Assertion (A) : If x and y are positive and (2x² - 5y²) : xy = 1 : 3, then x : y = 3 : 5.

Reason (R) : If four quantities a, b, c and d form a proportion, then Componendo and dividendo property states that :

(a + c) : (a - c) = (b - d) : (b + d)

1. A is true, R is false.

2. A is false, R is true.

3. Both A and R are true.

4. Both A and R are false.

Given,

$\Rightarrow (2x^2 - 5y^2) : xy = 1 : 3 \\[1em] \Rightarrow \dfrac{2x^2 - 5y^2}{xy} = \dfrac{1}{3} \\[1em] \Rightarrow \dfrac{3(2x^2 - 5y^2)}{xy} = 1 \\[1em] \Rightarrow \dfrac{6x^2 - 15y^2}{xy} = 1 \\[1em] \Rightarrow \dfrac{6x^2}{xy} - \dfrac{15y^2}{xy} = 1 \\[1em] \Rightarrow \dfrac{6x}{y} - \dfrac{15y}{x} = 1$

Let, $\dfrac{x}{y}$ = t, we get :

$\Rightarrow 6t - \dfrac{15}{t} = 1 \\[1em] \Rightarrow \dfrac{6t^2 - 15}{t} = 1 \\[1em] \Rightarrow 6t^2 - 15 = t \\[1em] \Rightarrow 6t^2 - t - 15 = 0 \\[1em] \Rightarrow 6t^2 - 10t + 9t - 15 = 0 \\[1em] \Rightarrow 2t(3t - 5) + 3(3t - 5) = 0 \\[1em] \Rightarrow (2t + 3)(3t - 5) = 0 \\[1em] \Rightarrow 2t + 3 = 0 \text{ or } 3t - 5 = 0 \\[1em] \Rightarrow 2t = -3 \text{ or } 3t = 5 \\[1em] \Rightarrow t = -\dfrac{3}{2} \text{ or } t = \dfrac{5}{3}.$

Since, x and y are positive.

∴ t = $\dfrac{x}{y} = \dfrac{5}{3}$

⇒ x : y = 5 : 3.

∴ Assertion (A) is false.

Given,

a, b, c and d form a proportion.

$\Rightarrow \dfrac{a}{b} = \dfrac{c}{d}$

Applying componendo and dividendo, we get :

$\Rightarrow \dfrac{a + b}{a - b} = \dfrac{c + d}{c - d}$

⇒ (a + b) : (a - b) = (c + d) : (c - d).

∴ Reason (R) is false.

Hence, Option 4 is the correct option.