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) About Me  (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
d%2F32 = d%2F36 + .5
Multiply equation by a common multiple of 32 & 36, 288 would do it
288*d%2F32 = 288*d%2F36 + 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) About Me  (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 eqnsystem%28Speed=Distance%2Ftime%29
We get, Distance=Speed%2Atime
We'll equate the 2 conditions: Normal Speed= Faster Speed
S%5B1%5Dt=S%5B2%5D%28t-0.5hr%29
32%2At=highlight%2836%29%28t-0.5%29---> S%5B2%5D= 4 mph faster
32t=36t-18
18=36t-32t----->18=4t---->cross%2818%294.5%2Fcross%284%29=cross%284%29t%2Fcross%284%29
t=4.5hrs
Therefore, going back to Normal Speed:
D=S%5B1%5D%2At=32%2A4.5=highlight%28144miles%29, Distance bet. towns
Going back at Faster Speed:
D=S%5B2%5D%28t-0.5%29=36%284.5-0.5%29=36%284%29=highlight%28144+miles%29, Distance bet. towns
Thank you,
Jojo