SOLUTION: . A man left A at 6:00 a.m. expecting to reach B at 9:00 a.m. 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

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Question 346579: . A man left A at 6:00 a.m. expecting to reach B at 9:00 a.m. 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 a.m. Find his speed before the delay and the distance between A and B.
thanks..
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A man left A at 6:00 a.m. expecting to reach B at 9:00 a.m.
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 a.m.
Find his speed before the delay and the distance between A and B.
:
Let d = distance from A to B
Given that he left at 6, and expected to be there at 9, therefore:
d%2F3 = original speed
:
He walked for 1 hr, then stopped for .5 hrs, therefore
he continued at a speed of (d%2F3 + 1) for 1.5 hrs
:
dist = speed * time
total dist = slow speed dist + faster speed dist
d = 1%28d%2F3%29 + 1.5%28d%2F3+%2B+1%29
d = d%2F3 + %281.5d%29%2F3+%2B+1.5%29
d = %282.5d%29%2F3+%2B+1.5%29
multiply both sides by 3
3d = 2.5d + 3(1.5)
3d - 2.5d = 4.5
.5d = 4.5
d = 4.5%2F.5
d = 9 miles from A to B
:
Find the original speed:
9%2F3 = 3 mph
:
:
Confirm this by finding the distance at each speed
1 * 3 = 3 miles
1.5 * 4 = 6 mile
----------------
total dist: 9 mi