An aeroplane is required to travel from a point A to another point...

Let $\vec{v}_p$ be the velocity of the plane with respect to the ground, $\vec{v}_w$ be the velocity of the wind, and $\vec{v}_{pw}$ be the velocity of the plane with respect to the wind.

We know that $\vec{v}_p = \vec{v}_{pw} + \vec{v}_w$, which can be rearranged as $\vec{v}_{pw} = \vec{v}_p - \vec{v}_w$.

The direction of $\vec{v}_p$ is 30^{}^{{}^{\circ}} east of north, and the direction of $\vec{v}_w$ is due north. The angle between $\vec{v}_p$ and $\vec{v}_w$ is 30^{}^{{}^{\circ}}.

Using the cosine rule for the vector triangle:

v_{pw}^2 = v_p^2 + v_w^2 - 2 v_p v_w \cos(30^{}^{{}^{\circ}})

Substituting the given values ($v_{pw} = 150 \text{ m/s}$, $v_w = 20 \text{ m/s}$):

$150^2 = v_p^2 + 20^2 - 2 v_p (20) \dfrac{\sqrt{3}}{2}$

$22500 = v_p^2 + 400 - 20\sqrt{3} v_p$

$v_p^2 - 20\sqrt{3} v_p - 22100 = 0$

Solving the quadratic equation for $v_p$ (taking the positive root):

$v_p = \dfrac{20\sqrt{3} + \sqrt{(20\sqrt{3})^2 - 4(1)(-22100)}}{2}$

$v_p = \dfrac{20\sqrt{3} + \sqrt{1200 + 88400}}{2} = 10\sqrt{3} + \sqrt{22400}$

$v_p = 10(\sqrt{3} + \sqrt{224}) \approx 10(1.732 + 14.967) = 166.99 \text{ m/s}$

The distance to be covered is $d = 500 \text{ km} = 500 \times 10^3 \text{ m}$.

Time taken $t = \dfrac{d}{v_p} = \dfrac{500000}{166.99} \approx 2994.2 \text{ s}$

Converting the time into minutes:

$t = \dfrac{2994.2}{60} \approx 49.9 \text{ min}$

Answer: $49.9$