SOLUTION: luoise walked for 2 hours then ran for 1.5 hours if she ran twice as long as she walks and the total trip was 20 miles the how fast did she run?
I am working on trying to set th
Algebra ->
Customizable Word Problem Solvers
-> Travel
-> SOLUTION: luoise walked for 2 hours then ran for 1.5 hours if she ran twice as long as she walks and the total trip was 20 miles the how fast did she run?
I am working on trying to set th
Log On
Question 107013: luoise walked for 2 hours then ran for 1.5 hours if she ran twice as long as she walks and the total trip was 20 miles the how fast did she run?
I am working on trying to set this up i have x to equal walking spd and 2x for running speed I know the rate is 3.5 hours together woulit be 3.5 = (x + 2x) / 20 ??? Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! I am working on trying to set this up i have x to equal walking spd and 2x for running speed I know the rate is 3.5 hours together woulit be 3.5 = (x + 2x) / 20 ???
AT LEAST YOU ARE GIVING IT A SHOT BUT I THINK YOU ARE TRYING TO MAKE IT HARDER THAN IT REALLY IS. TRY TO FOLLOW MY APPROACH.
distance(d) equals rate(r) times time(t) or d=rt; r=d/t and t=d/r
I'm assuming you mean "-----twice as fast---not twice as long--"
AS YOU HAVE DONE, WE'LL LET:
Let x=walking speed
Then 2x=running speed
Distance she walks=x*2 or 2x (rate she walks times time)
Distance she runs=2x*1.5 or 3x (rate she runs times time)
Now we know (actually we are told) that the distance she walks plus the distance she runs equals 20 mi, so:
2x+3x=20 collect like terms
5x=20 divide both sides by 5
x=4 mi/hr-----------------------------walking speed
2x=2*4=8 mi/hr-----------------------------running speed
CK
2*4+1.5*8=20
8+12=20
20=20
