Question

The distance between two stations A and B is 300km. Two motor cyclists start simultaneously from A and B in the opposite directions and the distance between them after 3 hours is 30 km. If the speed of one motor cyclist is less than that of the other by 10km/hr, find the speed of each motor cyclist.

Answer 1

Let the speeds of the motorcyclists be x and y kmph.

If x is the speed of the motorcyclist driving faster, then x = y+10.

Since the motorcyclists are driving in opposite directions, and after 3 hours, the distance between them is 20km, the combined distance traveled is 270km.

In 3 hours, the distance traveled by the faster motorcyclist is 3x and the distance traveled by the slower motorcyclist is 3y.

But x = y+10.

Thus, the distance traveled by the two motorcyclists is 3y+30 and 3y.

The total distance traveled by the two motorcyclists is 270, because the distance between them is 30 and at the beginning, the distance between them was 300km.

Thus, 270 = 3y+30+3y = 6y+30.

Therefore, 240 = 6y or y = 240/6 = 40kmph.

And x = 40+10 = 50.

Check. After 3 hours, first motorcyclist travels 50 x 3 = 150km and distance to travel = 300 - 150 = 150km

And second motorcyclist would have traveled 40 * 3 = 120km and distance to travel is 180km.

And the difference between the two motorcyclists is 180km - 150km = 30km.

Thus, the speeds are 50km and 40km respectively.

Another way to do this.

The distance traveled by first motorcyclist is 3x, and distance to travel is 300-3x.

Distance traveled by second motorcyclist is 3y.

Since x = y+10, distance to travel by first motorcyclist = 300-3*(y+10)

Also, the difference between the motorcyclists is 30km. Hence,

300-3*(y+10) - 3y = 30

300-3y-30-3y=30

300-6y-30-30=0

240-6y=0

6y=240 or
y=40.
and x = 50

Thus, the speeds of the motorcyclists are 40kmph and 50kmph respectively.