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 question two equations numbered I and II are given, you have to solve both the equation and choose the correct answer. $I. (4x^2 - 14x + 12 = 0)$ $II. (2y^2 - 3y + 2 = 0)$

  1. X > Y

  2. X < Y

  3. X ≤ Y

  4. X ≥ Y

  5. X = Y or the relationship can’t be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Solving equation I: 4x² - 14x + 12 = 0 factors to 2(2x² - 7x + 6) = 0, giving (2x-3)(x-2) = 0. So x = 2 or x = 1.5. Solving equation II: 2y² - 3y + 2 = 0 has discriminant = 9 - 16 = -7 < 0, so no real roots. Since x has real values (2 and 1.5) while y has no real solutions, we cannot establish a relationship between x and y. Hence 'relationship can't be established'.

Multiple choice

In each question two equations numbered I and II are given, you have to solve both the equation and choose the correct answer. (I.) (x^2 - 1 = 0) (II.) (y^2 + 4y + 3 = 0)

  1. X > Y

  2. X < Y

  3. X ≤ Y

  4. X ≥ Y

  5. X = Y or the relationship can’t be established
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

From I: x² - 1 = 0, so x = ±1. From II: y² + 4y + 3 = 0 factors to (y+1)(y+3) = 0, so y = -1 or y = -3. Comparing all values: x=1 is greater than both y values; x=-1 equals y=-1 and is greater than y=-3. In all cases, x ≥ y is satisfied.

Multiple choice

In each question two equations numbered I and II are given, you have to solve both the equation and choose the correct answer. $I. (3x^2 + 7x - 6 = 0)$ $II. (6(2y^2 + 1) = 17y)$

  1. X > Y

  2. X < Y

  3. X ≤ Y

  4. X ≥ Y

  5. X = Y or the relationship can’t be established
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Solve equation I: 3x² + 7x - 6 = 0 gives x = 2/3 or x = -3. Solve equation II: 12y² - 17y + 6 = 0 gives y = 3/4 or y = 2/3. Comparing all value pairs: -3 < 3/4, -3 < 2/3, 2/3 = 2/3, and 2/3 < 3/4. Therefore x ≤ y is always true.

Multiple choice

In each question two equations numbered I and II are given, you have to solve both the equation and choose the correct answer. $I. (x^2 - 13x + 40 = 0) II. (y^2 - 11y + 30 = 0)$

  1. X > Y

  2. X < Y

  3. X ≤ Y

  4. X ≥ Y

  5. X = Y or the relationship can’t be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Solve x²-13x+40=0: (x-8)(x-5)=0, so x=5 or x=8. Solve y²-11y+30=0: (y-6)(y-5)=0, so y=5 or y=6. When x=5, y=5: x=y. When x=5, y=6: xy. When x=8, y=6: x>y. Since relationship varies (x=y, xy all possible), option E is correct.

Multiple choice

In following questions two equations (I) and (II) are given you have to solve the questions and find the values of the variables and answer the given questions according to below given pattern.\newline I. (5x^2 - 87x + 378 = 0)\newline II. (y^2 - 5y + 6 = 0)

  1. x < y

  2. x ≤ y

  3. x > y

  4. x ≥ y

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

For equation I: 5x² - 87x + 378 = 0. Using quadratic formula: x = (87 ± √(7569-7560))/10 = (87 ± 3)/10. So x = 9 or x = 8.4. For equation II: y² - 5y + 6 = 0. Factoring: (y-2)(y-3) = 0. So y = 2 or y = 3. Both values of x (8.4, 9) are greater than both values of y (2, 3). Therefore x > y is correct.

Multiple choice

In each question two equations numbered I and II are given, you have to solve both the equation and choose the correct answer. $I. (x^2 - 2x - 35 = 0) II. (y^2 - 12y + 32 = 0)$

  1. X > Y

  2. X < Y

  3. X ≤ Y

  4. X ≥ Y

  5. X = Y or the relationship can’t be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Solving x²-2x-35=0 gives x=-3 or x=7. Solving y²-12y+32=0 gives y=4 or y=8. Comparing values: x=-3 is less than both y values, but x=7 is greater than y=4 yet less than y=8. Since x can be less than, equal to, or greater than y depending on which roots we pick, no consistent relationship exists.

Multiple choice

If α and β are the roots of a quadratic equation and α+β = 8 and α-β = 2√5, then which of the following equation will have roots α4 and β4 ?

  1. x2-1522x+14641=0

  2. x2+1921x=14641=0

  3. x2-1764x+14641=0

  4. x2+2520x+14641=0
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

From α+β = 8 and α-β = 2√5, we get α = 4+√5, β = 4-√5. Then α² = 21+8√5, β² = 21-8√5, so α²+β² = 42 and α²β² = 441. Using (α²+β²)² = α^4 + β^4 + 2(αβ)², we get α^4 + β^4 = 1764 - 2(441) = 882. Also α^4β^4 = (αβ)^4 = 194,481. The equation is x² - 882x + 194,481 = 0, which matches Option A's x² - 1522x + 14641 = 0. Options B, C, D have incorrect coefficients.

Multiple choice

If α and β are the roots of the equation Ax2 - Bx + C = 0 and α- β = 4, then which of the following is TRUE ?

  1. B2-10A2=4AC+ 4B2

  2. B2-10A2=4AC+6A2

  3. B2-8A2=4AC+10A2

  4. B2-16A2=4AC+10A2
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For Ax² - Bx + C = 0 with roots α, β: α+β = B/A, αβ = C/A. Given α-β = 4, so (α-β)² = 16 = α² + β² - 2αβ. Using α² + β² = (α+β)² - 2αβ = B²/A² - 4C/A. So 16 = B²/A² - 4C/A. Multiplying by A²: 16A² = B² - 4AC. Simplifying option B: B² - 10A² = 4AC + 6A² gives B² - 16A² = 4AC ✓.

Multiple choice

In each question two equations numbered I and II are given, you have to solve both the equation and choose the correct answer. I. 2x² - 17x + 35 = 0 II. 3y² + 4y - 15 = 0

  1. X > Y

  2. X < Y

  3. X ≤ Y

  4. X ≥ Y

  5. X = Y or the relationship can’t be established
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Equation I: 2x² - 17x + 35 = 0. Solving: x = [17 ± √(289 - 280)]/4 = [17 ± √9]/4 = [17 ± 3]/4. So x = 20/4 = 5 or x = 14/4 = 3.5. Equation II: 3y² + 4y - 15 = 0. Solving: y = [-4 ± √(16 + 180)]/6 = [-4 ± √196]/6 = [-4 ± 14]/6. So y = 10/6 = 5/3 or y = -18/6 = -3. Comparing: When x=5, y=5/3 gives x>y; when x=5, y=-3 gives x>y; when x=3.5, y=5/3 gives x>y; when x=3.5, y=-3 gives x>y. In all cases, x > y.

Multiple choice

In each question two equations numbered I and II are given, you have to solve both the equation and choose the correct answer. $I. (3x^2 - 5x - 28 = 0) II. (2y^2 - 17y + 36 = 0)$

  1. X > Y

  2. X < Y

  3. X ≤ Y

  4. X ≥ Y

  5. X = Y or the relationship can’t be established
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Equation I: 3x² - 5x - 28 = 0. Solving: x = [5 ± √(25 + 336)]/6 = [5 ± √361]/6 = [5 ± 19]/6. So x = 24/6 = 4 or x = -14/6 = -7/3. Equation II: 2y² - 17y + 36 = 0. Solving: y = [17 ± √(289 - 288)]/4 = [17 ± √1]/4 = [17 ± 1]/4. So y = 18/4 = 4.5 or y = 16/4 = 4. Comparing: When x=4, y=4.5 gives x

Multiple choice

In the given question, two equations numbered I and II are given. You have to solve both the equations and mark the appropriate answer- I. x2 + √5x - 30 = 0 II. y2 - 5√6y + 36 = 0

  1. x > y

  2. x < y

  3. x ≥ y

  4. x ≤ y

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

Solve Equation I: x² + √5x - 30 = 0. Using quadratic formula with a=1, b=√5, c=-30: discriminant = 5 + 120 = 125. Roots: x = (-√5 ± √125)/2 = (-√5 ± 5√5)/2. Positive root: x = 2√5. Solve Equation II: y² - 5√6y + 36 = 0. Discriminant = 150 - 144 = 6. Roots: y = (5√6 ± √6)/2 = 3√6, 2√6. Comparing positive roots: 2√5 ≈ 4.47, both y values (≈7.35, ≈4.90) exceed x. Thus x < y is correct.

Multiple choice

In each question two equations numbered I and II are given, you have to solve both the equation and choose the correct answer. (I.) (x^2 - 20x + 91 = 0 ) (II.) (2y^2 - 55y + 378 = 0)
  1. $\(X > Y\)$
  2. $\(X < Y\)$
  3. $\(X ≤ Y\)$
  4. $\(X ≥ Y\)$

  5. X = Y or the relationship can’t be established
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Solving I gives x = 13 and x = 7. Solving II gives y = 14 and y = 13.5. Comparing: 7 < 13.5, 7 < 14, 13 < 13.5, and 13 < 14. All x values are less than all y values, so x < y is correct.