Quantitative Aptitude: Quadratic Equations Questions Set 58

Directions(1-10): Find the value of x and y and then compare their values.

1. I. x² + 3x – 4 = 0
   II. y² + 10y + 24 = 0
   
   y > x
   x > y
   No relation
   x ≥ y
   y ≥ x
   Option D
   𝐈. x² + 3x – 4 = 0
   x² + 4x – x – 4 = 0
   x(x + 4) – 1 (x + 4) = 0
   (x − 1) (x + 4) = 0
   x = 1, −4
   II. y² + 10y + 24 = 0
   y² + 4y + 6y + 24 = 0
   y(y + 4) + 6(y + 4) = 0
   (y + 6) (y + 4) = 0
   y = −4, −6
   x ≥ y
   
   
    



2. I. 48/x² − 14/x + 1 = 0
   𝐈𝐈. 45/y² + 1/y = 2
   
   y ≥ x
   x ≥ y
   y > x
   x > y
   No relation
   Option D
   𝐈. 48/x² − 14/x + 1 = 0
   ⇒ x^2 − 14x + 48 = 0
   ⇒ x^2 − 8x − 6x + 48 = 0
   ⇒ x(x − 8) − 6(x − 8) = 0
   ⇒ (x − 8)(x − 6) = 0
   x = 8, 6
   𝐈𝐈. 45/y² + 1/y = 2
   ⇒ 2y^2 − y − 45 = 0
   ⇒ 2y^2 − 10y + 9y − 45 = 0
   ⇒ 2y(y − 5) + 9(y − 5) = 0
   ⇒ (2y + 9)(y − 5) = 0
   y = 5, − 9/2
   x > y
   
   
    



3. I. x² + 2x – 35 = 0
   II. y² + 15y + 56 = 0
   
   No relation
   x > y
   y ≥ x
   x ≥ y
   y > x
   Option D
   I.x² + 2x – 35 = 0
   => x² + 7x – 5x – 35 = 0
   => x (x + 7) – 5 (x + 7) = 0
   => (x – 5) (x + 7) = 0
   x = 5, –7
   II. y² + 15y + 56 = 0
   => y² + 7y + 8y + 56 = 0
   =>y (y + 7) + 6 (y + 7) = 0
   =>(y + 8) (y + 7) = 0
   y = – 8, – 7
   x ≥ y
   
   
    



4. I.2x² + 7x + 5 = 0
   II. 3y² + 12y + 9 = 0
   
   x > y
   y ≥ x
   No relation
   x ≥ y
   y > x
   Option C
   I.2x² + 7x + 5 = 0
   => 2x² + 2x + 5x + 5 = 0
   => 2x (x + 1) + 5 (x + 1) = 0
   => (2x + 5) (x + 1) = 0
   𝑥 = – 5/2 , – 1
   II. 3y² + 12y + 9 = 0
   =>3y² + 9y + 3y + 9 = 0
   => 3y (y + 3) +3 (y + 3) = 0
   =>(3y + 3) (y + 3) = 0
   y = –1, – 3
   No relation can be established.
   
   
    



5. I.(x – 12)² = 0
   II. y² – 21y + 108 = 0
   
   y > x
   x > y
   y ≥ x
   No relation
   x ≥ y
   Option E
   I.(x – 12)² = 0
   => x – 12 = 0
   x = 12
   II. y² – 21y + 108 = 0
   => y² – 12y – 9y + 108 = 0
   =>y (y – 12) – 9 (y – 12) = 0
   =>(y – 9) (y – 12) = 0
   y = 9, 12
   x ≥ y
   
   
    



6. I.x² – 17x + 72 = 0
   II. y² – 27y + 180 = 0
   
   x ≥ y
   y ≥ x
   y > x
   x > y
   No relation
   Option C
   I.x² – 17x + 72 = 0
   x² - 9x – 8x + 72 = 0
   x(x – 9) – 8 (x – 9) = 0
   (x – 8) (x – 9) = 0
   x = 8, 9
   II. y² – 27y + 180 = 0
   y² – 12y – 15y + 180 = 0
   y(y – 12) – 15 (y – 12) = 0
   (y – 15) (y – 12) = 0
   y = 15, 12
   y > x
   
   
    



7. I.2x² – 5x + 3 = 0
   II. 3y² – 4y + 1 = 0
   
   y > x
   No relation
   x > y
   x ≥ y
   y ≥ x
   Option D
   I.2x² – 5x + 3 = 0
   => 2x² – 2x – 3x + 3 = 0
   =>2x (x – 1) – 3(x – 1) = 0
   => (x – 1) (2x – 3) = 0
   x = 1, 3/2
   II. 3y² – 4y + 1 = 0
   =>3y² – 3y – y + 1 = 0
   =>3y(y – 1) –1 (y – 1) = 0
   => (3y – 1) (y – 1) = 0
   =>𝑦 = 1/3 , 1
   x≥y
   
   
    



8. I. (x − 2)² = 9
   𝐈𝐈. (2y + 8)² = 16
   
   y ≥ x
   y > x
   No relation
   x > y
   x ≥ y
   Option C
   𝐈. x² − 16x + 64 = 0
   => x² − 8x – 8x + 64 = 0
   =>x(x − 8) − 8(x − 8) = 0
   => (x − 8) (x − 8) = 0
   x = 8, 8
   𝐈𝐈. y² − 16y + 63 = 0
   =>y² − 7y – 9y + 63 = 0
   => y(y − 7) − 9(y − 7) = 0
   => (y – 9) (y – 7) = 0
   y = 9, 7
   No relation can be established between x & y
   
   
    



9. I. x² − 16x + 64 = 0
   𝐈𝐈. y² − 16y + 63 = 0
   
   y > x
   x ≥ y
   No relation
   x > y
   y ≥ x
   Option D
   𝐈. (x − 2)² = 9
   ⇒ (x − 2) = ± 3
   ⇒ x = 5, −1
   𝐈𝐈. (2y + 8)² = 16
   =>(2y + 8) = ± 4
   ⇒ y = −2, −6
   x > y
   
   
    



10. I. 25/x² − 15/x + 2 = 0
    𝐈𝐈. 40/y² + 1 = 13/y
    
    x > y
    No relation
    y > x
    y ≥ x
    x ≥ y
    Option D
    𝐈. 25/x² − 15 x + 2 = 0
    ⇒ 2x^2 − 15x + 25 = 0
    ⇒ 2x^2 − 10x − 5x + 25 = 0
    =>2x (x − 5) − 5(x − 5) = 0
    =>(2x − 5)(x − 5) = 0
    x = 5/2 , 5
    𝐈𝐈. 40/y² + 1 = 13/y
    ⇒ y^2 − 13y + 40 = 0
    ⇒ y^2 − 8y − 5y + 40 = 0
    ⇒ y(y − 8) − 5(y − 8) = 0
    =>(y − 5)(y − 8) = 0
    y = 5, 8
    y ≥ x