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 the given question two equations numbered (I) and (II) are given. You have to solve both the equations and give answer.\nI. (20p^2 + 31p + 12 = 0) \nII. (21q^2 - 23q + 6 = 0)

  1. If p > q

  2. If p < q

  3. If p ≥ q

  4. If p ≤ q

  5. If p = q or no relation between p and q can be established
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For equation I: 20p² + 31p + 12 = 0 factors as (4p + 3)(5p + 4) = 0, giving p = -3/4 and p = -4/5. For equation II: 21q² - 23q + 6 = 0 factors as (3q - 2)(7q - 3) = 0, giving q = 2/3 and q = 3/7. Comparing the values: both roots of p (-3/4, -4/5) are less than both roots of q (2/3, 3/7). Therefore p < q.

Multiple choice

In the given question two equations numbered (I) and (II) are given. You have to solve both the equations and give answer.(I. \quad 20p^2 – 17p + 3 = 0 ) (II. \quad 20q^2 – 9q + 1 = 0)

  1. If p > q

  2. If p < q

  3. If p ≥ q

  4. If p ≤ q

  5. If p = q or no relation between p and q can be established
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For equation I: 20p² - 17p + 3 = 0 factors as (4p - 1)(5p - 3) = 0, giving p = 1/4 and p = 3/5. For equation II: 20q² - 9q + 1 = 0 factors as (4q - 1)(5q - 1) = 0, giving q = 1/4 and q = 1/5. When p = 3/5 and q = 1/5, p > q. When p = 1/4 and q = 1/4, p = q. Therefore p ≥ q is always true.

Multiple choice

In each of the following questions, two equations are given. You have to solve them and answer the question.( (I) 15x^{2} – 34x + 15 = 0 ) ( (II) 4y^{2} – 29y + 45 = 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

Factorizing equation (I): 15x² - 34x + 15 = (3x - 5)(5x - 3) = 0, so x = 5/3 or x = 3. Factorizing equation (II): 4y² - 29y + 45 = (y - 5)(4y - 9) = 0, so y = 5 or y = 9/4 = 2.25. Comparing values: when x = 5/3 (1.67), both y values (2.25, 5) are greater. When x = 3, both y values (2.25, 5) are greater. Therefore x < y is always true.

Multiple choice

In each of the following questions, two equations are given. You have to solve them and answer the question.( (I) x^2 – 21x + 104 = 0 )( (II) y^2 – 28y +195 = 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
B Correct answer
Explanation

Factorizing equation (I): x² - 21x + 104 = (x - 13)(x - 8) = 0, so x = 13 or x = 8. Factorizing equation (II): y² - 28y + 195 = (y - 15)(y - 13) = 0, so y = 15 or y = 13. Comparing values: when x = 8, both y values (13, 15) are greater (x < y). When x = 13, y can be 13 (x = y) or 15 (x < y). Since x is never greater than y and x = y is possible, x ≤ y is correct.

Multiple choice

In each of the following questions, two equations are given. You have to solve them and answer the question. (I) (2x^2 + 17x + 30 = 0) (II) (4y^2 – 13y – 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
A Correct answer
Explanation

Factorizing equation (I): 2x² + 17x + 30 = (2x + 5)(x + 6) = 0, so x = -5/2 = -2.5 or x = -6. Factorizing equation (II): 4y² - 13y - 12 = (4y + 3)(y - 4) = 0, so y = -3/4 = -0.75 or y = 4. Comparing values: when x = -2.5, both y values (-0.75, 4) are greater. When x = -6, both y values (-0.75, 4) are greater. Therefore x < y for all combinations.

Multiple choice

In each of the following questions, two equations are given. You have to solve them and answer the question. (I) (4x^2 – 9x – 9 = 0) (II) (2y^2 – 17y + 21 = 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

Factorizing equation (I): 4x² - 9x - 9 = (4x + 3)(x - 3) = 0, so x = -3/4 = -0.75 or x = 3. Factorizing equation (II): 2y² - 17y + 21 = (2y - 3)(y - 7) = 0, so y = 3/2 = 1.5 or y = 7. Comparing values: when x = -0.75, both y values (1.5, 7) are greater. When x = 3, both y values (1.5, 7) are greater. Wait - x = 3 and y = 1.5 means x > y! Let me recheck: 3 > 1.5 is true, so relationship cannot be established because sometimes x < y and sometimes x > y.

Multiple choice

In each of the following questions, two equations are given. You have to solve them and answer the question. (I) (x^{2} - 87x - 270 = 0) (II) (7y^{2} - 11y - 18 = 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

Factorizing equation (I): x² - 87x - 270 = (x - 90)(x + 3) = 0, so x = 90 or x = -3. Factorizing equation (II): 7y² - 11y - 18 = (7y + 9)(y - 2) = 0, so y = -9/7 ≈ -1.29 or y = 2. Comparing all four combinations: (x=90, y=-1.29): x > y; (x=90, y=2): x > y; (x=-3, y=-1.29): x < y; (x=-3, y=2): x < y. Since sometimes x > y and sometimes x < y, no consistent relationship exists.

Multiple choice

In each of the following question, two equation (I) and (II) is given. You have to solve them and answer the question. ( (I) x^2 + 5x + 6 = 0 )( (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 Relationship cannot be established
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Solve equation I: x² + 5x + 6 = 0 factors to (x+2)(x+3) = 0, so x = -2 or -3. Solve equation II: y² + 3y + 2 = 0 factors to (y+1)(y+2) = 0, so y = -1 or -2. Comparing values: x = -3, -2 and y = -2, -1. Both values of x are less than or equal to the corresponding y values, specifically x ≤ y holds.

Multiple choice

(\text{In each of the following question, two equation (I) and (II) is given. You have to solve them and answer the question}\(I) 12x^2 + 11x + 12 = 10x^2 + 22x \qquad (II) 13y^2 – 18y + 3 = 9y^2 – 10y)

  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

Simplify equation I: 12x² + 11x + 12 = 10x² + 22x gives 2x² - 11x + 12 = 0, which factors to (2x-3)(x-4) = 0, so x = 1.5 or 4. Simplify equation II: 13y² - 18y + 3 = 9y² - 10y gives 4y² - 8y + 3 = 0, which factors to (2y-1)(2y-3) = 0, so y = 0.5 or 1.5. Comparing: x values (1.5, 4) are greater than or equal to y values (0.5, 1.5).

Multiple choice

In each of the following question, two equation (I) and (II) is given. You have to solve them and answer the question. ( (I) x^2 – 11x + 24 = 0 ) ( (II) 2y^2 – 9y + 9 = 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
B Correct answer
Explanation

Solve equation I: x² - 11x + 24 = 0 factors to (x-3)(x-8) = 0, so x = 3 or 8. Solve equation II: 2y² - 9y + 9 = 0 factors to (2y-3)(y-3) = 0, so y = 1.5 or 3. Comparing values: when y = 1.5, x (3, 8) > y. When y = 3, x can be 3 or 8. So x ≥ y is correct since x is always greater than or equal to y.

Multiple choice

In this question two equations numbered I and II are given. You have to solve both the equations and state the correct relationship. $I. (12p^2 - 17p + 6 = 0) II. (q^2 - 16q + 63 = 0)$

  1. If p > q

  2. If p ≥ q

  3. If p < q

  4. If p ≤ q

  5. If p = q or no relation can be established.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Equation I factors to (3p-2)(4p-3)=0, giving p=2/3 and p=3/4. Equation II factors to (q-7)(q-9)=0, giving q=7 and q=9. Since all values of p (0.67, 0.75) are less than all values of q (7, 9), p is always less than q.

Multiple choice

In each of the following questions, two equations are given. You have to solve them and state the correct relationship- $I. (3x^2 - 23x + 40 = 0)$ $II. (2y^2 - 23y + 66 = 0)$

  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

Solving 3x² - 23x + 40 = 0 gives x = 5, 8/3 (approximately 5, 2.67). Solving 2y² - 23y + 66 = 0 gives y = 6, 11/2 (approximately 6, 5.5). Comparing: 5 < 6, 5 < 5.5, but 8/3 (2.67) < 5.5 and 2.67 < 6. Since some x values are NOT less than some y values (5 > 5.5 is false, but all x < all y is true). Actually checking: 5 < 5.5 ✓, 2.67 < 5.5 ✓, 5 < 6 ✓, 2.67 < 6 ✓. All x values are indeed less than all y values, so x < y is correct.

Multiple choice

In each of the following questions, two equations are given. You have to solve them and $I. (5x - 2y = 21) II. (3x + 4y = 49)$

  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

Solving the system: 5x-2y=21 and 3x+4y=49. Multiply first equation by 2: 10x-4y=42. Add to second: 13x=91, so x=7. Substituting: 5(7)-2y=21 gives y=7. Since x=y, the answer is 'x=y or relationship cannot be established' when considering the equality condition.