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 question, two equations are I and II given. You have to solve them and give answer I. (x^2 - x - 12 = 0) II. (y^2 + 9y + 20 = 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: x^2 - x - 12 = 0 gives x = 4 or x = -3. Solving equation II: y^2 + 9y + 20 = 0 gives y = -4 or y = -5. Comparing all possible pairs: when x = 4, x > y (both y values are negative); when x = -3, x > -4 but x > -5. In all cases, x > y.

Multiple choice

In each of the following question, two equations are I and II given. You have to solve them and give answer. I. (6x^2 + x – 1 = 0) II. (8y^2 + 10y + 3 = 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
B Correct answer
Explanation

Solving equation I: 6x^2 + x - 1 = 0 factors to (3x - 1)(2x + 1) = 0, giving x = 1/3 or x = -1/2. Solving equation II: 8y^2 + 10y + 3 = 0 factors to (4y + 3)(2y + 1) = 0, giving y = -3/4 or y = -1/2. When y = -1/2, x can be -1/2 (x = y) or x = 1/3 (x > y). Therefore, x ≥ y.

Multiple choice

In each of the following questions, two equations are I and II given. You have to solve them and give answer $I. (2(x + y) = 16) II. (x^2 + y^2 = 32)$

  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

From equation I: 2(x + y) = 16, we get x + y = 8. Substituting into equation II: x^2 + y^2 = 32. Using (x + y)^2 = x^2 + 2xy + y^2, we get 64 = 32 + 2xy, so xy = 16. This means x and y are roots of t^2 - 8t + 16 = 0, which gives t = 4 (repeated root). Thus x = y = 4, so the relationship is x = y.

Multiple choice

In each of the following questions, two equations are I and II given. You have to solve them and give the answer. $I. (12x^2 - x - 1 = 0) II. (18y^2 - 9y + 1 = 0)$

  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
E Correct answer
Explanation

Solving equation I: 12x^2 - x - 1 = 0 factors to (4x + 1)(3x - 1) = 0, giving x = -1/4 or x = 1/3. Solving equation II: 18y^2 - 9y + 1 = 0 factors to (6y - 1)(3y - 1) = 0, giving y = 1/6 or y = 1/3. Comparing: when x = 1/3, it can equal y = 1/3 or be greater than y = 1/6; when x = -1/4, it's less than both y values. The relationship cannot be established.

Multiple choice

In each of the following questions, two equations (I) and (II) are given. You have to solve them and answer the question. I. (7x + 3y = 30) II. (6x – 2y = 12)

  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 the system: From equation I, 3y = 30 - 7x, so y = 10 - (7/3)x. From equation II, 6x - 2(10 - 7x/3) = 12 gives 6x - 20 + 14x/3 = 12, so (18x + 14x)/3 = 32, 32x/3 = 32, giving x = 3. Substituting: 7(3) + 3y = 30 gives 21 + 3y = 30, so y = 3. Since x = y, we have x = y, which satisfies option E (x = y or relationship cannot be established).

Multiple choice

(\text{In each of the following questions, two equations are given. You have to solve them} )(\text{(I)}) (5x^2 - 13x - 6 = 0) (\text{(II)}) (3y^2 + 8y + 4 = 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
A Correct answer
Explanation

Solve (I): 5x² - 13x - 6 = 0 gives x = 3 or x = -2/5. Solve (II): 3y² + 8y + 4 = 0 gives y = -2/3 or y = -2. Comparing: 3 > -2/3 and 3 > -2, also -2/5 > -2/3 and -2/5 > -2. So x > y always.

Multiple choice

(\text{In each of the following questions, two equations are given. You have to solve them})(\text{(I)}) (2x^2 + 17x + 30 = 0)(\text{(II)}) (6y^2 - 5y - 6 = 0)
  1. $\(x > y\)$
  2. $\(x < y\)$
  3. $\(x \ge y\)$
  4. $\(x \le y\)$

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

Solve (I): 2x² + 17x + 30 = 0 gives x = -5/2 or x = -6. Solve (II): 6y² - 5y - 6 = 0 gives y = 3/2 or y = -2/3. Comparing: -5/2 < 3/2, -5/2 < -2/3, -6 < 3/2, -6 < -2/3. So x < y always.

Multiple choice

In each question two equations numbered I and II are given, you have to solve both the equation and state the correct relationship. $I. (2x^2 - 21x + 54 = 0)$ $II. (y^2 - 14y + 49 = 0)$

  1. X > Y

  2. X < Y

  3. X ≤ Y

  4. X ≥ Y

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

Solve equation I: 2x² - 21x + 54 = 0 factors to (2x-9)(x-6) = 0, giving x = 4.5, 6. Solve equation II: y² - 14y + 49 = 0 is (y-7)² = 0, giving y = 7 (repeated root). Since both x values (4.5 and 6) are less than y = 7, 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 state the correct relationship. (I. ) (x^2 - 5x - 24 = 0 ) (II. ) (2y^2 + 19y + 35 = 0)
  1. $\(X > Y\)$
  2. $\(X < Y\)$
  3. $\(X ≤ Y\)$
  4. $\(X ≥ Y\)$

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

Solve equation I: x² - 5x - 24 = 0 gives x = 8, -3. Solve equation II: 2y² + 19y + 35 = 0 gives y = -5, -3.5. Since x has both positive and negative values while y is always negative, and specifically -3 > -3.5 but 8 > -5, the relationship cannot be consistently established.

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) (x^2 + 32x + 247 = 0) (II) (y^2 + 38y + 360 = 0)

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

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

Solving (I): x² + 32x + 247 = 0 gives x = -13, -19 (sum 32, product 247). Solving (II): y² + 38y + 360 = 0 gives y = -18, -20 (sum 38, product 360). Comparing values: -13 > -18 and -13 > -20, but -19 < -18 and -19 > -20. Since x values are sometimes greater and sometimes less than y values, no consistent relation exists.

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) x^{2} = 55225 ) ( (II) y^{4} = 887468 )

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

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

From (I): x = ±235. From (II): y = ±17.26 (fourth root of 887468). For positive roots, x = 235 > y = 17.26. For negative roots, x = -235 < y = -17.26. For mixed signs, comparison varies. Since x can be greater than or less than y depending on which roots we pick, no consistent relation exists.

Multiple choice

(\text{In each of the following questions, two equations are given. You have to solve them and Give answer})(\text{(I)}) (5x^2 + 69x = -234\\text{(II)}) (13y^2 + 660 = -217y)

  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

Solve both quadratic equations and compare x and y values. Equation I: 5x² + 69x + 234 = 0. Using quadratic formula or factorization, discriminant = 69² - 4(5)(234) = 4761 - 4680 = 81, so x = [-69 ± 9]/10, giving x = -6 or x = -7.8. Equation II: 13y² + 217y + 660 = 0. Discriminant = 217² - 4(13)(660) = 47089 - 34320 = 12769, giving two real roots. Since x can be -6 while y might be different, or vice versa, the relationship cannot be uniquely established without computing both exact y values.

Multiple choice

In each of the following questions, two equations are given. You have to solve them and Give answer. (I) (55x^2 + 309x = -378) (II) (26y^2 - 251y = 231)

  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
C Correct answer
Explanation

For equation (I): 55x² + 309x = -378 becomes 55x² + 309x + 378 = 0. Using the quadratic formula, x = [-309 ± √(309² - 4×55×378)]/(2×55) = [-309 ± √12321]/110 = [-309 ± 111]/110, giving x = -1.8 or x = -3.82. For equation (II): 26y² - 251y = 231 becomes 26y² - 251y - 231 = 0. Using the quadratic formula, y = [251 ± √(251² - 4×26×(-231))]/(2×26) = [251 ± √87025]/52 = [251 ± 295]/52, giving y = 10.5 or y = -0.846. Comparing the values: both x values (-1.8, -3.82) are less than both y values (-0.846, 10.5). Therefore, x < y.