Question 1082146
.
A diplomat travels from point X to Y. If he will be departing X at 8:00 AM, and travel 2 kilometers per hour, 
he will arrive at Y, 3 minutes earlier than his expected time of arrival. 
However if he will leave at 8:30 AM and travel {{{highlight(cross(2))}}} 3 kilometers per hour, 
he will arrive 6 minutes later than the expected time. What is the expected time of his arrival?
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<pre>
Let D be the unknown distance from X to Y.

Moving at 2 km/h, the diplomat will spend {{{D/2}}} hours.

Moving at 3 km/h, the diplomat will spend {{{D/3}}} hours.

The condition says 

{{{D/2 - D/3}}} = 30 minutes + (6-3) minutes = 33 minutes.    (It is the "time difference" equation)

         Try to understand on your own how these numbers come from the condition 
         and how and why do they combine in this expression (30 + (6-3)) minutes.


But we need to write both sides of the last equation in uniform units. So, this equation will take the form

{{{D/2 - D/3}}} = {{{33/60}}} of an hour.


To solve this equation, multiply both sides by 60. You will get

30D - 20D = 33  ---->  10D = 33  ---->  D = {{{33/10}}} = 3.3 kilometers.

Thus we found the distance. It is 3.3 kilometers.

Now, at the second scenario, the diplomat starts at 8:30 am,  moves {{{3.3/3}}} = 1.1 hour = 1 hour 6 minutes and arrives at 9:36 am, 
which is 6 minutes late.

Hence, the meeting is scheduled at 9:30 am.
</pre>

Solved.



If you want to see another solved problems of this kind, look into the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/travel/How-far-do-you-live-from-school.lesson>How far do you live from school?</A> 

in this site.



Also, you have this free of charge online textbook in ALGEBRA-I in this site

&nbsp;&nbsp;&nbsp;&nbsp;<A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A>.


The referred lessons are the part of this textbook under the section "<U>Word problems</U>", the topic "<U>Travel and Distance problems</U>".