SOLUTION: Driving from Tulsa to Detroit, Dean's average speed was 50 mph. If he had averaged 60 mph instead, this would have decreased the drive time by 3hr. How many miles was the drive f
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Question 692688: Driving from Tulsa to Detroit, Dean's average speed was 50 mph. If he had averaged 60 mph instead, this would have decreased the drive time by 3hr. How many miles was the drive from Tulsa to Detroit?
note: let x = the distance from Tulsa to Detroit, do not use other variables. Answer part a and b.
a) Fill in chart:
R ----- T ----- D
50
60
b) Write and solve a correct equation that could be used to solve for x.
What I've tried:
Chart:
r: 50 t: x/50 d: x
r: 60 t: x/60 - 3 d: x
my equation:
x/50 = x/60 -3
(lcd = 300)
300(x/50)=300(x/60)- 300(3)
6x = 5x - 900
x = -900
my answer: x =-900
I don't think it's right due to the negative answer but it's the best I can come up with.
Answer by MathTherapy(10552) (Show Source): You can put this solution on YOUR website!
Driving from Tulsa to Detroit, Dean's average speed was 50 mph. If he had averaged 60 mph instead, this would have decreased the drive time by 3hr. How many miles was the drive from Tulsa to Detroit?
note: let x = the distance from Tulsa to Detroit, do not use other variables. Answer part a and b.
a) Fill in chart:
R ----- T ----- D
50
60
b) Write and solve a correct equation that could be used to solve for x.
What I've tried:
Chart:
r: 50 t: x/50 d: x
r: 60 t: x/60 - 3 d: x
my equation:
x/50 = x/60 -3
(lcd = 300)
300(x/50)=300(x/60)- 300(3)
6x = 5x - 900
x = -900
my answer: x =-900
I don't think it's right due to the negative answer but it's the best I can come up with.
D (distance): x ; Rate (r) = 50 ; t (time) =
D (distance): x ; Rate (r) = 60 ; t (time) =
5x = 6x – 900 ------ Multiplying by LCD, 300
5x – 6x = - 900
- x = - 900
x, or distance = , or miles
You were correct right up until you tried to create the equation. You have to realize that the time traveled at 60 mph will BE LESS THAN the time traveled at 50 mph, or that the time traveled at 50 mph will be greater than the time taken to travel at 60 mph, by 3 hours, hence the equation: . This is where you went wrong. Furthermore, a negative distance could never be correct, as you have surmised.
You can do the check!!
Send comments and “thank-yous” to “D” at MathMadEzy@aol.com
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