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 (I) and (II) are given. You have to solve both the equations and give answer. $I. (8x + 13y = 62) II. (13x - 17y + 128 = 0)$

  1. x > y

  2. x < y

  3. x ≥ y

  4. x ≤ y

  5. x = y, or relation can’t be established
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Solving the system: 8x + 13y = 62 and 13x - 17y = -128. From the first equation, x = (62-13y)/8. Substituting into the second gives 13(62-13y)/8 - 17y = -128. Solving yields y = 6, and substituting back gives x = -2. Therefore, x < y.

Multiple choice

In each of the following questions two equations are given. You have to solve them and give answer - (I) (a^2 + a = 56) (II) (b^2 - 17b + 72 = 0)

  1. If a < b

  2. If a > b

  3. If a ≤ b

  4. If a ≥ b

  5. If a = b or relationship can not be determined
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Equation I: a² + a - 56 = 0 gives a = [-1 ± √(1 + 224)]/2 = [-1 ± 15]/2, so a = 7 or a = -8. Equation II: b² - 17b + 72 = 0 gives b = [17 ± √(289 - 288)]/2 = [17 ± 1]/2, so b = 9 or b = 8. For a = -8, a < both b values. For a = 7, a < both b values. Thus a < b always.

Multiple choice

In each of these question, two equations (I) and (II) are given. You have to solve both the equations and give answer.(I. 42x^{2} + 378x - 1512 = 0\II. 84y^{2} + 262y - 420 = 0)

  1. x > y

  2. x < y

  3. x ≤ y

  4. x ≥ y

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

Simplify first equation: 42x² + 378x - 1512 = 0 gives x² + 9x - 36 = 0, roots are x = 3, x = -12. Second equation: 84y² + 262y - 420 = 0. The roots are approximately y ≈ 0.337 and y ≈ -14.832. Since x = 3 > both y values but x = -12 < both y values, the relationship cannot be established.

Multiple choice

$I. (3x^2 + 14x + 15 = 0) II. (3y^2 – 13y + 14 = 0)$ Solve both the equations and determine the relationship between x and y.

  1. If x > y

  2. If x < y

  3. If x = y or the relationship between x and y can’t be established.

  4. If x ≥ y

  5. If x ≤ y
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Solving 3x² + 14x + 15 = 0 gives x = -5/3 or x = -3. Solving 3y² - 13y + 14 = 0 gives y = 7/3 or y = 2. Comparing all possible values: all x values (-1.67, -3) are less than all y values (2.33, 2). Therefore, x < y is correct.

Multiple choice

(\text{sin}\alpha \text{ and } \text{cos}\alpha \text{ are the roots of the equation } x^2 - abx + c = 0 \text{ then})
  1. $\(a^2c^2 - 2b - 1 = 0\)$
  2. $\(a^2 - ac + b^2 = 0\)$
  3. $\(a^2b^2 - 2c - 1 = 0\)$
  4. $\(a^2 - ab + ac = 0\)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given sin α and cos α are roots of x² - abx + c = 0. By Vieta's formulas: sum of roots = sin α + cos α = ab, product = sin α × cos α = c. We know sin²α + cos²α = 1. Using (sin α + cos α)² = sin²α + cos²α + 2sin α cos α, we get (ab)² = 1 + 2c, which simplifies to a²b² - 2c - 1 = 0. Option C is correct.

Multiple choice

In each question two equations numbered I and II are given. You have to solve both the equations and mark the appropriate answer.( )(\text{(I)}) (6p^2 + 13p – 169 = 0)(\text{(II)}) (q^2 + 8q – 65 = 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 between p and q.
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Solving equation (I) gives p = 6.5 or p = -13/3 ≈ -4.33. Solving equation (II) gives q = 5 or q = -13. Since p can be greater than q (when p = 6.5 and q = 5 or -13) but can also be less than q (when p = -4.33 and q = 5), no consistent relationship exists between p and q.

Multiple choice

In each question two equations numbered I and II are given. You have to solve both the equations and mark the appropriate answer. (\text{(I)}) (p^2 - 5p + 4 = 0 )(\text{(II)}) (q^2 + 11q - 12 = 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 between p and q.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Solving equation (I) gives p = 1 or p = 4. Solving equation (II) gives q = 1 or q = -12. When p = 1 and q = 1, p = q. When p = 4 and q = 1, p > q. When p = 4 and q = -12, p > q. When p = 1 and q = -12, p > q. Therefore p ≥ q is always true.

Multiple choice

In each question two equations numbered I and II are given. You have to solve both the equations and mark the appropriate answer. ()(\textbf{(I)}) (8p^2 + 50p + 57 = 0)(\textbf{(II)}) (6q^2 – q – 57 = 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 between p and q.
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Solving equation (I) gives p = -19/4 ≈ -4.75 or p = -3/2 = -1.5. Solving equation (II) gives q = 3 or q = -19/6 ≈ -3.17. Comparing values: when p = -4.75, p < q (both for q = 3 and q = -3.17). When p = -1.5, p > q = -3.17 but p < q = 3. Since p can be greater or less than q depending on which roots we compare, no consistent relationship exists.

Multiple choice

In the following questions , two equations are given. You have to solve both the equations and give answer as - (x^{2} = 289), (y^{3} = 4913)
  1. $\(x > y\)$
  2. $\(x < y\)$
  3. $\(x \ge y\)$
  4. $\(x \le y\)$
  5. $\(x = y\) or The relation cann’t be established$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

x² = 289 gives x = ±17 (both +17 and -17 are valid solutions). y³ = 4913 gives y = 17 (only one real cube root). Comparing the possible values: if x = +17, then x = y; if x = -17, then x < y. Therefore, x is always less than or equal to y, making option D (x ≤ y) correct.

Multiple choice

In each of the following questions two equations are given. You have to solve them and give answer - ( (I) -x^2 + 4x + 12 = 0 ) ( (II) 4y^2 – 4y + 1 = 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 determined
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

For -x² + 4x + 12 = 0, multiply by -1: x² - 4x - 12 = 0, so (x-6)(x+2) = 0, giving x = 6 or x = -2. For 4y² - 4y + 1 = 0, this is (2y-1)² = 0, so y = 0.5. Comparing: if x = 6, x > y (6 > 0.5); if x = -2, x < y (-2 < 0.5). Since x can be greater or less than y depending on which root we take, the relationship cannot be determined.

Multiple choice

In each question two equations numbered I and II are given. You have to solve both the equations and mark the appropriate answer. (I) (3p + 5q = 4) (II) (6p – 7q = 25)

  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 between p and q
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Solve the system of equations. From (I): 3p + 5q = 4. From (II): 6p - 7q = 25. Multiply (I) by 2: 6p + 10q = 8. Subtract this from (II): (6p - 7q) - (6p + 10q) = 25 - 8, which gives -17q = 17, so q = -1. Substitute q = -1 in (I): 3p + 5(-1) = 4, so 3p - 5 = 4, giving 3p = 9 and p = 3. Since p = 3 > q = -1, option A (p > q) is correct.

Multiple choice

In each of the given question below, there are two equations as I and II. You have to solve both equations and give answer- I. (x^2 + 3x – 4 = 0) II. (3y^2 – 10y + 8 = 0)

  1. If x > y

  2. If x < y

  3. If x ≥ y

  4. If x ≤ y

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

Solving equation I: x² + 3x - 4 = 0 factors to (x+4)(x-1) = 0, giving x = -4 or x = 1. Solving equation II: 3y² - 10y + 8 = 0 factors to (3y-4)(y-2) = 0, giving y = 4/3 or y = 2. Comparing all values: both y values (2, 4/3) are greater than both x values (-4, 1), so x < y is correct.

Multiple choice

In each of the following questions, two equations I and II have been given. Solve these questions and answer I. (30x^2 + 11x + 1 = 0) II. (42y^2 + 13y + 1 = 0)

  1. If x < y

  2. If x ≤ y

  3. If x = y or the relation cannot be established

  4. If x ≥ y

  5. If x > y
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Equation I: 30x² + 11x + 1 = 0 factors to (5x+1)(6x+1)=0, giving x = -1/5 ≈ -0.2 or x = -1/6 ≈ -0.167. Equation II: 42y² + 13y + 1 = 0 factors to (6y+1)(7y+1)=0, giving y = -1/6 ≈ -0.167 or y = -1/7 ≈ -0.143. Comparing the values: x's maximum is -0.167 which equals y's minimum, and x's minimum -0.2 is less than both y values. Therefore x ≤ y is always true.

Multiple choice

In each of the following questions, two equations I and II have been given. Solve these questions and answer I. 9x2 + 3x – 2 = 0 II. 8y2 + 6y + 1 = 0

  1. If x < y

  2. If x ≤ y

  3. If x = y or the relation cannot be established

  4. If x ≥ y

  5. If x > y
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Equation I: 9x² + 3x - 2 = 0 gives x = 1/3 ≈ 0.33 or x = -2/3 ≈ -0.67. Equation II: 8y² + 6y + 1 = 0 gives y = -1/4 = -0.25 or y = -1/2 = -0.5. When x = 0.33, x > both y values. When x = -0.67, x < both y values. Since x can be greater than OR less than y depending on which roots you compare, the relationship cannot be established.