x² - 365 = 364
⇒ x² = 364 + 365
∴ x = √729 = ± 27
and y - √324 = √81 ⇒ y - 18 = 9
∴ y = 27
So, y ≥ x or x ≤ y because y = 27 and x = 27 and - 27.

Correct Option: D

x² - 365 = 364
⇒ x² = 364 + 365
∴ x = √729 = ± 27
and y - √324 = √81 ⇒ y - 18 = 9
∴ y = 27
So, y ≥ x or x ≤ y because y = 27 and x = 27 and - 27.

4/√x + 7/√x = √x;
⇒ 11/√x = √x
∴ x = 11
and y² - (11)^5/2/√y = 0 ⇒ y² = (11)^5/2/(y)^1/2
⇒ y² x y^1/2 = (11)^5/2
⇒ (y)^5/2 = (11)^5/2
∴ y = 11
∴ x = y

Correct Option: E

4/√x + 7/√x = √x;
⇒ 11/√x = √x
∴ x = 11
and y² - (11)^5/2/√y = 0
⇒ y² = (11)^5/2/(y)^1/2
⇒ y² x y^1/2 = (11)^5/2
⇒ (y)^5/2 = (11)^5/2
∴ y = 11
∴ x = y

Let Mr . Arjun is on tour for N days.
Then, according to the question,
Difference in expenses per day for original and extended tour = 3
360/N - 360/(N + 4) = 3

Correct Option: B

Let Mr . Arjun is on tour for N days.
Then, according to the question,
Difference in expenses per day for original and extended tour = 3
360/N - 360/(N + 4) = 3
⇒ 1/N - 1/(N + 4) = 3/360 = 1/120
⇒ [N + 4 - N] / [ N x (N + 4)] = 1/120
⇒ N(N + 4) = 4 x 120 = 480
⇒ N² + 4N - 480 = 0
⇒ N² + 24 - 20x - 480 = 0
⇒ N(N + 24) - 20(N + 24) = 0
⇒ (N + 24) (N - 20) = 0
∴ N = 20
(we ignore -ve value of x, that is, -24 because days cannot be negative)

As per the given above question , we have
Let k₁ and k₂ be the roots of quadratic equation .
One root is k₁ = √5
So the other root is k₂ = − √5
∴ Sum of the roots = k₁ + k₂ = √5 + ( -√5 ) = 0
and product of the roots = k₁ × k₂ = − (√5) (- √5) = −5

Correct Option: C

As per the given above question , we have
Let k₁ and k₂ be the roots of quadratic equation .
One root is k₁ = √5
So the other root is k₂ = − √5
∴ Sum of the roots = k₁ + k₂ = √5 + ( -√5 ) = 0
and product of the roots = k₁ × k₂ = − (√5) (- √5) = −5
∴ Required equation is x² - (sum of the roots)x + (product of the roots) = 0
⇒ x² - 5 = 0.

Correct Option: C