Question 775661: a man left place A at 6:00 am expecting to reach place 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 1 mile per hour to reach place B at 9:00 am. find his speed before the delay and the distance between place A and place B.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! a man left place A at 6:00 am expecting to reach place 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 1 mile per hour to reach place B at 9:00 am.
find his speed before the delay and the distance between place A and place B.
:
let s = his normal walking speed
then
(s+1) = his faster walking speed
:
dist = 3s
Because of the delay, actual walking time only 2.5 hrs
:
dist at normal speed + dist at faster speed = total dist
1s + 1.5(s+1) = 3s
s + 1.5s + 1.5 = 3s
2.5s + 1.5 = 3s
1.5 = 3s - 2.5s
1.5 = .5s
s = 1.5 /.5
s = 3 mph is his normal walking speed
then
3(3) = 9 mi is the distance
:
:
Check
3 + 1.5(4) = 9
3 + 6 = 9
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