# 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