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 these questions, two equations are given. You have to solve these equations and find out the values of X and Y and give answers. I. (3x^2 - 5x - 12 = 0) II. (3y^2 - 19y + 28 = 0)

  1. If X < Y

  2. If X > Y

  3. If X ≤ Y

  4. If X ≥ Y

  5. If X=Y or no relation can be established.
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Equation I: 3x² - 5x - 12 = (3x+4)(x-3) = 0, so x = -4/3 or 3. Equation II: 3y² - 19y + 28 = 0 gives y = 4 or 7/3. Comparing: x=-4/3 < both y values, but x=3 < 4 while x=3 > 7/3. The inconsistent relationship means no definite relation can be established.

Multiple choice

In each of these questions, two equations are given. You have to solve these equations and find out the values of X and Y and give answers. I. (4x^2 - 9x - 9 = 0) II. (3y^2 - 19y - 14 = 0)

  1. If X < Y

  2. If X > Y

  3. If X ≤ Y

  4. If X ≥ Y

  5. If X = Y or no relation can be established.
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Equation I: 4x² - 9x - 9 = 0 gives x = 3 or -3/4 (discriminant 225, sqrt 15). Equation II: 3y² - 19y - 14 = 0 gives y = 7 or -2/3 (discriminant 529, sqrt 23). Checking comparisons: x=3 > -2/3 but x=3 < 7; x=-3/4 < 7 but x=-3/4 < -2/3. The mixed relationships mean no definite relation exists.

Multiple choice

Solve the following equations and find the values of (x) and (y): I. (4x^2 + 19x + 21 = 0) II. (15y^2 - 29y + 12 = 0)

  1. If X < Y

  2. If X > Y

  3. If X ≤ Y

  4. If X ≥ Y

  5. If X=Y or no relation can be established
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Equation I: 4x² + 19x + 21 = (4x+7)(x+3) = 0, so x = -7/4 or -3. Equation II: 15y² - 29y + 12 = 0 gives y = 4/3 or 3/5 (discriminant 121, sqrt 11). Since x values are -1.75 and -3, while y values are 0.6 and 1.33, every x is less than every y. Therefore X < Y is always true.

Multiple choice

In each of these questions, two equations are given. You have to solve these equations and find out the values of X and Y and give answers. I. (3x^2 + 6x - 9 = 0) II. (8y^2 + 46y - 25 = 0)

  1. If X < Y

  2. If X > Y

  3. If X ≤ Y

  4. If X ≥ Y

  5. If X = Y or no relation can be established.
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Equation I: 3x² + 6x - 9 = 3(x²+2x-3) = 3(x+3)(x-1) = 0, so x = -3 or 1. Equation II: 8y² + 46y - 25 = 0 gives y = 0.5 or -6.25 (discriminant 2916, sqrt 54). When x=-3: x > -6.25 but x < 0.5. When x=1: x > -6.25 and x > 0.5. The inconsistent relationship means no definite relation can be established.

Multiple choice

In the following questions, two equations (x) and (y) are given. You have to solve both the equations and give answer. (I.) (2x^2 - x - 66 = 0) (II.) (3y^2 - 5y - 42 = 0)

  1. x > y

  2. x ≥ y

  3. x < y

  4. x ≤ y

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

Equation I: 2x² - x - 66 = 0. Factoring: (2x + 11)(x - 6) = 0, so x = -5.5 or x = 6. Equation II: 3y² - 5y - 42 = 0. Factoring: (3y + 7)(y - 6) = 0, so y = -7/3 ≈ -2.33 or y = 6. When x = 6, y = 6, we have x = y. When x = 6, y = -2.33, we have x > y. When x = -5.5, y = 6, we have x < y. When x = -5.5, y = -2.33, we have x < y. Since the relationship varies (sometimes x > y, sometimes x < y, sometimes x = y), the correct answer is 'the relation 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: x^2 + (59319)^{1/3} = (529 \div 23)\times8 - 24 ) (II: y^2 - 26y + 169 = 0)

  1. X > Y

  2. X ≥ Y

  3. X < Y

  4. X ≤ Y

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

Equation I: x² + 59319^(1/3) = (529 ÷ 23) × 8 - 24. 59319^(1/3) = 39. 529 ÷ 23 = 23. So x² + 39 = 23 × 8 - 24 = 184 - 24 = 160. Therefore x² = 121, so x = ±11. Equation II: y² - 26y + 169 = 0 gives (y - 13)² = 0, so y = 13. Comparing x = -11 or 11 with y = 13, we get 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 - 19x + 78 = 0) II: (y^2 + 14y - 51 = 0)

  1. X > Y

  2. X ≥ Y

  3. X < Y

  4. X ≤ Y

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

Equation I: x² - 19x + 78 = 0 factors to (x - 6)(x - 13) = 0, so x = 6 or 13. Equation II: y² + 14y - 51 = 0 factors to (y + 17)(y - 3) = 0, so y = 3 or -17. Comparing: if x = 6, it's > both y values; if x = 13, it's also > both y values. Therefore X > Y is always true.

Multiple choice

In the following questions, two equations I and II are given. You have to solve both the equations and give answer. (I: 35x^2 - 3x - 2 = 0 ) (II: 49y^2 + 7y - 6 = 0)

  1. X > Y

  2. X ≥ Y

  3. X < Y

  4. X ≤ Y

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

Equation I: 35x² - 3x - 2 = 0. Using quadratic formula: x = (3 ± √(9 + 280))/70 = (3 ± √289)/70 = (3 ± 17)/70. So x = 20/70 = 2/7 ≈ 0.286 or x = -14/70 = -1/5 = -0.2. Equation II: 49y² + 7y - 6 = 0 gives y = (-7 ± √(49 + 1176))/98 = (-7 ± √1225)/98 = (-7 ± 35)/98. So y = 28/98 = 2/7 ≈ 0.286 or y = -42/98 = -3/7 ≈ -0.429. The ranges overlap: x ∈ [-0.2, 0.286], y ∈ [-0.429, 0.286]. Since x can equal y (2/7), but not always be greater or less, the relationship 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: x^2 - 12x + 35 = 0 ) (II: y^2 - 8y + 15 = 0)

  1. X > Y

  2. X ≥ Y

  3. X < Y

  4. X ≤ Y

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

Equation I: x² - 12x + 35 = 0 factors to (x - 5)(x - 7) = 0, so x = 5 or 7. Equation II: y² - 8y + 15 = 0 factors to (y - 3)(y - 5) = 0, so y = 3 or 5. Comparing all combinations: if x = 5, it equals y when y = 5 but is greater than y = 3; if x = 7, it's greater than both y values. So X is always ≥ Y, meaning X ≥ Y is correct.

Multiple choice

In the following questions, two equations I and II are given. You have to solve both the equations and give answer. (I: 5x^2 + 19x + 12 = 0 ) (II: 32y^2 - 28y + 3 = 0)

  1. X > Y

  2. X ≥ Y

  3. X < Y

  4. X ≤ Y

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

Equation I: 5x² + 19x + 12 = 0. Discriminant = 361 - 240 = 121. x = (-19 ± 11)/10. So x = -8/10 = -0.8 or x = -30/10 = -3. Equation II: 32y² - 28y + 3 = 0. Discriminant = 784 - 384 = 400. y = (28 ± 20)/64. So y = 48/64 = 3/4 = 0.75 or y = 8/64 = 1/8 = 0.125. Comparing: x = -0.8 or -3, y = 0.125 or 0.75. All x values are less than all y values. Therefore X < Y.

Multiple choice

In each of the following questions, two equations are given. You have to solve them and give answer- ( I. 7x - 4y = 29 )( II. 5x+3y-50 = 0 )

  1. If x<y

  2. If x≤y

  3. If x=y or relationship cannot be established

  4. If x>y

  5. If x≥y
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Solving the system: From I, 7x = 29 + 4y, so x = (29 + 4y)/7. Substituting in II: 5(29 + 4y)/7 + 3y = 50. This gives 145 + 20y + 21y = 350, so 41y = 205 and y = 5. Then x = (29 + 20)/7 = 7. Since x = 7 and y = 5, we have 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. $(\text{(I)} x^2 – 3.3x – 2.6 = 0 )(\text{(II)} y^2 + 3y + 2 = 0)$

  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
E Correct answer
Explanation

Equation I: x² - 3.3x - 2.6 = 0. Using quadratic formula: x = [3.3 ± √(10.89 + 10.4)]/2 = [3.3 ± √21.29]/2 = [3.3 ± 4.614]/2. Solutions: x₁ ≈ 3.957, x₂ ≈ -0.657. Equation II: y² + 3y + 2 = 0 factors as (y+1)(y+2) = 0. Solutions: y = -1, -2. Comparison: x₁ (3.957) > y values, but x₂ (-0.657) > y values? Actually -0.657 > -1 and -0.657 > -2, so both x solutions are greater than both y solutions. However, the question asks about x and y as values from these equations. Since we have multiple possible values for both x and y, and the relationship depends on which specific values we compare, the relationship cannot be uniquely determined from the given information alone.

Multiple choice

In the following questions two equations I and II are given. You have to solve both the equations and give answer. $(I) (x^2 - 8.7x + 18.5 = 0) (II) (y^2 - 9 = 0)$

  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
A Correct answer
Explanation

Equation I: x² - 8.7x + 18.5 = 0. Using quadratic formula: x = [8.7 ± √(75.69 - 74)]/2 = [8.7 ± √1.69]/2 = [8.7 ± 1.3]/2. Solutions: x₁ = (8.7 + 1.3)/2 = 5, x₂ = (8.7 - 1.3)/2 = 3.7. Equation II: y² - 9 = 0, so y² = 9, giving y = ±3. Comparing: Both x values (3.7 and 5) are greater than both y values (3 and -3). Therefore x > y consistently.

Multiple choice

In each of the following questions two equations are given. Solve these equations and give answer : (i) (x^2 - 5(\sqrt{2} + \sqrt{5})x + 25\sqrt{10} = 0) (ii) (y^2 - \sqrt{2}(\sqrt{3} + 2\sqrt{2})y + 4\sqrt{6} = 0)

  1. If x < y

  2. If x > y

  3. If x ≥ y

  4. If x ≤ y

  5. If x = y or no relationship can be established between x and y
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For equation (i): Use quadratic formula or factor. The quadratic is x^2 - 5(√2+√5)x + 25√10 = 0. By Vieta's formulas, sum = 5(√2+√5), product = 25√10. The roots are 5√2 and 5√5 (approximately 7.07 and 11.18). For equation (ii): y^2 - √2(√3+2√2)y + 4√6 = 0. Roots are 2√2 and 2√3 (approximately 2.83 and 3.46). Both roots of x are greater than both roots of y, so x > y always. Option B is correct.

Multiple choice

In each of the following questions two equations are given. Solve these equations and give answer : (i) (x^2 - 15\sqrt{3}x + 162 = 0) (ii) (y^2 - 9\sqrt{2}y + 36 = 0)

  1. If x < y

  2. If x > y

  3. If x ≥ y

  4. If x ≤ y

  5. If x = y or no relationship can be established between x and y
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For equation (i): x^2 - 15√3 x + 162 = 0. The roots are x = [15√3 ± √(675 - 648)]/2 = [15√3 ± √27]/2 = [15√3 ± 3√3]/2. Roots are x = 18√3/2 = 9√3 ≈ 15.59 and x = 12√3/2 = 6√3 ≈ 10.39. For equation (ii): y^2 - 9√2 y + 36 = 0. Using quadratic formula: y = [9√2 ± √(162 - 144)]/2 = [9√2 ± √18]/2 = [9√2 ± 3√2]/2. Roots are y = 12√2/2 = 6√2 ≈ 8.49 and y = 6√2/2 = 3√2 ≈ 4.24. Both roots of x (≈15.59, ≈10.39) are greater than both roots of y (≈8.49, ≈4.24). So x > y always. Option B is correct.