Quantitative Aptitude: Quadratic Equations Set 9

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. 8/√x + 9/(√x +1) = 7,
   II. 9/√y – 3/√y = 2
   A) x > y
   B) x < y
   C) x ≥ y
   D) x ≤ y
   E) x = y or relation cannot be established
   
   View Answer

   Option B
   Solution:
   8/√x + 9/(√x +1) = 7
   [8(√x +1) + 9√x]/[√x * (√x +1)] = 7
   17√x + 8 = 7 (x + √x)
   7x – 10√x – 8 = 0
   7x – 14√x + 4√x – 8 = 0
   7√x (√x – 2) + 4 (√x – 2) = 0
   √x cannot be -4/7
   So √x = 2, so x = 4
   9/√y – 3/√y = 2
   (9 – 3)/√y = 2
   Gives √y = 3, so y = 9
   
2. I. 9/√x + 3/√x = √x + 1,
   II. 4y² + 5y – 6 = 0
   A) x > y
   B) x < y
   C) x ≥ y
   D) x ≤ y
   E) x = y or relation cannot be established
   
   View Answer

   Option A
   Solution:
   9/√x + 3/√x = √x + 1
   12/√x = √x + 1
   x + √x – 12 = 0
   x + 4√x – 3√x – 12 = 0
   √x(√x + 4) – 3 (√x + 4) = 0
   √x cannot be -4, So √x = 3 => x = 9
   4y² + 5y – 6 = 0
   4y² + 5y – 6 = 0
   Gives y = -2, 3/4
   Put all values on number line and analyze the relationship
   -2… 3/4… 9
3. I. 6x² + 13x + 6 = 0,
   II. 6y² – y – 2 = 0
   A) x > y
   B) x < y
   C) x ≥ y
   D) x ≤ y
   E) x = y or relation cannot be established
   
   View Answer

   Option B
   Solution:
   6x² + 13x + 6 = 0
   6x² + 9x + 4x + 6 = 0
   Gives x = -2/3, -3/2
   6y² – y – 2 = 0
   6y² + 3y – 4y – 2 = 0
   Gives y = -1/2, 2/3
   Put all values on number line and analyze the relationship
   -3/2… -2/3… -1/2… 2/3
4. I. 3x² + 14x – 5 = 0,
   II. 3y² – 11y + 6 = 0
   A) x > y
   B) x < y
   C) x ≥ y
   D) x ≤ y
   E) x = y or relation cannot be established
   
   View Answer

   Option B
   Solution:
   3x² + 14x – 5 = 0
   3x² + 15x – x – 5 = 0
   Gives x = -5, 1/3
   3y² – 11y + 6 = 0
   3y² – 9y – 2y + 6 = 0
   Gives y = 2/3, 3
   Put all values on number line and analyze the relationship
   -5… 1/3… 2/3… 3
5. I. 6x² + 5x – 1 = 0,
   II. 3y² – 10y + 3 = 0
   A) x > y
   B) x < y
   C) x ≥ y
   D) x ≤ y
   E) x = y or relation cannot be established
   
   View Answer

   Option B
   Solution:
   6x² + 5x – 1 = 0
   6x² + 6x – x – 1 = 0
   Gives x = -1, 1/6
   3y² – 10y + 3 = 0
   3y² – 9y – y + 3 = 0
   Gives y = 1/3, 3
   Put all values on number line and analyze the relationship
   -1… 1/6… 1/3… 3
6. I. 12x² – 5x – 3 = 0,
   II. 3y² – 11y + 6 = 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:
   12x² – 5x – 3 = 0
   12x² + 4x – 9x – 3 = 0
   Gives x = -1/3, 3/4
   3y² – 11y + 6 = 0
   3y² – 9y – 2y + 6 = 0
   Gives y = 2/3, 3
   Put all values on number line and analyze the relationship
   -1/3… 2/3… 3/4… 3
7. I. 6x² + 7x + 2 = 0,
   II. 15y² – 38y – 40 = 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:
   6x² + 7x + 2 = 0
   6x² + 4x + 3x + 2 = 0
   Gives x = -2/3, -1/2
   15y² – 38y – 40 = 0
   15y² + 12y – 50y – 40 = 0
   Gives y = -4/5, 10/3
   Put all values on number line and analyze the relationship
   -4/5… -2/3… -1/2… 10/3
8. I. 3x² – 25x + 52 = 0,
   II. 2y² – 7y + 3 = 0
   A) x > y
   B) x < y
   C) x ≥ y
   D) x ≤ y
   E) x = y or relation cannot be established
   
   View Answer

   Option A
   Solution:
   3x² – 25x + 52 = 0
   3x² – 12x – 13x + 52 = 0
   Gives x = 4, 13/3
   2y² – 7y + 3 = 0
   2y² – 6y – y + 3 = 0
   So y = 1/2, 3
   Put all values on number line and analyze the relationship
   1/2… 3… 4… 13/3
9. I. x² = 1156,
   II. y = √1156
   A) x > y
   B) x < y
   C) x ≥ y
   D) x ≤ y
   E) x = y or relation cannot be established
   
   View Answer

   Option D
   Solution:
   x² = 1156,
   So x = -34, 34
   y = √1156
   So y = 34
   Put all values on number line and analyze the relationship
   -34… 34
10. I. x² – √3969 = √6561,
    II. y² – √1296 = √4096
    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:
    x² – √3969 = √6561
    x² – 63 = 81
    x² = 144
    So x = -12, 12
    y² – √1296 = √4096
    y² – 36 = 64
    y² = 100
    So y = -10, 10
    Put all values on number line and analyze the relationship
    -12… -10….10…. 12