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: $(I. x^2 + 13x + 40 = 0 )$ $(II. y^2 + 7y + 10 = 0)$
  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
D Correct answer
Explanation

Equation I: x^2 + 13x + 40 = 0 → (x+8)(x+5)=0 → x = -8, -5. Equation II: y^2 + 7y + 10 = 0 → (y+5)(y+2)=0 → y = -5, -2. Comparing: x=-8 < y=-5,-2; x=-5 = y=-5, but x=-5 < y=-2. Since x is sometimes equal but never greater than y, x <= y is correct.

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. 5x^2 - 27x + 36 = 0 ) (II. y^2 - 2y + 2 = 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
E Correct answer
Explanation

Equation I: 5x^2 - 27x + 36 = 0 → (5x-12)(x-3)=0 → x = 2.4 or 3. Equation II: y^2 - 2y + 2 = 0 → discriminant = 4 - 8 = -4 < 0. No real roots for y. Since y has imaginary roots while x has real roots, no relation can be established between real x and non-real y.

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. 13x - 8y + 81 = 0 ) (II. 15x + 5y + 65 = 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
C Correct answer
Explanation

Solving the system: From I: 13x = 8y - 81 → x = (8y-81)/13. Substitute into II: 15(8y-81)/13 + 5y + 65 = 0 → (120y-1215+65y+845)/13 = 0 → 185y - 370 = 0 → y = 2. Then x = (16-81)/13 = -65/13 = -5. Since x=-5 < y=2, option C is correct.

Multiple choice

(\text{In each of these questions, two equations (I) and (II) are given. You have to solve both equations to give the answer :})\n(\text{(I)}\quad x^2 + 15x + 56 = 0 \ \text{(II)}\quad y^2 - 256 = 0)
  1. $\(x < y\)$
  2. $\(x > y\)$
  3. $\(x \ge y\)$
  4. $\(x \le y\)$

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

Solve equation (I): x² + 15x + 56 = 0 factors to (x+7)(x+8) = 0, giving x = -7 or x = -8. Solve equation (II): y² - 256 = 0 gives y² = 256, so y = ±16 (y = 16 or y = -16). When we compare the values, both possible x values (-7, -8) are less than 16 but greater than -16. This means no definite relationship exists between x and y. For instance, if x = -7 and y = -16, then x > y, but if x = -7 and y = 16, then x < y. Since the relationship cannot be uniquely determined, option E is correct.

Multiple choice

In the following question two equations numbered I and II are given. You have to solve both the equations and give answer thereof. $(I. ) (2x^2 - 13x + 21 = 0)$ $(II. ) (5y^2 - 22y + 21 = 0)$
  1. $\(x > y\)$
  2. $\(x \ge y\)$
  3. $\(x < y\)$
  4. $\(x \le y\)$

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

Equation I: 2x²-13x+21 = (2x-7)(x-3) = 0, so x = 3.5, 3. Equation II: 5y²-22y+21 = (5y-7)(y-3) = 0, so y = 1.4, 3. Compare: x=3.5 vs y=1.4 (x > y), x=3.5 vs y=3 (x > y), x=3 vs y=1.4 (x > y), x=3 vs y=3 (x = y). Since x ≥ y in all cases, option B correct.

Multiple choice

In the following question two equations numbered I and II are given. You have to solve both the equations and give answer thereof. I. (x^2 + 3x + 2 = 0) II. (2y^2 = 5y)
  1. $\(x > y\)$
  2. $\(x \ge y\)$
  3. $\(x < y\)$
  4. $\(x \le y\)$

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

Equation I: x²+3x+2 = (x+1)(x+2) = 0, so x = -1, -2. Equation II: 2y²-5y = y(2y-5) = 0, so y = 0, 2.5. Compare: All x values (-1, -2) are less than all y values (0, 2.5). Therefore x < y. Option C correct.

Multiple choice

In the following question two equations numbered I and II are given. You have to solve both the equations and give answer. $(\text{I.}\quad y^2 + 2y - 3 = 0)$ $(\text{II.}\quad 2x^2 - 7x + 6 = 0)$
  1. $\(x > y\)$
  2. $\(x \ge y\)$
  3. $\(x < y\)$
  4. $\(x \le y\)$

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

Solving equation I gives y² + 2y - 3 = 0, which factors to (y+3)(y-1) = 0, so y = -3 or y = 1. Solving equation II gives 2x² - 7x + 6 = 0, which factors to (2x-3)(x-2) = 0, so x = 1.5 or x = 2. Comparing the values, both x solutions (1.5 and 2) are greater than both y solutions (-3 and 1), therefore x > y is always true.

Multiple choice

In the following question two equations numbered I and II are given. You have to solve both the equations and give answer.(I. x^2 + 2x - 8 = 0 ) (II. y^2 - 2 = 7)
  1. $\(x > y\)$
  2. $\(x \ge y\)$
  3. $\(x < y\)$
  4. $\(x \le y\)$

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

Solving equation I gives x² + 2x - 8 = 0, which factors to (x+4)(x-2) = 0, so x = -4 or x = 2. Solving equation II gives y² - 2 = 7, so y² = 9 and y = -3 or y = 3. When comparing x = -4 with y = -3, we get x < y. When comparing x = 2 with y = 3, we get x < y. However, we could also have x = 2 compared to y = -3, giving x > y. Since the relationship is not consistent across all combinations, no definite relation can be established.

Multiple choice

In the following question two equations numbered I and II are given. You have to solve both the equations and give answer.(I. x^2 - 5x + 6 = 0 ) (II. y^2 + y - 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
B Correct answer
Explanation

Solving equation I gives x² - 5x + 6 = 0, which factors to (x-2)(x-3) = 0, so x = 2 or x = 3. Solving equation II gives y² + y - 6 = 0, which factors to (y+3)(y-2) = 0, so y = -3 or y = 2. Comparing all values: when x = 2, y can be -3 or 2, giving x ≥ y in both cases (2 > -3 and 2 = 2). When x = 3, y can be -3 or 2, giving x > y in both cases (3 > -3 and 3 > 2). Since x is always greater than or equal to y, x ≥ y is the correct answer.

Multiple choice

In the following question two equations numbered I and II are given. You have to solve both the equations and give answer. ( I. ) (x^2 - 10x + 24 = 0 ) ( II. ) (y^2 - 9y + 20 = 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
E Correct answer
Explanation

Solving x^2 - 10x + 24 = 0 gives (x - 4)(x - 6) = 0, so x = 4 or 6. Solving y^2 - 9y + 20 = 0 gives (y - 4)(y - 5) = 0, so y = 4 or 5. Comparing all pairs: (x=4, y=4) gives x = y; (x=4, y=5) gives x < y; (x=6, y=4) gives x > y; (x=6, y=5) gives x > y. Since x can be greater than, less than, or equal to y, the relationship cannot be established.

Multiple choice

In the following question two equations numbered I and II are given. You have to solve both the equations and give answer. $I. (9x^2 - 33x + 28 = 0)$ $II. (6y^2 - 25y + 25 = 0)$

  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
E Correct answer
Explanation

Solving equation I: 9x²-33x+28=0 factors as (3x-4)(3x-7)=0, so x=4/3 or x=7/3. Solving equation II: 6y²-25y+25=0 factors as (2y-5)(3y-5)=0, so y=5/2 or y=5/3. Comparing values: x=1.33 or 2.33, y=2.5 or 1.67. When x=4/3 and y=5/2, xy. Since x can be greater or less than y depending on which roots we choose, no definite relation can be established.

Multiple choice

If the HCF of (x^3 - 27) and (x^3 + 4x^2 + 12x + k) is a quadratic polynomial, then what is the value of k?

  1. 27

  2. 9

  3. 3

  4. -3
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

x^3 - 27 = (x-3)(x^2+3x+9). For HCF to be quadratic, x-3 must divide the second polynomial. By factor theorem: (3)^3 + 4(3)^2 + 12(3) + k = 0. 27 + 36 + 36 + k = 0. 99 + k = 0. k = -9. Let me verify: If k=-9, second polynomial = x^3 + 4x^2 + 12x - 9. Dividing by (x-3): Using synthetic division with root 3: coefficients 1, 4, 12, -9. Bring down 1. 3×1=3, 4+3=7. 3×7=21, 12+21=33. 3×33=99, -9+99=90. This gives remainder 90, not 0. Let me check if x+3 is a factor: (-3)^3 + 4(-3)^2 + 12(-3) + k = 0. -27 + 36 - 36 + k = 0. -27 + k = 0. k = 27. But answer says 9. Let me reconsider: maybe the HCF is (x^2+3x+9), not (x-3). For this, both polynomials must be divisible by x^2+3x+9. We know x^3-27 = (x-3)(x^2+3x+9). For x^3+4x^2+12x+k to have factor x^2+3x+9, we need x^3+4x^2+12x+k = (x+a)(x^2+3x+9) = x^3 + (a+3)x^2 + (3a+9)x + 9a. Comparing: a+3=4 so a=1. Then 3(1)+9=12 matches. And 9a = k = 9. So k=9 is correct!

Multiple choice

In the following question two equations numbered I and II are given. You have to solve both the equations and Give answer: I. (12a^2 + 11a + 12 = 10a^2 + 22a) II. (13b^2 – 18b + 3 = 9b^2 – 10b)

  1. If a > b

  2. If a ≤ b

  3. If a < b

  4. If a ≥ b

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

For equation I: 2a² - 11a + 12 = 0 factors to (2a-3)(a-4) = 0, giving a = 3/2 or a = 4. For equation II: 4b² - 8b + 3 = 0 factors to (2b-3)(2b-1) = 0, giving b = 3/2 or b = 1/2. Since a can be 3/2 or 4 and b can be 3/2 or 1/2, the relationship is a ≥ b (a is always greater than or equal to b).