Quantitative Aptitude: Quadratic Equations Set 17

Quadratic Equations SBI PO, IBPS PO, NIACL, NICl, LIC, Dena Bank PO PGDBF, BOI, 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. 3x² – 25x + 52 = 0,
   II. 3y² – 16y + 16 = 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 – 25x + 52 = 0
   3x² – 12x – 13x + 52 = 0
   Gives x = 4, 13/3
   3y² – 16y + 16 = 0
   3y² – 14y – 4y + 16 = 0
   Gives y = 4, 4/3
   
2. I. 3x² – 8x – 16 = 0,
   II. 2y² – 5y – 12 = 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 E
   Solution:
   3x² – 8x – 16 = 0
   3x² – 12x + 4x – 16 = 0
   Gives x = -4/3, 4
   2y² – 5y – 12 = 0
   2y² – 8y + 3y – 12 = 0
   Gives y = -3/2, 4
   
3. I. 3x² + 22x + 24 = 0,
   II. 2y² + 7y – 4 = 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 E
   Solution:
   3x² + 22 x + 24 = 0
   3x² + 18x + 4x + 24 = 0
   So x = -4/3, -6
   2y² + 7y – 4 = 0
   2y² + 8y – y – 4 = 0
   Gives y = -4, 1/2
   
4. I. 2x² – 9x + 4 = 0,
   II. 2y² – 17y + 36 = 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:
   2x² – 9x + 4 = 0
   2x² – 8x – x + 4 = 0
   So x = 4 , 1/2
   2y² – 17y + 36 = 0
   2y² – 8y – 9y + 36 = 0
   Gives y= 4, 9/2
   
5. I. 3x² + 7x – 6 = 0,
   II. 3y² – 19y + 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:
   Explanation:
   3x² + 7x – 6 = 0
   3x² + 9x – 2x – 6 = 0
   Gives x = -3, 2/3
   3y² – 19y + 20 = 0
   3y² – 15y – 4y + 20 = 0
   Gives y = 4/3, 5
   
6. I. 3x² – 4x – 4 = 0,
   II. 3y² + 16y + 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 A
   Solution:
   3x² – 4x – 4 = 0
   3x² – 6x + 2x – 4 = 0
   Gives x = -2/3, 2
   3y² + 16y + 20 = 0
   3y² + 6y + 10y + 20 = 0
   Gives y = -10/3, -2
   
7. I. 7x² + 19x – 6 = 0,
   II. 2y² + 13y + 21 = 0
   A) x > y
   B) x < y
   C) x ≥ y
   D) x ≤ y
   E) x = y or relationship cannot be determined
   
   View Answer

   Option C
   Solution:
   7x² + 19x – 6 = 0
   7x² + 21x – 2x – 6 = 0
   Gives x = -3, 2/7
   2y² + 13y + 21 = 0
   2y² + 6y + 7y + 21 = 0
   So y = -7/2, -3
   
8. I. x² + (4 + √2)x + 4√2 = 0
   II. 3y² – (1 + 3√3)y + √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)x + 4√2 = 0
   (x² + 4x) + (√2x + 4√2) = 0
   x (x + 4) + √2 (x + 4) = 0
   So x = -4, -√2 (-1.4)
   3y² – (1 + 3√3)y + √3 = 0
   (3y² – y) – (3√3y – √3) = 0
   y (3y – 1) – √3 (3y – 1) = 0
   So, y = 1/3, √3 (1.7)
   
9. I. 3x² + (3 + 2√2)x + 2√2 = 0
   II. 5y² + (2 + 5√2)y + 2√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 E
   Solution:
   3x² + (3 + 2√2)x + 3√2 = 0
   (3x² + 3x) + (2√2x + 2√2) = 0
   3x (x + 1) + 2√2 (x + 1) = 0
   So x = -1, -2√2/3
   5y² + (2 + 5√2)y + 2√2 = 0
   (5y² + 2y) + (5√2y + 2√2) = 0
   y (5y + 2) + √2 (5y + 2) = 0
   So, y = -2/5, -√2
   
10. I. 4x² – 12x + 5 = 0,
    II. 2y² + 3y – 20 = 0
    A) x > y
    B) x < y
    C) x ≥ y
    D) x ≤ y
    E) x = y or relation cannot be established
    
    View Answer

    Option E
    Solution:
    4x² – 12x + 5 = 0
    4x² – 2x – 10x + 5 = 0
    x = 1/2, 5/2
    2y² + 3y – 20 = 0
    2y² + 8y – 5y – 20 = 0
    So y = -3, 5/2
    

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