Answer:
160 km or 200 km
Step-by-step explanation:
Calling
d the distance between the two cities
t = 4 h the time the traveller takes for the first trip
t' the time taken for the return trip
We know that the return trip took the biker 30 minutes longer, so it took
[tex]t'=4h +30 min = 4.5 h[/tex]
to come back, where t' is the time taken to come back.
We also know that
- During the first trip, he travelled with speed v
- He travelled with speed v for the first 100 km of the return trip
- Then he travelled with speed (v-10) for the (d-100) km left, where we called d the distance between the two cities.
So we can write the following equations:
[tex]d=vt=4 v\\d=100+(v-10) t_2\\\\t'=\frac{100}{v} +t_2=4.5[/tex]
where [tex]t_2[/tex] is the time taken for the second part of the return trip (when he travels 10 km/h slower than the original speed).
So we have a system of 3 equations in 3 unknown variables. Isolating t2 from the last 2 equations:
[tex]t_2 = \frac{d-100}{v-10}\\t_2 = 4.5-\frac{100}{v}[/tex]
and equalizing them
[tex]\frac{d-100}{v-10}=4.5-\frac{100}{v}[/tex]
Then by using the first equation here, we have
[tex]\frac{4v-100}{v-10}=4.5-\frac{100}{v}\\(4v-100)v=4.5v(v-10)-100(v-10)\\4v^2-100v=4.5v^2-45v-100v+1000\\0.5v^2-45v+1000 = 0[/tex]
which has 2 solutions:
v = 40 km/h, v = 50 km/h
With the first solution we have
[tex]d=vt=(40 km/h)(4 h)=160km[/tex]
with the second solution we have
[tex]d=vt=(50 km/h)(4 h)=200km[/tex]
