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

     

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

  7. 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

     

  8. 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

     

  9. 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

     

  10. 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 .

     

  11. 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

     


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