ≥ y (14 is greater than or equal to both 14 and -14). Option B is correct.

Multiple choice

In each of the following questions, two equations are given. You have to solve them and answer the following - $I. (10x^2 - 55x + 70 = 0)$ $II. (y^2 - 2.25 = 0)$

  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
A Correct answer
Explanation

Solving equation I: 10x² - 55x + 70 = 0. Dividing by 5: 2x² - 11x + 14 = 0. Factorizing: (2x - 7)(x - 2) = 0, so x = 3.5 or x = 2. Solving equation II: y² - 2.25 = 0, so y² = 2.25, giving y = ±1.5. Since both values of x (2 and 3.5) are greater than both values of y (-1.5 and 1.5), we conclude x > y always.