Tuesday, June 25, 2019

Quantitative Aptitude: Quadratic Equations Questions Set 56

Directions(1-10): Find the value of x and y and compare their values, then choose a correct option.
  1. I. 6𝑥^2 + 77𝑥 + 121 = 0
    II. 𝑦^2 + 9𝑦 − 22 = 0

    No relation
    x ≥ y
    y ≥ x
    y > x
    x > y
    Option A
    I.6x^2 + 77x + 121 = 0
    6x ^2 + 66x + 11x + 121 = 0
    6x(x + 11) + 11(x + 11) = 0
    (6x + 11)(x + 11) = 0
    x = − 11/ 6 , −11
    II. Y^ 2 + 9y − 22 = 0
    y^ 2 + 11y − 2y − 22 = 0
    y(y + 11) − 2(y + 11)
    (y − 2)(y + 11) = 0
    y = 2, −11
    No relation

     


  2. I. 25x² + 35x + 12 = 0
    II. 10y² + 9y + 2 = 0
    y > x
    No relation
    x > y
    y ≥ x
    x ≥ y
    Option A
    I. 25x² + 35x + 12 = 0
    ⇒ 25x² + 20x + 15x + 12 = 0
    ⇒ (5x + 4) (5x + 3) = 0
    ⇒ 𝑥 = − 4/5 , − 3/5
    II. 10y² + 9y + 2 = 0
    ⇒ 10y² + 5y + 4y +2 = 0
    ⇒ (2y + 1) (5y + 2) = 0
    ⇒ y = − 1/2 , − 2/5
    y > x

     


  3. I. 3x² - 13x + 14 = 0
    II. 2y² - 5y + 3 =0
    y ≥ x
    x > y
    y > x
    No relation
    x ≥ y
    Option B
    I. 3x² - 13x + 14 = 0
    ⇒ 3x² - 6x – 7x + 14 = 0
    ⇒ (x – 2) (3x – 7) = 0
    ⇒ 𝑥 = 2, 7/3
    II. 2y² - 5y + 3 = 0
    ⇒ 2y² - 2y – 3y + 3 = 0
    ⇒ (y – 1) (2y – 3) = 0
    ⇒ y = 1, 3/2
    x > y

     


  4. I. 2x² - 9x + 10 = 0
    II. 2y² - 13y + 20 = 0
    y > x
    y ≥ x
    x > y
    x ≥ y
    No relation
    Option B
    I. 2x² - 9x + 10 = 0
    ⇒ (x – 2) (2x - 5) = 0
    ⇒ 𝑥 = 2, 5/2
    II. 2y² - 13y + 20 = 0
    ⇒ (y - 4) (2y - 5) = 0
    ⇒ 𝑦 = 4,5/2
    y ≥ x

     


  5. I. 𝑥^2 − 19𝑥 + 84 = 0
    II. 𝑦^2 − 25𝑦 + 156 = 0
    x > y
    y ≥ x
    x ≥ y
    y > x
    No relation
    Option B
    I. 𝑥^ 2 − 19𝑥 + 84 = 0
    𝑥^ 2 − 7𝑥 − 12𝑥 + 84 = 0
    (𝑥 − 7)(𝑥 − 12) = 0
    𝑥 = 7, 12
    II. 𝑦 ^2 − 25𝑦 + 156 = 0
    𝑦^2 − 13𝑦 − 12𝑦 + 156 = 0
    (𝑦 − 13)(𝑦 − 12) = 0
    ⇒ 𝑦 = 13, 12
    𝑥 ≤ y

     


  6. I. 𝑥^2 + 4𝑥 + 4 = 0
    II. 𝑦^2 − 8𝑦 + 16 = 0
    x ≥ y
    y > x
    y ≥ x
    x > y
    No relation
    Option B
    I. x 2 + 4x + 4 = 0
    (x + 2)^2 = 0
    ⇒ x = −2
    II. y 2 − 8y + 16 = 0
    ⇒ (y − 4)^2 = 0
    ⇒ y = 4
    y > x

     


  7. I. x² = 144
    II. y² - 24y + 144 = 0
    x > y
    y ≥ x
    No relation
    x ≥ y
    y > x
    Option B
    I. x² = 144
    ⇒ 𝑥 = ±12
    II. y² - 24y + 144 = 0
    ⇒ (y – 12) ² = 0
    ⇒ y – 12 = 0
    ⇒ y = 12
    𝑥 ≤ y

     


  8. I. 3x² - 13x – 10 = 0
    II. 3y² + 10y – 8 = 0
    y ≥ x
    x ≥ y
    x > y
    y > x
    No relation
    Option E
    I. 3x² - 13x – 10 = 0
    3x² - 15x + 2x – 10 = 0
    (x – 5) (3x + 2) = 0
    x= 5, − 2 /3
    II. 3y² + 10y – 8 = 0
    ⇒ 3y² + 12y – 2y – 8 = 0
    ⇒ (y + 4) (3y – 2) = 0
    ⇒ 𝑦 = −4, 2/3
    No relation

     


  9. I. 2x² - 21x + 52 = 0
    II. 2y² - 11y + 12 = 0
    y ≥ x
    y > x
    No relation
    x ≥ y
    x > y
    Option D
    I. 2x² - 21x + 52 = 0
    ⇒ 2x² - 8x – 13x + 52 = 0
    ⇒ (x – 4) (2x – 13) = 0
    ⇒ 𝑥 = 4, 13/2
    II. 2y² - 11y + 12 = 0
    ⇒ 2y² - 8y – 3y + 12= 0
    ⇒ (y – 4) (2y – 3) = 0
    ⇒ 𝑦 = 4, 3 /2
    x ≥ y

     


  10. I. 12x² + 7x + 1 = 0
    II. 6y² + 5y + 1 = 0
    y ≥ x
    No relation
    y > x
    x > y
    x ≥ y
    Option E
    I. 12x² + 7x + 1 = 0
    ⇒ 12x² + 4x + 3x + 1 = 0
    ⇒ (3x + 1) (4x + 1) = 0
    ⇒ 𝑥 = − 1/ 4 , − 1/ 3
    II. 6y² + 5y+ 1= 0
    ⇒ (2y + 1) (3y + 1) = 0
    ⇒ 𝑦 = − 1 /2 , − 1/ 3
    x ≥ y

     




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