Question 1208623
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A delivery van was travelling at an average speed of 50 km per hour for the first 20 minutes. 
The driver took a wrong turn and realized that he would be late by 10 minutes. 
He then increased his speed to 75 km per hour. When he arrived he was early by 6 minutes. How far did he travel.
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<pre>
The entire trip consists of two parts.

First part length is the product of the speed 50 km/h by the time of 20 minutes, which is 1/3 of an hour.

So, the first part length is  {{{50/3}}} = 16{{{2/3}}} kilometers.


Let "d" be the length of the second part of the trip.


About the second part of the trip, we know that if the driver travels at 50 km/h, he would be late by 10 minutes;
                                                                                   
                                                if the driver travels at 75 km/h, he would be earlier by 6 minutes.


It gives us this time equation for the difference of travel times

    {{{d/50}}} - {{{d/75}}} = {{{10/60}}} + {{{6/60}}},

or

    {{{d/50}}} - {{{d/75}}} = {{{16/60}}}.


Simplify it step by step and find "d"

    {{{(d/25)*(1/2-1/3)}}} = {{{16/60}}},

    {{{(d/25)*(1/6)}}} = {{{16/60}}},

    {{{d/25}}} = {{{16/10}}},

    d = {{{(25*16)/10}}} = {{{400/10}}} = 40.


Thus the second part length is 40 kilometers.


Now the <U>ANSWER</U> to the problem's question is the sum   16{{{2/3}}} + 40 = 56{{{2/3}}} kilometers.
</pre>

Solved.