the story of Marvin's plane

Let the speed of the plane be $V_1$ and the speed of the wind be $V_2$

$\large{\text{With wind}}$

Distance travelled $=1800 \text{km}$

Time taken $=3 \text{hours}$

Speed $=\dfrac{\text{Distance travelled}}{\text{Time taken}}=\dfrac{1800}{3}\text{km/h} = 600 \text{km/h} = V_1+V_2$

$\large{\text{Opposite wind}}$

Distance travelled $=1800 \text{km}$

Time taken $=4 \text{hours}$

Speed $=\dfrac{\text{Distance travelled}}{\text{Time taken}}=\dfrac{1800}{4}\text{km/h} = 450 \text{km/h} =V_1-V_2$

Now,

\( \begin{align*} V_1+V_2&= 600 \text{km/h} \\

V_1-V_2&= 450 \text{km/h} \\ \end{align*}\)

Subtracting $(2)$ from $(1)$ ,

$(V_1+V_2)-(V_1-V_2)= 600 \text{km/h} -450 \text{km/h}$

$\implies V_1+V_2-V_1+V_2)= 150 \text{km/h}$

$\implies V_2+V_2= 150 \text{km/h}$

$\implies 2{V_2}= 150 \text{km/h}$

$\implies V_2 = \dfrac{150}{2} \text{km/h}$

$\implies V_2 = \boxed{75 \text{km/h} }$

Let the velocities of the plane and wind be $v$ and $w$ respectively. Then we have:

$\begin{cases} \text{In wind direction:} & 3(v+w) = 1800 & \implies v + w = 600 &...(1) \\ \text{Against wind:} & 4(v-w) = 1800 & \implies v - w = 450 &...(2) \end{cases}$

$\begin{aligned} (1)-(2): \quad 2w & = 150 \\ \implies w & = \boxed{\text{75 kph}} \end{aligned}$