Quantitative Aptitude

Quadratic Equations

367 Questions
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Quadratic Equations Questions

Multiple choice

Directions: In this question, two equations numbered I and II are given. You have to solve both the equations and mark the correct answer. I. 3x2 - 7x + 2 = 0 II. 2y2 - 9y + 10 = 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
D Correct answer
Explanation

I: 3x^2 - 7x + 2 = 0 -> (3x-1)(x-2) = 0 -> x = 1/3, 2. II: 2y^2 - 9y + 10 = 0 -> (2y-5)(y-2) = 0 -> y = 2.5, 2. Comparing: x values are {0.33, 2}, y values are {2.5, 2}. In all cases, x <= y.

Multiple choice

Directions: In this question, two equations numbered I and II are given. You have to solve both the equations and mark the correct answer. I. 3x2 - 19x + 30 = 0 II. 5y2 - 29y + 42 = 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
D Correct answer
Explanation

Solving I: 3x^2 - 19x + 30 = 0 gives (3x - 10)(x - 3) = 0, so x = 3.33 or 3. Solving II: 5y^2 - 29y + 42 = 0 gives (5y - 14)(y - 3) = 0, so y = 2.8 or 3. Comparing values: x can be 3.33 or 3, y can be 2.8 or 3. Since 3.33 > 2.8, 3.33 > 3, 3 > 2.8, and 3 = 3, x is always >= y.

Multiple choice

Directions: In this question, two equations numbered I and II are given. You have to solve both the equations and mark the correct answer. I. 6x2 + 11x - 112 = 0 II. 2y2 - 15y + 28 = 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
C Correct answer
Explanation

I: 6x^2 + 11x - 112 = 0. Roots are x = 3.5, -5.33. II: 2y^2 - 15y + 28 = 0. Roots are y = 4, 3.5. Comparing: 3.5 <= 4 and 3.5 <= 3.5, -5.33 <= 4 and -5.33 <= 3.5. Thus, x <= y.

Multiple choice

Directions: In this question, two equations numbered I and II are given. You have to solve both the equations and mark the correct answer. I. x2 - 16x + 63 = 0 II. y2 - 2y - 35 = 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
C Correct answer
Explanation

Solving equation I: x^2 - 16x + 63 = 0 gives (x - 9)(x - 7) = 0, so x = 9 or 7. Solving equation II: y^2 - 2y - 35 = 0 gives (y - 7)(y + 5) = 0, so y = 7 or -5. Comparing values, x is always >= y.

Multiple choice

Directions: In the following question, equations (I) and (II) are given. You have to solve these equations and choose the correct option. I. 20x2 – 130x = –200 II. 9y2 + 6y + 1 = 0

  1. x > y

  2. x < y

  3. x ≥ y

  4. x ≤ y

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

Equation I: 20x^2 - 130x + 200 = 0 -> 2x^2 - 13x + 20 = 0. Roots are x = 4 and x = 2.5. Equation II: 9y^2 + 6y + 1 = 0 -> (3y+1)^2 = 0. Root is y = -1/3. Since 4 > -1/3 and 2.5 > -1/3, x > y.

Multiple choice

Directions: In the following question, two equations are given. You have to solve both the equations and choose the correct relationship accordingly. I. 7x2 + 16x - 15 = 0 II. 3y2 + 25y + 50 = 0

  1. x > y

  2. x ≥ y

  3. x < y

  4. x ≤ y

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

7x^2 + 16x - 15 = 0 => (7x - 5)(x + 3) = 0. Roots: x = 5/7, -3. 3y^2 + 25y + 50 = 0 => (3y + 10)(y + 5) = 0. Roots: y = -10/3, -5. Since 5/7 > -10/3 and -3 > -5, x > y.

Multiple choice

Directions: In the following question, two equations are given. You have to solve both the equations and choose the correct relationship accordingly. I. 2x2 - 53x + 351 = 0 II. 2y2 - 51y + 325 = 0

  1. x > y

  2. x ≥ y

  3. x < y

  4. x ≤ y

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

Solving 2x^2 - 53x + 351 = 0 gives roots 13 and 13.5. Solving 2y^2 - 51y + 325 = 0 gives roots 12.5 and 13. Comparing the sets, x >= y is the correct relationship.

Multiple choice

Directions: In the following question, two equations are given. You have to solve both the equations and choose the correct relationship accordingly. I. 15x2 + 14x + 3 = 0 II. 15y2 - y - 2 = 0

  1. x > y

  2. x ≥ y

  3. x < y

  4. x ≤ y

  5. x = y, or no relationship can be established
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Solving 15x^2 + 14x + 3 = 0: (3x+1)(5x+3) = 0, so x = -1/3, -3/5. Solving 15y^2 - y - 2 = 0: (3y+1)(5y-2) = 0, so y = -1/3, 2/5. Comparing: -1/3 = -1/3, -1/3 < 2/5, -3/5 < -1/3, -3/5 < 2/5. Thus, x <= y.

Multiple choice

Directions: In the following question, two equations are given. You have to solve both the equations and choose the correct relationship accordingly. I. 9x2 - 31x + 26 =0 II. 5y2 - 12y + 7 = 0

  1. x > y

  2. x ≥ y

  3. x < y

  4. x ≤ y

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

Solving 9x^2 - 31x + 26 = 0: roots are 2 and 13/9 (approx 1.44). Solving 5y^2 - 12y + 7 = 0: roots are 1 and 1.4. Since 2 > 1.4 and 1.44 > 1.4, x is generally greater than y.

Multiple choice

Directions: In the following question, two equations numbered I and II are given. You have to solve both the equations and give answer accordingly. I. 20x2 - 31x + 12 = 0 II. 4y2 + 5y - 6 = 0

  1. x > y

  2. x ≥ y

  3. x < y

  4. x ≤ y

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

I: 20x^2 - 31x + 12 = 0 -> (4x-3)(5x-4) = 0 -> x = 0.75, 0.8. II: 4y^2 + 5y - 6 = 0 -> (4y-3)(y+2) = 0 -> y = 0.75, -2. Comparing: x >= y is true since 0.75=0.75 and 0.8 > -2.

Multiple choice

Directions: In the following question, two equations numbered I and II are given. You have to solve both the equations and then choose the correct option. I. x2 − 18x + 45 = 0 II. 3y2 + 2y = 1

  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
A Correct answer
Explanation

Eq I: x^2 - 18x + 45 = 0 -> (x-15)(x-3) = 0, so x = 15 or 3. Eq II: 3y^2 + 2y - 1 = 0 -> (3y-1)(y+1) = 0, so y = 1/3 or -1. Comparing values, x is always greater than y.

Multiple choice

Directions: In the following question, two equations numbered I and II are given. You have to solve both the equations and then choose the correct option. I. x2 + 5x + 6 = 0 II. 11y2 + 18y + 7 = 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

I: x^2 + 5x + 6 = 0 => (x+2)(x+3) = 0 => x = -2, -3. II: 11y^2 + 18y + 7 = 0 => (11y+7)(y+1) = 0 => y = -7/11 (~ -0.63), -1. Since -2 < -1 and -3 < -0.63, x < y.

Multiple choice

Directions: In this question, two equations are given, numbered I and II. You have to solve both the equations and compare the two quantities. Choose the correct option accordingly. I. x2 - 9x + 18 = 0 II. y2 - 5y + 6 = 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

Solving I: x^2 - 9x + 18 = 0 gives (x-3)(x-6)=0, so x=3, 6. Solving II: y^2 - 5y + 6 = 0 gives (y-2)(y-3)=0, so y=2, 3. Comparing the sets {3, 6} and {2, 3}, we see that for any x and y, x >= y is always true.

Multiple choice

Directions: In the following question, two equations (i) and (ii) are given. You have to solve both the equations and mark the correct option accordingly. (i) x2 - 9x + 20 = 0 (ii) 2y2 + y - 3 = 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

Solving (i): x^2 - 9x + 20 = 0 -> (x-4)(x-5)=0, so x=4 or 5. Solving (ii): 2y^2 + y - 3 = 0 -> (2y+3)(y-1)=0, so y=1 or -1.5. In all cases, x > y.