SOLUTION: Sue is traveling 50 miles to an appointment she has to be to in 1 hour. After the first 30 miles she realizes she isn't going to make it and increases her speed by 15mph to get the
Algebra ->
Customizable Word Problem Solvers
-> Travel
-> SOLUTION: Sue is traveling 50 miles to an appointment she has to be to in 1 hour. After the first 30 miles she realizes she isn't going to make it and increases her speed by 15mph to get the
Log On
Question 360006: Sue is traveling 50 miles to an appointment she has to be to in 1 hour. After the first 30 miles she realizes she isn't going to make it and increases her speed by 15mph to get there on time. What was her speed for the first 30 miles? Answer by mananth(16949) (Show Source):
You can put this solution on YOUR website! Sue is traveling 50 miles to an appointment she has to be to in 1 hour. After the first 30 miles she realizes she isn't going to make it and increases her speed by 15mph to get there on time. What was her speed for the first 30 miles?
let her speed for 30 miles be xmph
time taken = 30/x
speed increased to x+15 mph
time taken = 20/(x+5)
..
(30/x)+20/(x+5)=1
30(x+5)+20x / x(x+5)=1
30x+150+20x = x^2+5x
50x+150-x^2-5x=0
-x^2+45x+150=0
x^2-45x-150=0
solve using quadratic formula
discriminant = 2625
x1= (45+sqrt(2625))/2
x1=48.11 mph
x2 will be negative so ignore
...
m.ananth@hotmail.ca
