# Quantitative Aptitude: Quadratic Equations Questions Set 50

Directions(1-10): Establish the correct relation between x and y and choose a correct option.

1. I.6x – 2 = (x)^1/2
II.2y = (3)^1/2y + 3

x=y or relation cannot be established.
x=<y
x<y
x>y
x>=y
Option C
From I: 6x – 2 = (x)^1/2
Squaring on both the sides, we get
36x^2 – 25x + 4 = 0
=> 36x^2 – 16x – 9x +4 = 0
=> 4x(9x-4)-1(9x-4) = 0
=> x= 4/9,1/4
From II: 2y = (3)^1/2y + 3
Squaring on both the sides, we get
4y^2 – 15y + 9 = 0
=> 4y^2 – 12y -3y + 9 = 0
=> y(4y-3)-3(4y-3) = 0
=> y = ¾,3
x<y

2. I.6x^2 -31x+18 = 0
II.2y^2 -21y+54 = 0

x>y
x<y
x=<y
x=y or relation cannot be established.
x>=y
Option C
From I: 6x^2 -31x+18 = 0
=> 6x^2 -27x -4x+18 = 0
=>3x(2x-9)-2(2x-9) = 0
=> (2x-9)(3x-2) = 0
=> x = 9/2,2/3
From II: 2y^2 -21y+54 = 0
=> 2y^2 – 12y – 9y + 54 =0
=> 2y(y-6)-9(y-6) = 0
=> (y-6)(2y-9) = 0
=> y=6 , 9/2
x=<y

3. I.5x^2 – 27x + 10 = 0
II.y^2 – 14y + 48 =0

x=y or relation cannot be established.
x=<y
x>=y
x<y
x>y
Option D
From I: 5x^2 – 27x + 10 = 0
=> 5x^2 – 25x – 2x + 10 = 0
=> 5x(x-5)-2(x-5) = 0
=>(x-5)(5x-2) = 0
=> x= 5,2/5
From II: y^2 – 14y + 48 =0
=> y^2 – 8y -6y + 48=0
=> y(y-8)-6(y-8) = 0
=> (y-8)(y-6) =0
=> y = 8,6
x<y

4. I.4x^2 -44x + 85 = 0
II.2y^2 +y – 28 = 0

x>=y
x=y or relation cannot be established.
x>y
x=<y
x<y
Option B
From I: 4x^2 -44x + 85 = 0
=> 4x^2 – 34x – 10x + 85 =0
=>2x(2x-17)-5(2x-17) = 0
=> (2x-17)(2x-5) = 0
=>x = 17/2,5/2
From II: 2y^2 +y – 28 = 0
=> 2y^2 + 8y – 7y – 28 =0
=> 2y(y+4)-7(y+4) = 0
=> (y+4)(2y-7) = 0
=> y = -4,7/2
x = y or relation cannot be established.

5. I.8x^2-10x+3 = 0
II.15y^2 -32y +16 = 0

x<y
x>=y
x>y
x=<y
x=y or relation cannot be established.
Option A
From I: 8x^2-10x+3 = 0
=> 8x^2 – 4x – 6x + 3 =0
=> 4x(2x-1)-3(2x-1) = 0
=> (4x-3)(2x-1) = 0
=>x = ¾,1/2
From II: 15y^2 -32y +16 = 0
=> 15y^2 – 12y – 20y + 16 = 0
=> 3y(5y-4)-4(5y-4) = 0
=> (3y-4)(5y-4) = 0
=> y = 4/3,4/5
x<y

6. I.x^2 +2x-63 = 0
II.y^2 +20y+99 = 0

x>y
x<y
x>=y
x=y or relation cannot be established.
x=<y
Option C
From I: x^2 +2x-63 = 0
=> x^2 -7x+9x-63 = 0
=> x(x-7)+9(x-7) = 0
=> (x-7)(x+9) = 0
=> x = 7,-9
From II: y^2 +20y+99 = 0
=> y^2 +9y+11y+99 = 0
=> y(y+9)+11(y+9) = 0
=> y = -9,-11
x>=y

7. I.21x^2 – 13x + 2 = 0
II. 33y^2 = 21 – 68y

x=y or relation cannot be established.
x>y
x<y
x=<y
x>=y
Option B
From I: 21x^2 – 13x + 2 = 0
=> 21x^2 -7x – 6x + 2 = 0
=> 7x(3x-1)-2(3x-1) = 0
=> x = 1/3,2/7
From II: 33y^2 = 21 – 68y
=> 33y^2 + 68y – 21 = 0
=> 33y^2 + 77y -9y – 21 = 0
=> 11y(3y+7)-3(3y+7) = 0
=> y = -7/3,3/11
x>y

8. I.16x+15y = 94
II.2x = 10 – y

x>=y
x>y
x<y
x=<y
x=y or relation cannot be established.
Option B
From I: x = (94-15y)/16
From II: 2*(94-15y)/16 = 10 – y
=> 188 – 30y = 160 – 16y
=> 14y = 28
=> y = 2
x = (94-15y)/16 = 4
x>y

9. I.9x+4y = 79
II.7x – 5y = 29

x=y or relation cannot be established.
x=<y
x>=y
x>y
x<y
Option D
Multiply I from 5 and II from 4, we get
x = 7
y = 4
x>y

10. I.12x^2 + 13x – 35 = 0
II.10y^2 = 21y – 9

x>=y
x=<y
x>y
x<y
x=y or relation cannot be established.
Option E
From I: 12x^2 + 13x – 35 = 0
=> 12x^2 + 28x – 15x – 35 = 0
=> 4x(3x+7)-5(3x+7) = 0
=> (3x+7)(4x-5) = 0
=> x= -7/3,5/4
From II: 10y^2 = 21y – 9
=> 10y^2 – 21y + 9 = 0
=> 10y^2 – 15y – 6y + 9 = 0
=> 5y(2y-3)-3(2y-3) = 0
=> (2y-3)(5y-3) = 0
=> y = 3/2,3/5
x =y or relation cannot be established.