Quantitative Aptitude: Quadratic Equations Set 27

Quadratic Equations Practice Sets for IBPS PO, IBPS RRB, LIC, UIICL, OICL, Bank of Baroda and other competitive exams.

Directions: In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly-

1. I. 4x² – x – 14 = 0,
   II. 2y² – 13y + 20 = 0
   A) If x > y
   B) If x < y
   C) If x ≥ y
   D) If x ≤ y
   E) If x = y or relation cannot be established
   
   View Answer

    Option B
   Solution: 
   4x² – x – 14 = 0
   4x² – 8x + 7x – 14 = 0
   Gives x = -7/4, 2
   2y² – 13y + 20 = 0
   2y² – 8y – 5y + 20 = 0
   Gives y = 5/2, 4
   
2. I. 2x² – 11x + 14 = 0,
   II. 3y² + 13y + 14 = 0
   A) If x > y
   B) If x < y
   C) If x ≥ y
   D) If x ≤ y
   E) If x = y or relation cannot be established
   
   View Answer

    Option A
   Solution: 
   2x² – 11x + 14 = 0
   2x² – 4x – 7x + 14 = 0
   Gives x = 2, 7/2
   3y² + 13y + 14 = 0
   3y² + 6y + 7y+ 14 = 0
   Gives y = -7/3, -2
   
3. I. 3x² + 14x + 15 = 0,
   II. 3y² – 13y – 30 = 0
   A) If x > y
   B) If x < y
   C) If x ≥ y
   D) If x ≤ y
   E) If x = y or relation cannot be established
   
   View Answer

    Option D
   Solution: 
   3x² + 14x + 15 = 0
   3x² + 9x + 5x + 15 = 0
   Gives x = -5/3, 6
   3y² – 13y – 30 = 0
   3y² – 18y + 5y – 30 = 0
   Gives y = -3, -5/3
4. I. 3x² + 28x + 60 = 0,
   II. 2y² – 3y – 20 = 0
   A) If x > y
   B) If x < y
   C) If x ≥ y
   D) If x ≤ y
   E) If x = y or relation cannot be established
   
   View Answer

    Option B
   Solution: 
   3x² + 28x + 60 = 0
   3x² + 18x + 10x + 60 = 0
   Gives x = -6, -10/3
   2y² – 3y – 20 = 0
   2y² – 8y + 5y – 20 = 0
   Gives y= -5/2, 4
   
5. I. 3x² – 8x – 35 = 0,
   II. 3y² + 37y + 104 = 0
   A) If x > y
   B) If x < y
   C) If x ≥ y
   D) If x ≤ y
   E) If x = y or relation cannot be established
   
   View Answer

    Option A
   Solution: 
   3x² – 8x – 35 = 0
   3x² – 15x + 7x – 35 = 0
   Gives x = 5, -7/3
   3y² + 37y + 104 = 0
   3y² + 24y + 13y + 104 = 0
   Gives y= -8, -13/3
6. I. 3x² – 5x – 78 = 0,
   II. 3y² + 28y + 65 = 0
   A) If x > y
   B) If x < y
   C) If x ≥ y
   D) If x ≤ y
   E) If x = y or relation cannot be established
   
   View Answer

    Option C
   Solution: 
   3x² – 5x – 78 = 0
   3x² – 18x + 13x – 78 = 0
   Gives x = -13/3, 6
   3y² + 28y + 65 = 0
   3y² + 15y + 13y + 65 = 0
   Gives y = -5, -13/3
   
7. I. 3x² – 7x – 40 = 0,
   II. 3y² + 26y + 48 = 0
   A) If x > y
   B) If x < y
   C) If x ≥ y
   D) If x ≤ y
   E) If x = y or relation cannot be established
   
   View Answer

    Option C
   Solution: 
   3x² – 7x – 40 = 0
   3x² – 15x + 8x – 40 = 0
   Gives x = -8/3, 5
   3y² + 26y + 48 = 0
   3y² + 18y + 8y + 48 = 0
   Gives y = -6, -8/3
   
8. I. x² + (4 + 2√2)x + 8√2 = 0
   II. 3y² + (3 + 3√2)y + 3√2 = 0
   A) If x > y
   B) If x < y
   C) If x ≥ y
   D) If x ≤ y
   E) If x = y or relation cannot be established
   
   View Answer

   Option B
   Solution:
   x² + (4 + 2√2)x + 8√2 = 0
   (x² + 4x) + (2√2x + 8√2) = 0
   x (x + 4) + 2√2 (x + 4) = 0
   So x = -4, -2√2 (-2.8)
   3y² + (3 + 3√2)y + 3√2 = 0
   (3y² + 3y) + (3√2y + 3√2) = 0
   3y (y + 1) + 3√2 (y + 1) = 0
   So, y = -1, -√2 (-1.41)
   
9. I. 6x² – (3 + 4√3)x + 2√3 = 0,
   II. 4y² – (2 + 4√3)y + 2√3 = 0
   A) x > y
   B) x < y
   C) x ≥ y
   D) x ≤ y
   E) x = y or relationship cannot be determined
   
   View Answer

   Option E
   Solution:
   6x² – (3 + 4√3)x + 2√3 = 0
   (6x² – 3x) – (4√3x – 2√3) = 0
   3x (2x- 1) – 2√3 (2x – 1) = 0,
   So x = 1/2 (0.5), 2√3/3 (1.16)
   4y² – (2 + 4√3)y + 2√3 = 0
   (4y² – 2y) – (4√3y – 2√3) = 0
   2y (2y – 1) – 2√3 (2y – 1) = 0
   So, y = 1/2 (0.5), √3 (1.73)
10. I. x² + (4 + 2√2)x + 8√2 = 0
    II. y² – (2 + 3√3)y + 6√3 = 0
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established
    
    View Answer

    Option B
    Solution:
    x² + (4 + 2√2)x + 8√2 = 0
    (x² + 4x) + (2√2x + 8√2) = 0
    x (x + 4) + 2√2 (x + 4) = 0
    So x = -4, -2√2 (-2.82)
    y² – (2 + 3√3)y + 6√3 = 0
    (y² – 2y) – (3√3y – 6√3) = 0
    y (y – 2) – 3√3 (y – 2) = 0
    So y = 2, 3√3 (5.2)
    

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