A train went 300 km from province A to province B.... See the details....

Please answer this:

A train went 300 km from Province A to Province B at an average rate of 180 kph. At what speed did it travel on the way back if its average speed for the whole trip was 120 kph?

And please with your solution...

total time= total distance/avg speed

for onward journey time = 300/180=5/3 h
for whole journey time = 600/120= 5 h

so time taken in return journey = 5 - 5/3=10/3 h

avg speed for return journey = 300/(10/3) =90 kph

Suppose that the train traveled an average speed of x km per hour on the way back.

The total distance traveled by the train is 600 km (300 km from A to B and then another 300 km from B back to A). The total time spent is:
300/180 + 300/x = 15/9 + 300/x (since t = d/v).

The average speed is given by the total distance divided by the total time, so the average speed of the trip is:
600/(15/9 + 300/x).

Since we want this to be 120 km per hour, we have:
600/(15/9 + 300/x) = 120.

Solving this equation yields x = 90. Therefore, the train traveled at an average speed of 90 km per hour on the way back.

I hope this helps!

Make a table.

direction ---- rate ---- time ----- distance
A to B ------- 180 ----- 300/180 --- 300
B to A ------- x -------- 300/x ------- 300

average speed = total distance/total time

600/((5/3) + (300/x)) = 120

600/((5x + 900)/3x) = 120

(5 * 3x)/(5x + 900) = 1

5 * 3x = 5x + 900

10x = 900

x = 90