Question 1039935
A salesman started walking from office A at 8:30 AM. at the rate of 2.50 kph. He arrived at office B 12 seconds late. 
Had he started at A at 8:00 AM and  walked at 1.50 kph , he would have arrived at B one minute before the appointed time. 
At what time was he supposed to be at B?
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<pre>
This problem is on the difference of times spent by the body (by the two bodies) traveled with different rates (speeds).

Before hands, input data should be converted to the minutes units:

2.5 {{{km/h}}} = {{{2.5/60}}} {{{km/minute}}},  1.5 {{{km/h}}} = {{{1.5/60}}} {{{km/minute}}},  12 seconds = {{{1/5}}} of a minute.

An equation is on the difference of times

{{{D/((1.5/60))}}} - {{{D/((2.5/60)))}}} = {{{30-1-(1/5)}}} = {{{28}}}{{{4/5}}} minutes = {{{144/5}}} minutes     (1)     or

{{{(60D)/1.5 - (60D)/2.5}}} = {{{144/5}}}.     (2)

where D is an unknown distance in kilometers.

Multiply both sides of (2) by 30 to rid of the denominators. You will get

20*60D - 12*60D = 144*6,   or  1200D - 720D = 864,  or  480D = 864,  or  D = {{{864/480}}} = {{{1.8}}} km.
Time to walk in the 1-st scenario is  {{{t[1]}}} = {{{1.8/((2.5/60))}}} = 43.2 minutes = {{{43}}}{{{1/5}}} minutes.

Time to walk in the 2-nd scenario is  {{{t[2]}}} = {{{1.8/((1.5/60))}}} = 72 minutes.

The difference  {{{t[2]-t[1]}}} = {{{72}}} - {{{43}}}{{{1/5}}} = {{{28}}}{{{4/5}}} minutes, exactly as stated in the condition.
</pre>

The lesson to learn from this solution: this problem is on the difference of times.


For similar problem see the lesson &nbsp;<A HREF=https://www.algebra.com/algebra/homework/word/travel/How-far-do-you-live-from-school.lesson>How far do you live from school?</A>&nbsp; in this site.