SOLUTION: 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 wo

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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?
Found 2 solutions by ankor@dixie-net.com, ikleyn:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
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?
:
We are using km/hr, therefore:
Change 12 sec to hrs: 12/3600 = .0033 hrs
Change 1 min to hrs 1/60 = .0167 hrs
:
t = walking time to arrive on time when leaving at 8:30 (2.5 km/hr)
then
(t+.0033) = actual walking time when he was 12 sec late
:
then leaving a half hr earlier
let (t+.5) = walking time to arrive on time when leaving at 8:00 (1.5 km/hr)
then
(t+.5-.0167) = (t+.4833) = actual walking time when he was 1 min early
:
Write a distance equation: dist = speed * time
2.5(t+.0033) = 1.5(t+.4833) (distances are the same)
2.5t + .00825 = 1.5t + .72495
2.5t - 1.5t = .72495 - .00825
t = .7167 hrs
Change to min
60 * .7167 = 43 minutes
8:30 + :43 = 9:13 time he is suppose to be there

Answer by ikleyn(52776) About Me  (Show Source):
You can put this solution on YOUR website!
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?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 

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%2Fh = 2.5%2F60 km%2Fminute,  1.5 km%2Fh = 1.5%2F60 km%2Fminute,  12 seconds = 1%2F5 of a minute.

An equation is on the difference of times

D%2F%28%281.5%2F60%29%29 - D%2F%28%282.5%2F60%29%29%29 = 30-1-%281%2F5%29 = 284%2F5 minutes = 144%2F5 minutes     (1)     or

%2860D%29%2F1.5+-+%2860D%29%2F2.5 = 144%2F5.     (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%2F480 = 1.8 km.
Time to walk in the 1-st scenario is  t%5B1%5D = 1.8%2F%28%282.5%2F60%29%29 = 43.2 minutes = 431%2F5 minutes.

Time to walk in the 2-nd scenario is  t%5B2%5D = 1.8%2F%28%281.5%2F60%29%29 = 72 minutes.

The difference  t%5B2%5D-t%5B1%5D = 72 - 431%2F5 = 284%2F5 minutes, exactly as stated in the condition.

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

For similar problem see the lesson  How far do you live from school?  in this site.