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 l and II are given. Solve both the equations and mark the appropriate answer. $I. (x^2 + 2x - 35 = 0) II. (y^2 - 2y - 48 = 0)$
  1. $\(x > y\)$
  2. $\(x < y\)$
  3. $\(x \ge y\)$
  4. $\(x \le y\)$

  5. x = y or relation between x and y can not be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Equation I: x² + 2x - 35 = (x+7)(x-5) = 0, so x = -7 or 5. Equation II: y² - 2y - 48 = (y-8)(y+6) = 0, so y = 8 or -6. Comparing: x can be > y (5 > -6), x < y (-7 < 8), or x = y (no match). No consistent relation.

Multiple choice

In the given question, two equations numbered l and II are given. Solve both the equations and mark the appropriate answer. $I. (x^2 - 13x - 30 = 0) II. (y^2 + 11y + 24 = 0)$
  1. $\(x > y\)$
  2. $\(x < y\)$
  3. $\(x \ge y\)$
  4. $\(x \le y\)$

  5. x = y or relation between x and y can not be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Equation I: x² - 13x - 30 = (x-15)(x+2) = 0, so x = 15 or -2. Equation II: y² + 11y + 24 = (y+3)(y+8) = 0, so y = -3 or -8. Comparing: x can be > y (15 > -3, -2 > -3), x < y (15 < -8 is false, -2 < -8 is false), or x = y (no match). No consistent relation.

Multiple choice

In each of these questions, two equations (a) and (b) are given. You have to solve both the equations and choose the correct option. ( (a) x^2 – x - 110 =0 ) ( (b) y^2 + 24 y +140 =0 )
  1. $x > y$
  2. $x ≤ y$
  3. $x ≥ y$
  4. $x < y$

  5. x = y or relationship between x and y can't be established
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For x² - x - 110 = 0, factors are (x-11)(x+10) = 0, giving x = 11 or x = -10. For y² + 24y + 140 = 0, factors are (y+14)(y+10) = 0, giving y = -10 or y = -14. Since x can be 11 or -10, and y can be -14 or -10, the relationship x ≥ y holds for all combinations.

Multiple choice

In each of these questions, two equations (a) and (b) are given. You have to solve both the equations and choose the correct option.( (a) x^2 – 50x +589 =0 ) ( (b) y^2 -19y -416 =0 )

  1. x > y

  2. x ≤ y

  3. x ≥ y

  4. x < y

  5. x = y or relationship between x and y can't be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

For x² - 50x + 589 = 0, factors are (x-31)(x-19) = 0, giving x = 31 or x = 19. For y² - 19y - 416 = 0, factors are (y-32)(y+13) = 0, giving y = 32 or y = -13. When x = 31 and y = 32, we have x < y, but when x = 31 and y = -13, we have x > y. Since no single relationship holds consistently, the answer is that the relationship cannot be established.

Multiple choice

In each of these questions, two equations (a) and (b) are given. You have to solve both the equations and choose the correct option. (a) (x^2 - 53x + 462 = 0) (b) (y^2 - 88y + 1932 = 0)

  1. x > y

  2. x ≤ y

  3. x ≥ y

  4. x < y

  5. x = y or relationship between x and y can't be established
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For x² - 53x + 462 = 0, factors are (x-42)(x-11) = 0, giving x = 42 or x = 11. For y² - 88y + 1932 = 0, factors are (y-42)(y-46) = 0, giving y = 42 or y = 46. When x = 42 and y = 42, we have x = y. When x = 11 and y = 42 or 46, we have x < y. Since x is never greater than y, x ≤ y is correct.

Multiple choice

In each of these questions, two equations (a) and (b) are given. You have to solve both the equations and choose the correct option. (a) (x^{3} = -9261) (b) (y^{2} = 441)
  1. $x > y$
  2. $x ≤ y$
  3. $x ≥ y$
  4. $x < y$
  5. $x = y or relationship between x and y can't be established$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Solving x³ = -9261 gives x = -21 (cube root of negative is negative). Solving y² = 441 gives y = ±21. Comparing: x = -21 and y can be 21 or -21. When y = 21: x < y. When y = -21: x = y. Therefore x ≤ y is always true.

Multiple choice

In each of these questions, two equations (a) and (b) are given. You have to solve both the equations and choose the correct option. a) (x^2 - x - 20 = 0) b) (y^2 + y - 30 = 0)

  1. x > y

  2. x ≤ y

  3. x ≥ y

  4. x < y

  5. x = y or relationship between x and y can't be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Solve equation (a): x² - x - 20 = 0 factors to (x-5)(x+4) = 0, so x = 5 or x = -4. Solve equation (b): y² + y - 30 = 0 factors to (y+6)(y-5) = 0, so y = -6 or y = 5. Comparing values: when x=5, y can be 5 (x=y) or -6 (x>y); when x=-4, y can be 5 (xy). Since we have x=y possible and also x>y and x

Multiple choice

In each of these questions, two equations (a) and (b) are given. You have to solve both the equations and choose the correct option. a) (x^2 – 13.5x + 38 = 0) b) (y^2 – 1.5y – 10 = 0)

  1. x > y

  2. x ≤ y

  3. x ≥ y

  4. x < y

  5. x = y or relationship between x and y can't be established
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Solve equation (a): x² - 13.5x + 38 = 0. Using quadratic formula or factoring: x = (13.5 ± √(13.5² - 4×38))/2 = (13.5 ± √(182.25 - 152))/2 = (13.5 ± √30.25)/2 = (13.5 ± 5.5)/2. So x = 19/2 = 9.5 or x = 8/2 = 4. Solve equation (b): y² - 1.5y - 10 = 0 gives y = (1.5 ± √(2.25 + 40))/2 = (1.5 ± √42.25)/2 = (1.5 ± 6.5)/2. So y = 8/2 = 4 or y = -5/2 = -2.5. All x values (9.5, 4) are greater than or equal to all y values (4, -2.5). Since x=4 and y=4 is possible (x=y) and x=9.5 > all y values, x ≥ y is correct.

Multiple choice

In each of these questions, two equations (a) and (b) are given. You have to solve both the equations and choose the correct option. (a) 5x^2 - 13x - 28 = 0 ) (b) 3y^2 - 20y + 32 = 0

  1. x > y

  2. x ≤ y

  3. x ≥ y

  4. x < y

  5. x = y or relationship between x and y can't be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Solving (a): 5x²-13x-28=0 gives x=4 or x=-7/5. Solving (b): 3y²-20y+32=0 gives y=4 or y=8/3. When x=4, y=4 gives x=y. When x=-7/5, y=4 gives x

Multiple choice

In each of these questions, two equations (a) and (b) are given. You have to solve both the equations and choose the correct option. a) (x^2 - 1.2x - 1.6 = 0) b) (y^2 + 2.3y + 1.2 = 0)

  1. x > y

  2. x ≤ y

  3. x ≥ y

  4. x < y

  5. x = y or relationship between x and y can't be established
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Solving x² - 1.2x - 1.6 = 0 gives x = 2, -0.8. Solving y² + 2.3y + 1.2 = 0 gives y = -0.8, -1.5. Since x = 2 > y = -1.5 but x = -0.8 = y = -0.8, the relationship is x ≥ y.

Multiple choice

In each of these questions, two equations (a) and (b) are given. You have to solve both the equations and choose the correct option. a) (4x^2 + x - 14 = 0) b) (y^2 - 15y + 56 = 0)

  1. x > y

  2. x ≤ y

  3. x ≥ y

  4. x < y

  5. x = y or relationship between x and y can't be established
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Solving 4x² + x - 14 = 0 gives x = 1.75, -2. Solving y² - 15y + 56 = 0 gives y = 7, 8. Both positive roots of x (1.75) are less than both positive roots of y (7, 8), so x < y.

Multiple choice

In each of these questions, two equations (a) and (b) are given. You have to solve both the equations and choose the correct option. a) (x^2 - 32x + 247 = 0) b) (3y^2 + 13y + 12 = 0)

  1. x > y

  2. x ≤ y

  3. x ≥ y

  4. x < y

  5. x = y or relationship between x and y can't be established
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Solving equation (a): x^2-32x+247=0 gives x=(32±√(1024-988))/2=(32±√36)/2=(32±6)/2, so x=19 or x=13. Solving equation (b): 3y^2+13y+12=0 gives y=(-13±√(169-144))/6=(-13±5)/6, so y=-4/3 or y=-3. Since x values (13, 19) are always greater than y values (-4/3, -3), x > y is always true. Option A is correct.

Multiple choice

In each of these questions, two equations (a) and (b) are given. You have to solve both the equations and choose the correct option. a) (x^2 - 5x + 4 = 0) b) (y^2 + 11y - 12= 0)

  1. x > y

  2. x ≤ y

  3. x ≥ y

  4. x < y

  5. x = y or relationship between x and y can't be established
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Solving equation (a): x²-5x+4=0 factors to (x-1)(x-4)=0, so x=1 or x=4. Solving equation (b): y²+11y-12=0 factors to (y+12)(y-1)=0, so y=-12 or y=1. Comparing all values: x=1 equals y=1; x=4 is greater than both y=-12 and y=1; x=1 is greater than y=-12. Therefore, x is always greater than or equal to y, making x≥y the correct relationship.

Multiple choice

In each of the following question, two equations are I and II given. You have to solve them and give answer $I. (5x^2 + 3x = 14)\nII. (5y^2 + 11y = 36)$

  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

Solving equation I: 5x² + 3x - 14 = 0 factors as (5x - 7)(x + 2) = 0, so x = 7/5 = 1.4 or x = -2. Solving equation II: 5y² + 11y - 36 = 0 factors as (5y - 9)(y + 4) = 0, so y = 9/5 = 1.8 or y = -4. When x = 1.4, y = 1.8 gives x < y, but when x = -2, y = -4 gives x > y. Since the relationship changes based on which roots we compare, the relationship cannot be established.

Multiple choice

In each of the following questions, two equations are I and II given. You have to solve them and give answer I. $(2x^2 - 17x + 36 = 0)$ II. $(7y^2 - 29y + 30 = 0)$

  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

Solve equation I: 2x² - 17x + 36 = 0 factors to (2x - 9)(x - 4) = 0, giving x = 4.5 or x = 4. Solve equation II: 7y² - 29y + 30 = 0 factors to (7y - 14)(y - 2.14...) ≈ y = 2 or y = 2.14. Both x values (4, 4.5) are greater than both y values (2, 2.14), so x > y is correct.