SOLUTION: A man left A at 6:00 am expecting to reach B at 9:00 am. But after walking one hour, he was delayed for half an hour and so he had to increase his rate by 1 mile per hour to reach

Algebra ->  College  -> Linear Algebra -> SOLUTION: A man left A at 6:00 am expecting to reach B at 9:00 am. But after walking one hour, he was delayed for half an hour and so he had to increase his rate by 1 mile per hour to reach       Log On


   



Question 903387: A man left A at 6:00 am expecting to reach B at 9:00 am. But after walking one hour, he was delayed for half an hour and so he had to increase his rate by 1 mile per hour to reach B at 9:00 am. Find his speed before the delay and the distance between A and B.
Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
______________rate_________time_________distance
_______________r___________3____________r*3=d
before interr___r___________1_____________r*1
after interr____r+1_________1_____________(r+1)*1
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Sum of before and after interruption d distance.
r%2B%28r%2B1%29=d
2r%2B1=d

Expected regular travel r%2A3=d;

2r%2B1=r%2A3, equating the distance
1=1%2Ar
highlight%28r=1%29-----walking speed, before interruption.

r%2A3=d
d=3r
d=3%2A1
highlight%28d=3%29-----travel distance.