Answer
$\left.\text { When moving in the same direction as the jet stream (of speed } v_{s}\right),$ the time is
$$
t_{1}=\frac{d}{v_{j a}+v_{s}}
$$
where $d=4000 \mathrm{km}$ is the distance and $v_{\mathrm{ja}}$ is the speed of the jet relative to the air $(1000 \mathrm{km} / \mathrm{h}) .$ When moving against the jet stream, the time is
$$
t_{2}=\frac{d}{v_{j a}-v_{s}}
$$
where $t_{2}-t_{1}=\frac{70}{60} \mathrm{h} .$ Combining these equations and using the quadratic formula to solve gives $$v_{s}=143 \mathrm{km} / \mathrm{h}$$
Work Step by Step
$\left.\text { When moving in the same direction as the jet stream (of speed } v_{s}\right),$ the time is
$$
t_{1}=\frac{d}{v_{j a}+v_{s}}
$$
where $d=4000 \mathrm{km}$ is the distance and $v_{\mathrm{ja}}$ is the speed of the jet relative to the air $(1000 \mathrm{km} / \mathrm{h}) .$ When moving against the jet stream, the time is
$$
t_{2}=\frac{d}{v_{j a}-v_{s}}
$$
where $t_{2}-t_{1}=\frac{70}{60} \mathrm{h} .$ Combining these equations and using the quadratic formula to solve gives $$v_{s}=143 \mathrm{km} / \mathrm{h}$$
