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 question two equations numbered I and II are given, you have to solve both the equation and choose the correct answer. I. (X^2 = 1225) II. (5Y^2 - 20Y - 105 = 0)

  1. X > Y

  2. X < Y

  3. X ≤ Y

  4. X ≥ Y

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

Solving equation I: X² = 1225 gives X = ±35. Solving equation II: 5Y² - 20Y - 105 = 0 simplifies to Y² - 4Y - 21 = 0, which factors to (Y-7)(Y+3) = 0, giving Y = 7 or Y = -3. Comparing: when X = 35, X > Y (since Y is at most 7); but when X = -35, X < Y. The relationship can't be established consistently.

Multiple choice

In each question two equations numbered I and II are given, you have to solve both the equation and choose the correct answer. $I. (x^2 - 27x + 182 = 0)$ $II. (y^2 - 36y + 323 = 0)$

  1. X > Y

  2. X < Y

  3. X ≤ Y

  4. X ≥ Y

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

Equation I: x²-27x+182=0 factors as (x-13)(x-14)=0, so x=13 or 14. Equation II: y²-36y+323=0 factors as (y-17)(y-19)=0, so y=17 or 19. In all cases (x=13,14 and y=17,19), X

Multiple choice

In each of the following questions two equations (I) and (II) are given. Solve these equations and give the answer: (I) (x^2 - 9.2x + 19.2 = 0) (II) (y^2 - 5.5y + 7.5 = 0)

  1. x > y

  2. x ≥ y

  3. x < y

  4. x ≤ y

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

For equation I: x² - 9.2x + 19.2 = 0. Discriminant = 9.2² - 4(19.2) = 7.84. Roots: x = (9.2 ± 2.8)/2 = 6, 3.2. For equation II: y² - 5.5y + 7.5 = 0. Discriminant = 5.5² - 4(7.5) = 0.25. Roots: y = (5.5 ± 0.5)/2 = 3, 2.5. All x values (3.2, 6) are greater than all y values (2.5, 3). Therefore x > y.

Multiple choice

For what value of (k), will the roots of the equation (kx^2 - 5x + 6 = 0) be in the ratio of 2 : 3?

  1. 0

  2. 1

  3. -1

  4. 2
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For roots in ratio 2:3, let roots be 2a and 3a. Using Vieta's formulas: Sum = 5/k = 5a, so a = 1/k. Product = 6/k = 6a^2 = 6(1/k^2). Therefore 6/k = 6/k^2, which gives k = 1. Verification: with k=1, equation is x^2 - 5x + 6 = 0, giving roots 2 and 3, which are in ratio 2:3. Option A (k=0) is invalid - equation would be -5x + 6 = 0 with only one root. Option C gives roots 2 and 1.5 (ratio 4:3). Option D gives roots 1.5 and 2 (ratio 3:4).

Multiple choice

In the following questions, two equations Quantity I and Quantity II are given. You have to solve both the equations and give answer. \text{Quantity I:} 7x^2 – 34x - 5 = 0 \text{Quantity II:} 2y^2 – 3y - 14 = 0

  1. Quantity I > Quantity II

  2. Quantity I ≥ Quantity II

  3. Quantity I < Quantity II

  4. Quantity II ≤ Quantity II

  5. Quantity I = Quantity II or the relation cannot be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Quantity I: 7x² - 34x - 5 = 0 factors to (7x+1)(x-5)=0, giving x = -1/7 or x = 5. Quantity II: 2y² - 3y - 14 = 0 factors to (2y+7)(y-2)=0, giving y = -3.5 or y = 2. Comparing roots: max(x) = 5, max(y) = 2, so x > y for maximum. But min(x) = -1/7 ≈ -0.14, min(y) = -3.5, so x > y for minimum too. Actually comparing individual roots: 5 > 2 (yes), -1/7 > -3.5 (yes). So x > y always? Wait, we need to compare both roots. If x = -1/7 and y = 2, then x < y. If x = 5 and y = -3.5, then x > y. No consistent relationship. Option E is correct.

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) 8p2 + 50p + 57 = 0 (II) 6q2 – q – 57 = 0

  1. If p > q If p > q

  2. If p ≥ q If p ≥ q

  3. If p < q If p < q

  4. If p ≤ q If p ≤ q

  5. If p = q or no relation can be established between p and q. यदि p = q या p और q के मध्य कोई सम्बन्ध स्थापित नहीं किया जा सकता है
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Solving 8p² + 50p + 57 = 0 gives p = -1.5 or p = -4.75 (both negative). Solving 6q² - q - 57 = 0 gives q = 19/6 ≈ 3.17 or q = -3. When p = -1.5 and q = -3, we have p > q. When p = -4.75 and q = 3.17, we have p < q. Since the relationship changes depending on which roots we compare, no single relation always holds between p and q.

Multiple choice

In each question two equations numbered I and II are given, you have to solve both the equation and choose the correct answer. $I. (14x^2 + 77x - 91 = 0)$ $II. (y^2 = y)$

  1. X > Y

  2. X < Y

  3. X ≤ Y

  4. X ≥ Y

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

Equation I: 14x² + 77x - 91 = 0 factors to 7(2x - 1)(x + 13) = 0, giving x = 1/2 or x = -13/2. Equation II: y² = y means y(y - 1) = 0, giving y = 0 or y = 1. When x = 0.5, x < y for y=1 and x > y for y=0. When x = -6.5, x < y for both y=0 and y=1. The relationship is inconsistent, so E is correct.

Multiple choice

In each question two equations numbered I and II are given, you have to solve both the equation and choose the correct answer.\nI. (15x^2 - 46x + 35 = 0)\nII. (4y^2 - 15y + 14 = 0)

  1. X > Y

  2. X < Y

  3. X ≤ Y

  4. X ≥ Y

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

Equation I: 15x² - 46x + 35 = 0 factors to (5x - 7)(3x - 5) = 0, giving x = 7/5 = 1.4 or x = 5/3 ≈ 1.67. Equation II: 4y² - 15y + 14 = 0 factors to (4y - 7)(y - 2) = 0, giving y = 7/4 = 1.75 or y = 2. For x = 1.4, x < y (both y values are 1.75, 2). For x = 1.67, x < y (both y values). Therefore X < Y always, making B correct.

Multiple choice

A and B attempted to solve a quadratic equation. A made a mistake in writing down the constant term and ended up in roots (4, 3). B made a mistake in writing down the coefficient of x and get roots (3, 2). The correct roots of the equation are-

  1. -4, -3

  2. 6, 1

  3. 4, 3

  4. -6, -1
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

A's roots (4, 3) give the equation x² - 7x + constant_A = 0, so the correct coefficient of x is -7. B's roots (3, 2) give the equation x² + (coefficient_B)x + 6 = 0, so the correct constant term is 6. The correct equation is x² - 7x + 6 = 0, which factors as (x-6)(x-1) = 0, giving roots 6 and 1.

Multiple choice

In the following questions two equations I and II are given. You have to solve both the equations and give answer. ( (I) 625x^2 – 95x – 6 = 0 ) ( (II) 40y^2 + 29y – 56 = 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

For equation I: 625x² - 95x - 6 = 0. Using quadratic formula: x = (95 ± √(9025 + 15000))/1250 = (95 ± √24025)/1250 = (95 ± 155)/1250. Solutions: x = 250/1250 = 0.2 or x = -60/1250 = -0.048. For equation II: 40y² + 29y - 56 = 0. Solutions: y = (-29 ± √(841 + 8960))/80 = (-29 ± √9801)/80 = (-29 ± 99)/80. Solutions: y = 70/80 = 0.875 or y = -128/80 = -1.6. Since x can be greater than y (0.2 < 0.875 is false, but comparing -0.048 to -1.6: -0.048 > -1.6), and x can be less than y (0.2 < 0.875), 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. $( (I) x^2 - 26x + 69 = 0 )( (II) 4y^2 - 19y + 21 = 0 )$

  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
B Correct answer
Explanation

Solving the first equation gives x = 23 or x = 3. Solving the second equation gives y = 3 or y = 1.75. When x = 23, it is greater than both y values. When x = 3 and y = 3, they are equal. Therefore x is always greater than or equal to y.

Multiple choice

In the following questions two equations I and II are given. You have to solve both the equations and give answer. $( (I) \ 12x^2 + 41x + 35 = 0 )( (II) \ 6y^2 + 31y + 28 = 0 )$

  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 gives x = -1.67 or x = -1.75. Solving the second equation gives y = -1.17 or y = -4. When comparing values: x = -1.67 is between y values (-1.17 > x > -4). This means sometimes x > y and sometimes x < y, making the relationship undetermined.

Multiple choice

In the following questions two equations I and II are given. You have to solve both the equations and give answer. $( (I) x^2 – 15x – 54 = 0 )( (II) 2y^2 + 17y +35 = 0 )$

  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
A Correct answer
Explanation

Solving the first equation gives x = 18 or x = -3. Solving the second equation gives y = -3.5 or y = -5. Comparing all pairs: 18 > -3.5, 18 > -5, -3 > -3.5, and -3 > -5. In every case, x is greater than y.

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 + 15x + 22 = 0 ) ( (II) y^3 - 0.331 = 1 )

  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
C Correct answer
Explanation

Solving the quadratic equation gives x = -2 or x = -5.5. Solving the cubic equation y³ = 1.331 gives y = 1.1. Comparing these: -2 < 1.1 and -5.5 < 1.1. Both possible x values are less than y.

Multiple choice

In the following question, the given equations are given, after solving the equation, find the correct answer. $I. (x^2 - 1.1x + 0.18 = 0) II. (y^2 - 3.5y + 2.5 = 0)$
  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

For equation I: x² - 1.1x + 0.18 = 0 factors to (x-0.2)(x-0.9) = 0, giving x = 0.2 or 0.9. For equation II: y² - 3.5y + 2.5 = 0 factors to (y-1.25)(y-2) = 0, giving y = 1.25 or 2. Comparing all values: 0.2 < 1.25, 0.2 < 2, 0.9 < 1.25, 0.9 < 2. Therefore x < y always.