Direction: In the following questions two equations numbered I and II are given. You have to solve both the equations and give answer
( 1 ) if x > y
( 2 ) if x ≥ y
( 3 ) if x < y
( 4 ) if x ≤ y
( 5 ) if x = y or the relationship cannot be established.

1. Ⅰ. 5x + 2y = 31
Ⅱ. 3x + 7y = 36

1. Given that :- Ⅰ. 5x + 2y = 31 .....( 1 )
Ⅱ. 3x + 7y = 36 .....( 2 )

##### Correct Option: A

Given that :- Given that :- Ⅰ. 5x + 2y = 31 .....( 1 )
Ⅱ. 3x + 7y = 36 .....( 2 )
Solving these two linear equations, we get x = 5, y = 3 .
x > y is correct answer .

1. Ⅰ. 2x2 + 11x + 14 = 0
Ⅱ. 4y2 + 12y + 9 = 0

1. From the given equations , we have
Ⅰ. 2x2 + 11x + 14 = 0
⇒ 2x2 + 7x + 4x + 14 = 0
⇒ x( 2x + 7 ) + 2( 2x + 7 ) = 0

Ⅱ. 4y2 + 12y + 9 = 0
⇒ ( 2y + 3 )2 = 0

##### Correct Option: C

From the given equations , we have
Ⅰ. 2x2 + 11x + 14 = 0
⇒ 2x2 + 7x + 4x + 14 = 0
⇒ x( 2x + 7 ) + 2( 2x + 7 ) = 0
⇒ ( x + 2 )( 2x + 7 ) = 0

 ⇒ x = -7 or -2 2

Ⅱ. 4y2 + 12y + 9 = 0
⇒ ( 2y + 3 )2 = 0
⇒ ( 2y + 3 )( 2y + 3 ) = 0
 ⇒ y = -3 , -3 2 2
From above both equations it is clear that x < y is correct answer .

1. Ⅰ. x2 - 7x + 12 = 0
Ⅱ. y2 - 12y + 32 = 0

1. As per the given above equations , we have
From equation Ⅰ. x2 - 7x + 12 = 0
⇒ x2 - 4x - 3x + 12 = 0

Ⅱ. y2 - 12y + 32 = 0
⇒ y2 - 8y - 4y + 32 = 0

##### Correct Option: D

As per the given above equations , we have
From equation Ⅰ. x2 - 7x + 12 = 0
⇒ x2 - 4x - 3x + 12 = 0
⇒ x( x - 4) - 3( x - 4) = 0
⇒ ( x - 4) ( x - 3) = 0
⇒ x = 4 or, 3
Ⅱ. y2 - 12y + 32 = 0
⇒ y2 - 8y - 4y + 32 = 0
⇒ y( y - 8) - 4( y - 8) = 0
⇒ ( y - 8) ( y - 4) = 0
⇒ y = 4 or, 8
Thus , x ≤ y is required answer .

1. Ⅰ. x2 = 729
Ⅱ. y = √729

1. According to question ,we have
From equation Ⅰ.
x2 = 729
⇒ x = ± √729 = ± 27

From equation Ⅱ.
y = √729

##### Correct Option: D

According to question ,we have
From equation Ⅰ. x2 = 729
⇒ x = ± √729 = ± 27
From equation Ⅱ. y = √729
⇒ y = 27
Hence , x ≤ y is required answer .

1. The equation x2 − px + q = 0, p, q ∈ R has on real root if :

1. According to question , we can say that
The equation x2 − px + q = 0, p, q ∈ R has no real root if B2 < 4AC

##### Correct Option: B

According to question , we can say that
The equation x2 − px + q = 0, p, q ∈ R has no real root
On comparing with quadratic eq. Ax2 + Bx + C = 0 , we get
∴ A =1, B= - p, C = q
Now ,we have B2 < 4AC
⇒ ( - P )2 < 4 x 1 x q
⇒ p2 < 4q.
Hence , option B is correct answer .