SOLUTION: A train averaged 120km/h for the first 80% of a trip and 90km/h for the whole trip. Find it's average speed for the last 20% of the trip.

Let D be the entire distance.

The train covered first 80% of the entire distance, i.e. 0.8*D, at the average speed of 120 . 
Hence, it spent  hours to cover this part. 

Let u be the speed of the train on the second (20%) part of the trip.
Then the train spent  hours to cover the second part. 

Thus the full time for the trip was

t =  +     (1)

hours. We are given that the average speed was 90 . 
It means that the whole distance divided by the full time is 90 . 

It gives you an equation

 = 90.

Cancel the factor D in the numerator and the denominator. You will get

 = 90.

It is your equation to find the speed u.

Let us simplify it step by step:

 = 90  ----->   = 3  ----->  4u = 3*(0.8*u + 24)  ----->  4u = 2.4u + 72  ----->  4u - 2.4 u = 72  ----->  1.6u = 72,

u =  = 45 . 

It is the speed of the train on the second part of the trip.

The solution is completed.

Answer. The speed of the train on the second part of the trip was 45 .

Let distance traveled on entire journey be D

Then distance traveled on 1st 80% = .8D

Time taken to travel 1st 80% of journey = 

Let speed during last 20% of trip be S

Since distance traveled on last 20% = .2D, then

Time taken to travel last 20% of journey = 



Total distance: D 

Total time: 

Average speed = Total distance, divided by total time, OR



 ----- Cross-multiplying





.6DS + 18D = DS ------- Multiplying by LCD, S

D(.6S + 18) = D(S) ---- Factoring out GCF, D

.6S + 18 = S

18 = S - .6S

18 = .4S

S, or speed on 2nd leg of journey = , or  km/h