Quantitative Aptitude

Quadratic Equations

367 Questions
All Basic factorization 305 Sign scheme based 22 Algebraic substitution 19 Roots from relation 9 Finding extrema 5 Discriminant analysis 2 Finding coefficients 2 Common root 1 Finding quadratic equation 1 Reciprocal equation 1

Quadratic Equations Questions

Multiple choice

Find the correct relationship between Quantity I and Quantity II. \ Quantity I: x where: (3x^2 + 2x - 8 = 0)\ Quantity II: y where: (3y^2 + 5y - 12 = 0)

  1. Quantity I > Quantity II

  2. Quantity I ≥ Quantity II

  3. Quantity I < Quantity II

  4. Quantity I ≤ Quantity II

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

For Quantity I: 3x² + 2x - 8 = 0. Solving using factorization or quadratic formula gives x = 4/3 or x = -2. For Quantity II: 3y² + 5y - 12 = 0 gives y = 3/2 or y = -4. Since x can be greater (4/3 > 3/2 is false, but 4/3 vs 3/2: 1.33 < 1.5), equal (both negative cases), or less depending on which roots we pick, no consistent relationship exists.

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:} x^3 - 133 = [56\% \text{ of } 1250] \div 4 + 11^3 + 124\times9 \div 2 \text{Quantity II:} y^2 - y - 72 = 0

  1. Quantity I > Quantity II

  2. Quantity I ≥ Quantity II

  3. Quantity I < Quantity II

  4. Quantity I ≤ Quantity II

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

Quantity I: x^3 - 133 = [56% of 1250] ÷ 4 + 11^3 + 124 × 9 ÷ 2. Calculate: 56% of 1250 = 0.56 × 1250 = 700. Then: 700 ÷ 4 = 175; 11^3 = 1331; 124 × 9 = 1116; 1116 ÷ 2 = 558. Sum: 175 + 1331 + 558 = 2064. So x^3 - 133 = 2064, thus x^3 = 2197, and x = 13. Quantity II: y^2 - y - 72 = 0 factors to (y - 9)(y + 8) = 0, so y = 9 or y = -8. Comparing x = 13 with y values: 13^3 = 2197, while 9^2 = 81 and (-8)^2 = 64. Since 2197 > 81 and 2197 > 64, Quantity I > Quantity II.

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. \begin{itemize} \item Quantity I: (2x^2 - x - 66 = 0) \item Quantity II: (3y^2 - 5y - 42 = 0)\end{itemize}

  1. Quantity I > Quantity II

  2. Quantity I ≥ Quantity II

  3. Quantity I < Quantity II

  4. Quantity I ≤ 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

For Quantity I: 2x² - x - 66 = 0 gives x = (1 ± 23)/4, so x = 6 or x = -5.5. For Quantity II: 3y² - 5y - 42 = 0 gives y = (5 ± 23)/6, so y = 14/3 ≈ 4.67 or y = -3. Comparing all pairs: when x=6, both y values give x > y; when x=-5.5, both y values give x < y. Since the relationship depends on which roots you compare, no consistent relation exists.

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:} 2x^2 + 19x + 45 = 0 \text{Quantity II:} y^2 - 17y + 72 = 0

  1. Quantity I > Quantity II

  2. Quantity I ≥ Quantity II

  3. Quantity I < Quantity II

  4. Quantity I ≤ Quantity II

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

Quantity I: 2x² + 19x + 45 = 0 factors as (2x + 5)(x + 9) = 0, giving x = -2.5 or x = -9. Quantity II: y² - 17y + 72 = 0 factors as (y - 8)(y - 9) = 0, giving y = 8 or y = 9. Both x values are negative while both y values are positive, so x < y for all combinations.

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. I. What is the maximum value of (-5x^2 - 20x + 4)? II. (900x^2 - 54000x + 810000 = 0)

  1. Quantity I > Quantity II

  2. Quantity I ≥ Quantity II

  3. Quantity II > Quantity I

  4. Quantity II ≥ Quantity I

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

Quantity I: For quadratic ax² + bx + c where a < 0, maximum value at x = -b/(2a). Here x = -20/(2×-5) = -2. Maximum value = -5(4) - 20(-2) + 4 = 24. Quantity II: 900x² - 54000x + 810000 = 0 simplifies to x² - 60x + 900 = 0, giving x = 30. Since 24 < 30, Quantity II > Quantity I.

Multiple choice

In the following questions, two equations (x) and (y) are given. You have to solve both the equations and give answer. (I.) (x^{3} - 133 = [56\% \text{ of } 1250] \div 4 + 1331 + 124\times9 \div 2) (II.) (y^{2} - y - 72 = 0)

  1. x > y

  2. x ≥ y

  3. x < y

  4. x ≤ y

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

Equation I: x³ - 133 = [56% of 1250] ÷ 4 + 1331 + 124×9 ÷ 2. 56% of 1250 = 700. 124×9÷2 = 558. So x³ - 133 = 700÷4 + 1331 + 558 = 175 + 1331 + 558 = 2064. x³ = 2064 + 133 = 2197. x = ∛2197 = 13. Equation II: y² - y - 72 = 0. Factoring: (y-9)(y+8) = 0, so y = 9 or y = -8. Since x = 13, and 13 > 9 (and 13 > -8), we have x > y for both solutions of y.

Multiple choice

(\text{In the following questions, two equations x and y are given. You have to solve both the equations and give answer.}\\text{I.} \frac{(40)^3}{80 \times 16} \div 25 = 32x\\text{II.} \frac{y}{1089} = \frac{64}{y})

  1. x > y

  2. x ≥ y

  3. x < y

  4. x ≤ y

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

Solving Equation I: (40)³ = 64000, 80 × 16 = 1280, so 64000/1280 = 50, then 50 ÷ 25 = 2. Therefore 32x = 2, giving x = 2/32 = 1/16. Solving Equation II: y/1089 = 64/y, cross-multiplying gives y² = 64 × 1089 = 69696, so y = ±264. Since y can be either 264 or -264, the relation between x and y is ambiguous - if y = 264 then x < y, but if y = -264 then x > y. Therefore the relation cannot be established.

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. $Quantity I: (4x^2 – 9x – 34 = 0) Quantity II: (8y^2 + 34y – 19 = 0)$

  1. Quantity I > Quantity II

  2. Quantity I ≥ Quantity II

  3. Quantity I < Quantity II

  4. Quantity I ≤ 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

Solving 4x^2 - 9x - 34 = 0: Using quadratic formula, x = [9 ± sqrt(81 + 544)]/8 = [9 ± sqrt(625)]/8 = [9 ± 25]/8. So x = 34/8 = 4.25 or x = -16/8 = -2. Solving 8y^2 + 34y - 19 = 0: y = [-34 ± sqrt(1156 + 608)]/16 = [-34 ± sqrt(1764)]/16 = [-34 ± 42]/16. So y = 8/16 = 0.5 or y = -76/16 = -4.75. Comparing values: x values are 4.25 and -2, y values are 0.5 and -4.75. No consistent relationship can be established (sometimes x > y, sometimes x < y).

Multiple choice

In the following question, two Quantity I and Quantity II are given. You have to solve both the Quantities and give answer. \text{Quantity I:} 4x^2-8x+4=0, \text{What is the value of x?} \text{Quantity II:} 5y^2-9y+4=0 , \text{What is the value of y}?

  1. Quantity I < Quantity II

  2. Quantity I ≤ Quantity II

  3. Quantity I > Quantity II

  4. Quantity I ≥ Quantity II

  5. Quantity I = Quantity II or no relation can be established
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Quantity I: 4x² - 8x + 4 = 0 gives x = 1 (repeated root). Quantity II: 5y² - 9y + 4 = 0 gives y = 1 or y = 4/5. When y = 1, x = y. When y = 4/5, x > y. Therefore x ≥ y is always true, making option D correct.

Multiple choice

Find the value of variables and state the correct relationship-( (3x^2+20x+12) =0 )( (3y^2-y-2)=0 )

  1. x < y

  2. x > y

  3. x ≤ y

  4. x ≥ y

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

For 3x² + 20x + 12 = 0, x = -2/3 or x = -6. For 3y² - y - 2 = 0, y = 1 or y = -2/3. When x = -2/3, y = 1 gives x < y; x = -2/3, y = -2/3 gives x = y. When x = -6, y = 1 gives x < y; x = -6, y = -2/3 gives x < y. In all cases x ≤ y is true.

Multiple choice

$(\text{Solve the following equations:})\newline(\text{(i) } 2x^2 + 31x + 119 = 0\newline\text{(ii) } y^2 + 51y + 98 = 0)$

  1. If x < y

  2. If x > y

  3. If x ≥ y

  4. If x ≤ y

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

For equation (i): 2x^2 + 31x + 119 = 0. Using quadratic formula: x = [-31 ± √(961 - 952)]/4 = [-31 ± √9]/4 = [-31 ± 3]/4. Roots are x = -34/4 = -8.5 and x = -28/4 = -7. For equation (ii): y^2 + 51y + 98 = 0. Roots are y = [-51 ± √(2601 - 392)]/2 = [-51 ± √2209]/2 = [-51 ± 47]/2. Roots are y = -98/2 = -49 and y = -4/2 = -2. So x can be -7 or -8.5, while y can be -2 or -49. Sometimes x > y (when x=-7 and y=-49), sometimes x < y (when x=-8.5 and y=-2). No definite relationship. Option E is correct.

Multiple choice

Two equations (I) and (II) are given in each question. On the basis of these equations you have to decide the relation between ‘x’ and ‘y’ and give answer.(I: 12x^2 + x - 13 = 0 )(II: 6y^2 + 11y + 3 = 0)

  1. x < y

  2. x > y

  3. x ≤ y

  4. x ≥ y

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

Solving 12x² + x - 13 = 0 gives x = 1 or x = -13/12. Solving 6y² + 11y + 3 = 0 gives y = -0.5 or y = -1. When x = 1, x > y for both y values. When x = -13/12 ≈ -1.083, we have: compared to y = -0.5, x < y; compared to y = -1, x < y. Since x can be greater or less than y depending on the values chosen, the relationship cannot be established.

Multiple choice

Two equations (I) and (II) are given in each question. On the basis of these equations you have to decide the relation between ‘x’ and ‘y’ and give answer. $I: (x^2 + 6\sqrt{3}x - 48 = 0) II: (y^2 - 8\sqrt{2}y + 30 = 0)$

  1. x > y

  2. x ≥ y

  3. x < y

  4. x ≤ y

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

Equation I: x² + 6√3x - 48 = 0 gives x = 4√3 or -2√3. Equation II: y² - 8√2y + 30 = 0 has discriminant = 128 - 120 = 8, giving two real roots. Solving gives x > y as the relationship.

Multiple choice

Two equations (I) and (II) are given in each question. On the basis of these equations you have to decide the relation between ‘x’ and ‘y’ and give answer. I: 63x-194√x+143=0 II: 99y-255√y+150=0

  1. x > y

  2. x ≥ y

  3. x < y

  4. x ≤ y

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

Let a = √x, b = √y. Equation I gives 63a² - 194a + 143 = 0 with roots a = 11/7, 13/9. Equation II gives 99b² - 255b + 150 = 0 with roots b = 5/3, 10/11. This gives x values 121/49, 169/81 and y values 25/9, 100/121. No definite relationship exists.