SOLUTION: This is from my text book. If you could show me how to set it up to solve that would be great!! Problem: Donna is late for a sales meeting after traveling from one town to anothe
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Question 187200: This is from my text book. If you could show me how to set it up to solve that would be great!! Problem: Donna is late for a sales meeting after traveling from one town to another at a speed of 32mph. If she had traveled 4mph faster, she could have made the trip in 1/2 hr less time. How far apart are the towns?
Found 2 solutions by ankor@dixie-net.com, jojo14344:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Donna is late for a sales meeting after traveling from one town to another at
a speed of 32mph. If she had traveled 4mph faster, she could have made the trip
in 1/2 hr less time. How far apart are the towns?
:
Let d = distance between the town
;
Write a time equation: Time = dist/speed
:
Actual time = faster time + .5 hr
= + .5
Multiply equation by a common multiple of 32 & 36, 288 would do it
288* = 288* + 288(.5)
9d = 8d + 144
9d - 8d = 144
d = 144 mi between towns
;
;
Check solution by finding the times:
144/32 = 4.5 hr
144/36 = 4.0 hr (half hr less)
Answer by jojo14344(1513) (Show Source): You can put this solution on YOUR website!
Just to show also of one way of looking the problem:
Based on our working eqn
We get,
We'll equate the 2 conditions: Normal Speed= Faster Speed
---> = 4 mph faster
----->---->
Therefore, going back to Normal Speed:
, Distance bet. towns
Going back at Faster Speed:
, Distance bet. towns
Thank you,
Jojo
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