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

Directions: In the following question, two equations (I) and (II) are given. You have to solve both the equations and give answer. (I) 61x2 - 123x + 2 = 0 (II) 2y2 + 3y + 1 = 0

  1. x > y

  2. x < y

  3. x ≥ y

  4. x ≤ y

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

I: 61x^2 - 123x + 2 = 0. Roots are approx 2 and 0.016. II: 2y^2 + 3y + 1 = 0 => (2y+1)(y+1) = 0. Roots are -0.5 and -1. Since 2 > -0.5 and 0.016 > -1, x > y.

Multiple choice

Which of the following equations is not quadratic?

  1. (x + 1)2 = 2(x - 3)

  2. x2 - 2x = -2(3 - x)

  3. (x - 2)(x + 1) = (x - 1)(x + 3)

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A quadratic equation must have an x^2 term. In option C: (x-2)(x+1) = x^2 - x - 2, and (x-1)(x+3) = x^2 + 2x - 3. The x^2 terms cancel out, leaving a linear equation.

Multiple choice

The number of integers k for which the equation x3 - 27x + k = 0 has at least two distinct integer roots is

  1. 1

  2. 2

  3. 3

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

x^3 - 27x + k = 0. For at least two distinct integer roots, the function must have integer values for x. Testing values, k must be such that the cubic factors into (x-a)(x-b)(x-c). For integer roots, the product of roots is -k and sum is 0. Possible integer roots for x^3-27x+k=0: if roots are a, b, c, then a+b+c=0 and ab+bc+ca=-27. If a=3, b=3, c=-6, k = -(3*3*-6) = 54. If a=-3, b=-3, c=6, k = -(-3*-3*6) = -54. Two values.

Multiple choice

If x1 and x2 are the roots of the equation x2 - 2x + 4 = 0, what is the equation with roots 3x1 - 2 and 3x2 - 2?

  1. x2 - 2x - 28 = 0

  2. x2 + 2x + 28 = 0

  3. x2 + 2x - 28 = 0

  4. x2 - 2x + 28 = 0

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

Roots x1, x2 satisfy x^2 - 2x + 4 = 0. x1+x2 = 2, x1x2 = 4. New roots y1 = 3x1-2, y2 = 3x2-2. Sum = 3(x1+x2) - 4 = 3(2) - 4 = 2. Product = (3x1-2)(3x2-2) = 9x1x2 - 6(x1+x2) + 4 = 9(4) - 6(2) + 4 = 36 - 12 + 4 = 28. Equation: x^2 - (sum)x + product = 0 => x^2 - 2x + 28 = 0.

Multiple choice

Directions: In this question, two equations (I) and (II) are given. You have to solve both the equations and give the answer. (I) 2p2 + p - 3 = 0 (II) 6q2 = 15q - 9

  1. p > q

  2. p < q

  3. p ≥ q

  4. p ≤ q

  5. p = q, or no relation can be established between 'p' and 'q'.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Equation I: 2p^2 + p - 3 = 0. (2p+3)(p-1) = 0. p = 1, -1.5. Equation II: 6q^2 - 15q + 9 = 0. 2q^2 - 5q + 3 = 0. (2q-3)(q-1) = 0. q = 1.5, 1. Comparing roots: p can be 1 or -1.5, q can be 1.5 or 1. Thus p <= q.

Multiple choice

Directions: In this question, two equations (I) and (II) are given. You have to solve both the equations and give answer as (a) if x > y (b) if x > y (c) if x y (e) if x = y, or the relationship between x and y cannot be established I. x2 + 5x + 6 = 0 II. y2 + 11y + 30 = 0

  1. a

  2. b

  3. c

  4. d

  5. e
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The first equation has roots -2 and -3, while the second has roots -5 and -6. Every possible x value is greater than every possible y value, so x > y.

Multiple choice

Directions: The following question is provided with two equations. You are required to solve the equations and identify the correct answer from the alternatives given below. I. x2 + 7x + 6 = 0 II. y2 + 13y + 42 = 0

  1. x > y

  2. x ≥ y

  3. x < y

  4. x ≤ y

  5. x = y or no relation can be determined
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Solving I: x^2 + 7x + 6 = 0 gives (x+6)(x+1) = 0, so x = -6, -1. Solving II: y^2 + 13y + 42 = 0 gives (y+6)(y+7) = 0, so y = -6, -7. Comparing values: x can be -6 or -1, y can be -6 or -7. If x=-6, y=-6 (x=y); if x=-1, y=-6 or -7 (x>y). Thus x >= y.

Multiple choice

In each of the following questions, two equations are given. You have to solve them and $( (I) \quad x^2 + 17x + 72 = 0 ) ( (II) \quad y^2 + 15y + 56 = 0 )$

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

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

Solve equation (I): x^2 + 17x + 72 = 0 factors to (x+8)(x+9) = 0, giving x = -8 or x = -9. Solve equation (II): y^2 + 15y + 56 = 0 factors to (y+7)(y+8) = 0, giving y = -7 or y = -8. When x = -9, x < y for both y values. When x = -8, x = y when y = -8, and x < y when y = -7. Since x can be equal to y or less than y in all cases, x ≤ y is correct.

Multiple choice

In each of the following questions, two equations are given. You have to solve them and $( (I) \quad x^2 + 12x + 27 = 0 \ (II) \quad y^2 + 11y + 30 = 0 )$

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

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

Solve equation (I): x^2 + 12x + 27 = 0 factors to (x+3)(x+9) = 0, giving x = -3 or x = -9. Solve equation (II): y^2 + 11y + 30 = 0 factors to (y+5)(y+6) = 0, giving y = -5 or y = -6. When x = -3, x > y for both y values. When x = -9, x < y for both y values. Since x can be greater or less than y depending on which root you choose, no single relationship can be established.