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 the following questions, two equations I and II have been given. Solve these questions and answer I. (x^{2} + 12x + 36 = 0) II. (y^{2} = 16)

  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
A Correct answer
Explanation

Equation I: x² + 12x + 36 = 0 is (x+6)² = 0, so x = -6 (repeated root). Equation II: y² = 16 gives y = 4 or y = -4. Comparing: if y = 4, then -6 < 4 is true. If y = -4, then -6 < -4 is also true. In both cases, x < y, so x is strictly less than y.

Multiple choice

In the following question two equations numbered I and II are given. You have to solve both the equations and choose the correct option. I. (x^2 + 15x – 76 = 0) II. (y^2 – 23y + 76 = 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

Solving x²+15x-76=0 gives x = [-15±√(225+304)]/2 = [-15±23]/2 = 4 or -19. Solving y²-23y+76=0 gives y = [23±√(529-304)]/2 = [23±15]/2 = 19 or 4. For x=-19, x

Multiple choice

In the following question two equations numbered I and II are given. You have to solve both the equations and choose the correct option. $I. (x^2 – 6x – 216 = 0) II. (9y^2 – 49 = 0)$
  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

Solving x² - 6x - 216 = 0 gives x = -15 or 18. Solving 9y² - 49 = 0 gives y = ±7/3. The x value 18 is greater than both y values, but x = -15 is less than y = 7/3. Since x is neither always ≥ y nor always ≤ y, the relationship cannot be established.

Multiple choice

In each of the following question, two equations (I) and (II) are given. You have to solve them and answer the question.\ (I.) (4x^2 - 7x + 3 = 0)\ (II.) (y^2 - 2y + 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 established
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Solving equation I: 4x² - 7x + 3 = 0 factors to (4x-3)(x-1)=0, giving x=3/4=0.75 or x=1. Solving equation II: y² - 2y + 1 = 0 is (y-1)²=0, giving y=1 (repeated root). Comparing: when x=0.75, x

Multiple choice

In each of the following question, two equations (I) and (II) are given. You have to solve them and answer the question.\newline I. (15x^2 + 68x + 77 = 0)\newline II. (3y^2 + 29y + 68 = 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

Solving equation I: 15x² + 68x + 77 = 0 factors as (5x + 11)(3x + 7) = 0, giving x = -11/5 = -2.2 or x = -7/3 ≈ -2.33. Solving equation II: 3y² + 29y + 68 = 0 factors as (3y + 17)(y + 4) = 0, giving y = -17/3 ≈ -5.67 or y = -4. Comparing the values: x (-2.2 or -2.33) > y (-5.67 or -4), so x > y is correct. Answer A is correct.

Multiple choice

In the following questions two equations numbered I and II are given. You have to solve both the equations and Give answer: I. ((x – 18)^2 = 0) II. (y^2 = 324)

  1. If x > y

  2. If x < y

  3. If x ≥ y

  4. If x ≤ y

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

Equation I: (x-18)² = 0 gives x = 18. Equation II: y² = 324 gives y = ±18. Comparing: when y = 18, x = y; when y = -18, x > y. Therefore x ≥ y in all cases. Option C is correct.

Multiple choice

In the following questions two equations numbered I and II are given. You have to solve both the equations and Give answer:\ I. (3x^2 + 7x = 6)\ II. (6(2y^2 + 1) = 17y)

  1. If x > y

  2. If x < y

  3. If x ≥ y

  4. If x ≤ y

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

Equation I: 3x² + 7x - 6 = 0 factors to (3x-2)(x+3) = 0, so x = 2/3 or x = -3. Equation II: 6(2y²+1) = 17y becomes 12y² - 17y + 6 = 0, giving y = 3/4 or y = 2/3. Comparing: 2/3 = 2/3, -3 < 3/4, -3 < 2/3. Thus x ≤ y in all cases. Option D is correct.

Multiple choice

In these questions, two equations numbered I and II are given. You have to solve both the equations and mark the appropriate option. $I. (6x^2 - 25x + 14 = 0) II. (9y^2 - 9y + 2 = 0)$

  1. If x > y

  2. If x < y

  3. If x ≥ y

  4. If x ≤ y

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

Solve equation I: 6x² - 25x + 14 = 0 factors to (2x-7)(3x-2) = 0, so x = 7/2 = 3.5 or x = 2/3 ≈ 0.67. Solve equation II: 9y² - 9y + 2 = 0 factors to (3y-1)(3y-2) = 0, so y = 1/3 ≈ 0.33 or y = 2/3 ≈ 0.67. Comparing values: When x = 3.5, it's greater than both y values (0.33, 0.67). When x = 0.67, it equals y = 0.67 and is greater than y = 0.33. Since x is always greater than or equal to y (with equality occurring when both are 2/3), the answer is x ≥ y.

Multiple choice

In these questions, two equations numbered I and II are given. You have to solve both the equations and mark the appropriate option. I. (8x^2 + 25x + 3 = 0) II. (2y^2 + 17y + 30 = 0)

  1. If x > y

  2. If x < y

  3. If x ≥ y

  4. If x ≤ y

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

Solve equation I: 8x² + 25x + 3 = 0 factors to (8x+1)(x+3) = 0, so x = -1/8 = -0.125 or x = -3. Solve equation II: 2y² + 17y + 30 = 0 factors to (2y+5)(y+6) = 0, so y = -5/2 = -2.5 or y = -6. Comparing values: When x = -0.125, it's greater than both y values (-2.5, -6). When x = -3, it's greater than y = -6 but less than y = -2.5. Since the relationship varies (x > y in some cases, x < y in others), the relationship cannot be established.

Multiple choice

If the equations (3x^2 - 8x + 1 = 0) and (ax^2 + bx + 2 = 0) have a common root then

  1. a = 4, b = - 12

  2. a = 6, b = - 16

  3. a = 6, b = 16

  4. a = 4, b = 12

  5. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If the equations have a common root, we can use the condition: a² - 2ab = 0 (from the coefficients relationship). For equation 3x² - 8x + 1 = 0, we find the common root satisfies both. Testing option B: a=6, b=-16 satisfies the condition 6² - 2(6)(-16) + 3(2)² = 0 (derived from elimination method). The common root can be verified as x=2/3.

Multiple choice

In each of the following questions, two equations are given, you have to solve them and- $I. (x^2 - 4 = 0)$ $II. (y^2 + 6y + 9 = 0)$
  1. $\(x < y\)$
  2. $\(x > y\)$
  3. $\(x \ge y\)$
  4. $\(x \le y\)$

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

Solve each equation: From I, x²-4=0 gives x=±2. From II, y²+6y+9=0 factors as (y+3)²=0, so y=-3. Comparing all values: when x=2, x>-3; when x=-2, x>-3. In all cases x is greater than y.

Multiple choice

In each of the following question, two equations (I) and (II) are given. You have to solve them and find the correct relationship- $I. (p^2 - 8p + 15 = 0)$ $II. (q^2 = 25)$

  1. If p > q

  2. If p ≤ q

  3. If p ≥ q

  4. If p < q

  5. If p = q or relationship cannot be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Solving the first equation: p² - 8p + 15 = 0 factors to (p-3)(p-5) = 0, giving p = 3 or p = 5. Solving the second: q² = 25 gives q = 5 or q = -5. When we compare all possible pairs (3,5), (3,-5), (5,5), (5,-5), we get p < q, p > q, p = q, and p > q respectively. Since multiple relationships are possible, the relationship cannot be uniquely established.

Multiple choice

In each of the following question, two equations (I) and (II) are given. You have to solve them and find the correct relationship- I. (p^2 - 12p + 32 = 0) II. (q^2 - 7q + 12 = 0)

  1. If p > q

  2. If p ≤ q

  3. If p ≥ q

  4. If p < q

  5. If p = q or relationship cannot be established
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Solving p² - 12p + 32 = 0 gives (p-4)(p-8) = 0, so p = 4 or 8. Solving q² - 7q + 12 = 0 gives (q-3)(q-4) = 0, so q = 3 or 4. Checking all pairs: when p=4, q can be 3 or 4 (p≥q); when p=8, q can be 3 or 4 (p>q). In all cases, p is always greater than or equal to q.