Question

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.

Answer 1

let the "certain" speed be x km/h
then the distance between the two cities is 4x km

time for return trip
= 100/x + (4x-100)/(x-10)

100/x + (4x-100)/(x-10) - 4 = 1/2
100/x + (4x-100)/(x-10) = 9/2
multiply each term by 2x(x-10), the LCD
200(x-10) + 2x(4x-100) = 9x(x-10)
200x - 2000 + 8x^2 - 200x = 9x^2 - 90x
x^2 - 90x + 2000 = 0
(x-50)(x-40) = 0
x = 50 or x = 40

The distance between the two cities is either 200 km or 160 km, depending on the speed.
How can that be??


Case 1: x = 50
then the distance between the two cities is 4(50) or 200 km
and the speed during the first part was 50 km/h
check:
time for first trip = 200/50 = 4 hrs
time for return trip
= 100/50 + 100/40 = 4.5
so it took 1/2 hour longer
Answer is correct

Case 2: x = 40
then the distance between the two cities is 4(40) or 160 km
and the speed during the first part was 40 km/h
check:
time for first trip = 100/40 + 60/30
= /40 = 4 hours
time for return trip = 100/40 + 60/30
= 4.5
so it took 1/2 hour longer
This answer is also correct

Answer 2

Read the question.

Answer 3

200, 160

Answer 4

69, 420 pp long