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 .

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

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 .

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 .

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 .