Quantitative Aptitude: Quadratic Equations Questions Set 57

Directions(1-10): Find the values of x and y and compare their values, then choose a correct option.

  1. I. 9x² – 45x + 56 = 0
    II. 4y² – 17y + 18 = 0
    No relation
    x ≥ y
    x > y
    y > x
    𝑦 ≥ x
    Option C
    I. 9x² – 45x + 56 = 0
    ⇒ 9x² – 24x – 21x + 56 = 0
    ⇒ 3x (3x – 8) – 7 (3x – 8) = 0
    ⇒ (3x – 8) (3x – 7) = 0
    ⇒ x = 8/3, 7/3
    II. 4y² – 17y + 18 = 0
    ⇒ 4y² – 8y – 9y + 18 = 0
    ⇒ (y – 2) (4y – 9) = 0
    ⇒ y = 2, 9/4
    x > y

     

  2. I. x² – 72 = x
    II. y² = 64
    𝑦 ≥ x
    x > y
    y > x
    No relation
    x ≥ y
    Option D
    I. x² –x – 72 = 0
    x² – 9x + 8x – 72 = 0
    x (x – 9) +8 (x – 9) = 0
    x = 9 or -8
    II. y² = 64
    y = ± 8
    No relation

     

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

     

  4. I. 4x² + 16x + 15 = 0
    II. 2y² + 3y + 1 = 0
    𝑦 ≥ x
    x > y
    y > x
    x ≥ y
    No relation
    Option C
    I. 4x² + 16x + 15 = 0
    ⇒ 4x² + 10x + 6x + 15 = 0
    ⇒ 2x (2x + 5) + 3 (2x + 5) = 0
    ⇒ (2x + 5) (2x + 3) = 0
    ⇒ x= -5/2, -3/2
    II. 2y² + 3y + 1 = 0
    ⇒ 2y² + 2y + y + 1 = 0
    ⇒ (y + 1) (2y + 1) = 0
    ⇒ y = -1, -1/2
    y > x

     

  5. I. 2x² + 11x + 14 = 0
    II. 2y² + 15y + 28 = 0
    y > x
    𝑦 ≥ x
    x > y
    x ≥ y
    No relation
    Option D
    I. 2x² + 11x + 14 = 0
    ⇒ 2x² + 4x + 7x + 14= 0
    ⇒ (x + 2) (2x + 7) = 0
    ⇒ x = -2, -7/2
    II. 2y² + 15y + 28= 0
    ⇒ 2y² + 8y + 7y + 28 = 0
    ⇒ (y + 4) (2y + 7) = 0
    ⇒ y = –4, –7/2
    x ≥ y

     

  6. I. 4x² – 25x + 39 = 0
    II. 18y² – 15y + 3 = 0
    x ≥ y
    No relation
    y > x
    x > y
    𝑦 ≥ x
    Option D
    I. 4x² – 25x + 39 = 0
    4x² – 13x – 12x + 39 = 0
    x (4x – 13) – 3 (4x – 13) = 0
    𝑥 = 13/ 4 𝑜𝑟 3
    II. 18y² – 15y + 3 = 0
    18y² – 9y – 6y + 3= 0
    9y (2y – 1) – 3 (2y – 1) = 0
    𝑦 = 1 /2 𝑜𝑟 1/ 3
    x > y

     

  7. I. 6x² + 23x + 20 = 0
    II. 6y² +31y + 35 = 0
    𝑦 ≥ x
    No relation
    x ≥ y
    y > x
    x > y
    Option B
    I. 6x² + 23x + 20 = 0
    6x² + 15x + 8x + 20 = 0
    3x (2x + 5) +4(2x + 5) = 0
    𝑥 = −5/ 2 𝑜𝑟 −4 /3
    II. 6y² + 31y + 35 = 0
    6y² + 21y + 10y + 35 = 0
    3y (2y + 7) + 5(2y + 7) = 0
    𝑦 = −7 /2 𝑜𝑟 −5/ 3
    No relation

     

  8. I. 30x² + 11x + 1 = 0
    II. 42y² + 13y + 1 = 0
    𝑦 ≥ x
    x ≥ y
    x > y
    No relation
    y > x
    Option A
    I. 30x² + 11x +1 = 0
    30x² + 5x + 6x + 1= 0
    5x (6x + 1) +1 (6x + 1) = 0
    𝑥 = − 1 /6 or − 1 /5
    II. 42y² + 13y + 1 = 0
    42y² + 6y + 7y + 1 = 0
    6y (7y + 1) +1 (7y + 1) = 0
    𝑦 = − 1/ 7 𝑜𝑟 − 1 /6
    𝑦 ≥ x

     

  9. I. 3x + 5y = 28
    II. 8x – 3y = 42
    x > y
    y > x
    No relation
    x ≥ y
    𝑦 ≥ x
    Option A
    I. 3x + 5y = 28 …(i)
    II. 8x – 3y = 42 …..(ii)
    Multiplying (i) by 3 and (ii) by 5
    𝑥 = 6
    y = 2
    x > y

     

  10. I. 9x² – 36x + 35 = 0
    II. 2y² – 15y – 17 = 0
    x ≥ y
    No relation
    x > y
    y > x
    𝑦 ≥ x
    Option B
    I. 9x² – 36x + 35 = 0
    ⇒ 9x² – 21x – 15x + 35 = 0
    ⇒ 3x (3x – 7) -5 (3x – 7) = 0
    ⇒ (3x- 7) (3x – 5) = 0
    ⇒ 𝑥 = 5 /3 , 7/ 3
    II. 2y² – 15y – 17 = 0
    ⇒ 2y² – 17y + 2y – 17 = 0
    ⇒ (y + 1) (2y – 17) = 0
    ⇒ y= -1, 17 /2
    No relation

     


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