Aptitude Questions: Quadratic Equations Set 19

Hello Aspirants. Welcome to Online Quantitative Aptitude Section in AffairsCloud.com. Here we are creating sample questions inQuadratic Equations which is common for all the competitive exams. We have included Some questions that are repeatedly asked in bank exams !!!

Follow the link To solve Quadratic Equations with the help of Number Line

Bank PO Level Questions :-

1. x² – 1 = 0, y² + 4y + 3 = 0
   A. X > Y
   B. X < Y
   C. X ≥ Y
   D. X ≤ Y
   E. X = Y or relation cannot be established
   

   Answer & Explanation

   C. X ≥ Y
   Explanation:
   x² = 1
   x = ± 1
   y² + 4y + 3 = 0
   y² + y + 3y + 3 = 0
   y = -1, -3
   Put on number line
   -3 -1 -1 1
   
   
2. x² – 10x + 24 = 0, y² – 14y + 48 = 0
   A. X > Y
   B. X < Y
   C. X ≥ Y
   D. X ≤ Y
   E. X = Y or relation cannot be established
   

   Answer & Explanation

   D. X ≤ Y
   Explanation:
   x² – 10x + 24 = 0
   x² – 6x – 4x + 24 = 0
   x = 4, 6
   y² – 14y + 48 = 0
   y² – 6y – 8y + 48 = 0
   y = 6, 8
   Put on number line
   4 6 6 8
   
   
3. 2x² – 13x + 20 = 0, 2y² – 7y + 6 = 0
   A. X > Y
   B. X < Y
   C. X ≥ Y
   D. X ≤ Y
   E. X = Y or relation cannot be established
   

   Answer & Explanation

   A. X > Y
   Explanation:
   2x² – 13x + 20 = 0
   2x² – 8x – 5x + 20 = 0
   x = 2.5, 4
   2y² – 7y + 6 = 0
   2y² – 3y – 4y + 6 = 0
   y = 1.5, 2
   Put on number line
   1.5 2 2.5 4
   
   
4. (15/√x) + (9/√x) = 11√x, (√y/4) + (5√y/12)= (1/√y)
   A. X > Y
   B. X < Y
   C. X ≥ Y
   D. X ≤ Y
   E. X = Y or relation cannot be established
   

   Answer & Explanation

   A. X > Y
   Explanation:
   (15/√x) + (9/√x) = 11√x
   24/√x = 11√x
   x = 24/11 = 2.18
   (√y/4) + (5√y/12)= (1/√y)
   (8√y/12) = (1/√y)
   y = 1.5
   
   
5. x⁴ – 227 = 398, y² + 321 = 346
   A. X > Y
   B. X < Y
   C. X ≥ Y
   D. X ≤ Y
   E. X = Y or relation cannot be established
   

   Answer & Explanation

   E. X = Y or relation cannot be established
   Explanation:
   x⁴ – 227 = 398
   x⁴ = 625
   Take square root on both sides
   x² = 25
   x = 5, -5
   y² + 321 = 346
   y² = 25
   y² = 25
   y = ±5
   
   
6. x² + x – 20 = 0, y² + y – 30 = 0
   A. X > Y
   B. X < Y
   C. X ≥ Y
   D. X ≤ Y
   E. X = Y or relation cannot be established
   

   Answer & Explanation

   E. X = Y or relation cannot be established
   Explanation:
   x² + x – 20 = 0
   x² + 5x – 4x + 20 = 0
   x = -5, 4
   y² + y – 30 = 0
   y² +6y -5y – 30 = 0
   y = -6, 5
   Put on number line
   -6 -5 4 5
   
   
7. x² – 365 = 364, y – √324 = √81
   A. X > Y
   B. X < Y
   C. X ≥ Y
   D. X ≤ Y
   E. X = Y or relation cannot be established
   

   Answer & Explanation

   D. X ≤ Y
   Explanation:
   x² – 365 = 364
   x² = 729
   x = ± 27
   y – √324 = √81
   y – 18 = 9
   y = 27
   Put on number line
   -27 27 27
   
   
8. 9x – 15.45 = 54.55 + 4x, √(y+155) – 6 = 7
   A. X > Y
   B. X < Y
   C. X ≥ Y
   D. X ≤ Y
   E. X = Y or relation cannot be established
   

   Answer & Explanation

   E. X = Y or relation cannot be established
   Explanation:
   9x – 15.45 = 54.55 + 4x
   9x – 4x = 54.55 + 15.45
   5x = 70 => x= 14
   √(y+155) – 6 = 7
   √(y+155) = 13
   Squaring on both sides
   y + 155 = 169
   y = 14
   
   
9. x² + 11x + 30 = 0, y² + 7y + 12 = 0
   A. X > Y
   B. X < Y
   C. X ≥ Y
   D. X ≤ Y
   E. X = Y or relation cannot be established
   

   Answer & Explanation

   B. X < Y
   Explanation:
   x² + 11x + 30 = 0
   x² + 6x + 5x + 30 = 0
   x = -6, -5
   y² + 7y + 12 = 0
   y² + 4y + 3y + 12 = 0
   y = -4, -3
   Put on number line
   -6 -5 -4 -3
   
   
10. y² – x² = 32, y – x = 2
    A. X > Y
    B. X < Y
    C. X ≥ Y
    D. X ≤ Y
    E. X = Y or relation cannot be established
    

    Answer & Explanation

    B. X < Y
    Explanation:
    y² – x² = 32
    (y + x)(y – x) = 32
    From equation(2) y – x = 2
    (y + x)2 = 32
    y + x = 16 –(a)
    y – x = 2 —(b)
    2y = 18 => y = 9
    9 – x = 2
    x = 7