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