It took 4 hours for a biker to travel from one city to another going at a certain speed. on the return trip, the biker traveled at the same speed for the first 100-km and then for the rest of the trip he traveled at a speed which was 10 km/hour slower than the original speed, and thus the return trip took him 30 min longer. find the distance between the two cities.

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

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:

where 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:

and equalizing them

Then by using the first equation here, we have

which has 2 solutions:

v = 40 km/h, v = 50 km/h

With the first solution we have

with the second solution we have

73.0769230769% of the games were won

step-by-step explanation:

you divide the denominator by the numerator on both sides and if the quotients are equal then the fractions are equivalent