Question 1012510
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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.
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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 {{{km/h}}}. 
Hence, it spent {{{(0.8*D)/120}}} 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 {{{(0.2*D)/u}}} hours to cover the second part. 

Thus the full time for the trip was

t = {{{(0.8*D)/120}}} + {{{(0.2*D)/u}}}    (1)

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

It gives you an equation

{{{D/((0.8*D)/120 + (0.2*D)/u)}}} = 90.

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

{{{1/(0.8/120 + 0.2/u)}}} = 90.

It is your equation to find the speed u.

Let us simplify it step by step:

{{{(120*u)/(0.8*u + 0.2*120)}}} = 90  ----->  {{{(4u)/(0.8*u + 24)}}} = 3  ----->  4u = 3*(0.8*u + 24)  ----->  4u = 2.4u + 72  ----->  4u - 2.4 u = 72  ----->  1.6u = 72,

u = {{{72/1.6}}} = 45 {{{km/h}}}. 

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

The solution is completed.

<U>Answer</U>. The speed of the train on the second part of the trip was 45 {{{km/h}}}.
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