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 are given. You have to solve them and state the correct relationship- $I. (x^2 + 14x + 49 = 0) II. (y^2 + 9y = 0)$

  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

Equation I: x² + 14x + 49 = 0 factors as (x + 7)² = 0, giving x = -7. Equation II: y² + 9y = 0 factors as y(y + 9) = 0, giving y = 0 or y = -9. When x = -7 and y = 0, x < y. When x = -7 and y = -9, x > y. Since both relationships are possible, no single relationship can be established.

Multiple choice

In the following questions two equations I and II are given. You have to solve both the equations and give answer. $(I) (x^2 + 4x - 21 = 0)$ $(II) (y^2 + 2y - 35 = 0)$

  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 determined
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Solving (I): x²+4x-21=0 gives x = -7, 3. Solving (II): y²+2y-35=0 gives y = -7, 7. When x = -7, y = -7 gives x = y. When x = 3 and y = -7, x > y. When x = 3 and y = 7, x < y. Since the relationship cannot be uniquely determined, the answer is E.

Multiple choice

In the following questions two equations I and II are given. You have to solve both the equations and give answer. $( (I) \ 2x^2 + 13x + 20 = 0 \ (II) \ 2y^2 - 7y - 4 = 0 )$

  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 determined
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Solving (I): 2x²+13x+20=0 gives x = -2.5, -4. Solving (II): 2y²-7y-4=0 gives y = -0.5, 4. For all combinations: -2.5 < -0.5, -2.5 < 4, -4 < -0.5, -4 < 4. Therefore x < y always, so answer is A.

Multiple choice

In the following question two equations I and II are given. You have to solve both the equations and give answer. $( (I) \ (x + 3)^{2} - 25 = 0 )$ $( (II) \ y^{2} - 14y + 49 = 0 )$

  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 determined
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Solving (I): (x+3)²-25=0 gives x = -8, 2. Solving (II): y²-14y+49=0 gives (y-7)²=0, so y = 7 (repeated root). For all combinations: -8 < 7, 2 < 7. Therefore x < y always, so answer is A.

Multiple choice

In each of the following questions, two equations (I) and (II) are given. You have to solve them and give answer. $( (I) \quad 2x^2 – 17x + 36 = 0 \ (II) \quad 15y^2 – 16y + 4 = 0 )$

  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

For equation (I): 2x²-17x+36=0 gives x=4.5 or x=4. For equation (II): 15y²-16y+4=0 gives y=2/3 or y=2/5. Since all values of x (4, 4.5) are greater than all values of y (0.4, 0.67), we conclude x > y.

Multiple choice

In each of the following questions, two equations (I) and (II) are given. You have to solve them and give answer. $(I) (7x^2 - 41x + 30 = 0)$ $(II) (7y^2 - 26y + 24 = 0)$

  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

For equation (I): 7x²-41x+30=0 gives x=5 or x=6/7≈0.86. For equation (II): 7y²-26y+24=0 gives y=2 or y=12/7≈1.71. When x=5, x>y. When x=6/7≈0.86, x

Multiple choice

In each of the following questions, two equations (I) and (II) are given. You have to solve them and give answer. $(I) ( (8x^2 + 21) – (6x^2 + 23) = 0 ) (II) ( (7y^2 – 14) – (10y^2 – 17) = 0 )$

  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

Simplifying equation (I): (8x²+21)-(6x²+23)=2x²-2=0 gives x²=1, so x=±1. Simplifying equation (II): (7y²-14)-(10y²-17)=-3y²+3=0 gives y²=1, so y=±1. Possible comparisons: when both are positive, x=y=1; when x=1,y=-1, x>y; when x=-1,y=1, x

Multiple choice

In each of the following questions two equations are given. You have to solve both the equation and state the correct relationship. $I. (4x^2 - 11x + 6 = 0)$ $II. (4y^2 - 17y + 18 = 0)$

  1. If x > y

  2. If x < y

  3. If x ≥ y

  4. If x ≤ y

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

Solving equation I: 4x²-11x+6=0 factors to (4x-3)(x-2)=0, so x=3/4 or x=2. Solving equation II: 4y²-17y+18=0 factors to (4y-9)(y-2)=0, so y=9/4 or y=2. Comparing values: x can be 3/4 or 2, y can be 9/4 or 2. When x=3/4, x

Multiple choice

In each of the following questions two equations are given. You have to solve both the equation and state the correct relationship. $I. (x^2 - 22x - 23 = 0)$ $II. (y^2 - 12y - 13 = 0)$

  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

Solving equation I: x²-22x-23=0 factors to (x-23)(x+1)=0, so x=23 or x=-1. Solving equation II: y²-12y-13=0 factors to (y-13)(y+1)=0, so y=13 or y=-1. Comparing values: when x=23, x>y (13). When x=-1 and y=-1, x=y. When x=-1 and y=13, x

Multiple choice

In each of the following questions two equations are given. You have to solve both the equation and state the correct relationship. $I. (4x^2 - 8x + 3 = 0)$ $II. (12y^2 - 7y + 1 = 0)$

  1. If x > y

  2. If x < y

  3. If x ≥ y

  4. If x ≤ y

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

Solving equation I: 4x²-8x+3=0 factors to (2x-3)(2x-1)=0, so x=3/2 or x=1/2. Solving equation II: 12y²-7y+1=0 factors to (3y-1)(4y-1)=0, so y=1/3 or y=1/4. Comparing values: x=3/2 or 1/2, y=1/3 or 1/4. In all cases, x > y. Therefore, x > y.

Multiple choice

In each of the following questions two equations are given. You have to solve both the equation and state the correct relationship. $I. (14x^2 + 15x + 4 = 0)$ $II. (14y^2 + 29y + 12 = 0)$

  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

Solving equation I: 14x²+15x+4=0 factors to (7x+4)(2x+1)=0, so x=-4/7 or x=-1/2. Solving equation II: 14y²+29y+12=0 factors to (7y+4)(2y+3)=0, so y=-4/7 or y=-3/2. Comparing values: when x=-4/7, x≥y (equal to -4/7, greater than -3/2). When x=-1/2, x≥y (greater than both -4/7 and -3/2). Therefore, x ≥ y.

Multiple choice

In each of the following questions two equations are given. You have to solve both the equation and state the correct relationship. $I. (4y^2 + 9y + 2 = 0)$ $II. (15x^2 - 32x + 16 = 0)$

  1. If x > y

  2. If x < y

  3. If x ≥ y

  4. If x ≤ y

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

Solving equation I: 4y²+9y+2=0 factors to (4y+1)(y+2)=0, so y=-1/4 or y=-2. Solving equation II: 15x²-32x+16=0 factors to (5x-4)(3x-4)=0, so x=4/5 or x=4/3. Comparing values: x is always positive (4/5 or 4/3), y is always negative (-1/4 or -2). Therefore, x > y.

Multiple choice

In each of the following questions two equations are given. You have to solve both the equation and find out the values of x and y and Give answer - I. (3x^2 – 13x + 14 = 0) II. (y^2 – 7y + 12 = 0)

  1. If x > y

  2. If x < y

  3. If x ≥ y

  4. If x ≤ y

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

For equation I: 3x² - 13x + 14 = 0 factors to (3x - 7)(x - 2) = 0, giving x = 2 or x = 7/3. For equation II: y² - 7y + 12 = 0 factors to (y - 3)(y - 4) = 0, giving y = 3 or y = 4. Comparing all values: both x values (2, 2.33) are less than both y values (3, 4), so x < y is always true.

Multiple choice

In the following questions two equations I and II are given. You have to solve both the equations and give answer. $I. (x^2 - 9 = 0) II. (y^2 - 16 = 0)$

  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 determined.
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Solving the first equation: x² = 9, so x = ±3. Solving the second: y² = 16, so y = ±4. The possible pairs are (3, 4), (3, -4), (-3, 4), (-3, -4). This gives x > y (when x=3, y=-4), x < y (when x=-3, y=4 or x=3, y=4), and x < y (when x=-3, y=4). Since x can be both greater than and less than y depending on which roots we choose, the relationship cannot be uniquely determined.

Multiple choice

In the following questions two equations I and II are given. You have to solve both the equations and give answer. $I. (x^2 – 3.5x + 3 = 0) II. (y^2 – 5.5y + 7 = 0)$

  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 determined.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Solving x² - 3.5x + 3 = 0 using quadratic formula: x = (3.5 ± √0.25)/2 = (3.5 ± 0.5)/2, giving x = 2 or x = 1.5. Solving y² - 5.5y + 7 = 0: y = (5.5 ± √2.25)/2 = (5.5 ± 1.5)/2, giving y = 3.5 or y = 2. Comparing all pairs: x ∈ {1.5, 2} and y ∈ {2, 3.5}. The maximum value of x is 2 and the minimum value of y is 2, so x is always less than or equal to y.