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

     


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