Quantitative Aptitude: Quadratic Equations Questions Set 46

Directions (1-5): In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly-

1. I. x = (4)^^1/2/ (9)^^1/2
   II. 9y² -12y +4 =0
   x>=y
   x>y
   y>=x
   Relation cannot be established.
   y>x
   Option D
   I.x = (4)^1/2/ (9)^^1/2
   =>x = 2/3
   II.9y² -12y +4 =0
   => (3y – 2)² = 0
   => y = 2/3
   Hence, x= y or relation cannot b established.

    

2. I. 20x² – 79x + 77 = 0
   II. 4y² + 9y – 28 = 0
   x>=y
   y>x
   y>=x
   x>y
   Relation cannot be established.
   Option A
   I. 20x² – 79x + 77 = 0
   => 5x(4x-7) -11 (4x-7) = 0
   => (4x-7)(5x-11)=0
   => x = 11/5 , 7/4
   II. 4y² + 9y – 28 = 0
   => 4y(y+4) – 7(y+4) = 0
   => (y + 4 ) (4y – 7) = 0
   => y = -4 & 7/4
   Hence, x >= y

    

3. I. x² + 13x + 42 =0
   II. y² = 36
   y>x
   Relation cannot be established
   y>=x
   y>x
   x>y
   Option C
   I. x² + 13x + 42 =0
   =>(x+7)(x+6)=0
   => x = -7 & -6
   II. y² = 36
   => y = -6 & +6
   Hence, y >=x

    

4. I. 3x² + 2x -1 =0
   II. 2y² + 7y + 6 =0
   x>y
   x>=y
   y>x
   y>=x
   Relation cannot be established .
   Option A
   I. 3x² + 2x -1 =0
   =>(3x-1)(x+1)=0
   => x = 1/3 & -1
   II. 2y² + 7y + 6 =0
   => (2y+3)(y+2)=0
   => y = -3/2 & -2
   Hence, x>y

    

5. I. 6x² + 5x +1 = 0
   II. 20y² + 9y = -1
   Relation cannot be established.
   y>=x
   y>x
   x>=y
   x>y
   Option C
   I. 6x² + 5x +1 = 0
   =>(3x+1)(2y+1)=0
   =>x = -1/3 & -1/2
   II. 20y² + 9y = -1
   =>(4y+1)(5y+1) =0
   =>y = -1/4 & -1/5
   Hence, y>x

    

Directions (6-10): Find out the relationship between m and n and choose a required option.

6. I. 9m² + 45m + 26 =0
   II. 7n² – 59n – 36 = 0
   n>m
   n>=m
   m>n
   m>=n
   Relation cannot be established .
   Option A
   I. 9m² + 45m + 26 =0
   => m = – (6/9) & – (39/6)
   II. 7n² – 59n – 36 = 0
   => n = -(4/7) & 9
   Hence, n > m

    

7. I. m – (2401)^1/2 = 0
   II. n² – 2401 = 0
   n>=m
   m>n
   Relation cannot be established .
   n>m
   m>=n
   Option E
   I. m – (2401)^1/2 = 0
   => m = 49
   II. n² – 2401 = 0
   => n = -49 & +49
   Hence, m>=n

    

8. I. m⁷ – (28*7)^7.5/ (m)^1/2 = 0
   II. (n)^1/3 = 27
   m>=n
   m>n
   n>m
   n>=m
   Relation cannot be established.
   Option C
   I. m⁷ – (28*7)^7.5/ (m)^1/2 = 0
   => m⁷* (m)^1/2 – (28*7)^7.5 = 0
   => m^7.5 – (28*7)^7.5 = 0
   => m = 196
   II. (n)^1/3 = 27
   => n = (27)³
   Hence, n>m

    

9. I. (24-10m)^1/2 = 3 – 4m
   II. 6n² – 5n – 25 =0
   Relation cannot be established .
   n>m
   m>=n
   m>n
   n>=m
   Option A
   I. (24-10m)^1/2 = 3 – 4m
   => (24 – 10m) = (3 – 4m)²
   => 16m² – 14m – 15 =0
   => (2m-3)(8m+5)=0
   =>m = 3/2 & -5/8
   II. 6n² – 5n – 25 =0
   =>(3n+5)(2n-5)=0
   =>n = -5/3 & 5/2
   Relation can’t be established .

    

10. I. (625m²)^1/2 – 375 =0
    II. (2116)^1/2 n – 736 = 0
    n>m
    m>n
    n>=m
    m>=n
    Relation cannot be established .
    Option A
    I. (625m²)^1/2– 375 =0
    => 25m – 375 = 0
    => m = 375/25 = 15
    II. (2116)^1/2 n – 736 = 0
    => 46 n – 736 = 0
    => n = 16
    Hence,n>m