SOLUTION: Sue walks and jogs to school. She walks at a speed of 5kmh and jogs at a speed of 9kmh. It is 8km to school. She makes it in 1 hour. How far does she jog?
I know that d=rt but I
Algebra ->
Customizable Word Problem Solvers
-> Travel
-> SOLUTION: Sue walks and jogs to school. She walks at a speed of 5kmh and jogs at a speed of 9kmh. It is 8km to school. She makes it in 1 hour. How far does she jog?
I know that d=rt but I
Log On
Question 170955This question is from textbook Introductory Algebra
: Sue walks and jogs to school. She walks at a speed of 5kmh and jogs at a speed of 9kmh. It is 8km to school. She makes it in 1 hour. How far does she jog?
I know that d=rt but I'm drawing a blank on this one.
Thanks This question is from textbook Introductory Algebra
You can put this solution on YOUR website! just set up an equation that adds up the distances traveled:
let x denote the time that it was used walking, then the time for joging cam be denoted as (1-x) because it was one hour of total trip, then use the formula that distance = velocity * time; then the equation looks like:
then to find out how much she jogs, we multiply 9 * (1-0.25) because that is the time used for joggin and the answer is 6.75
You can put this solution on YOUR website! ok.......this isnt written very well but what they are getting at is she walked a portion and jogged a portion of the trip(8km) to school. d=rt. lets call the distance she walked x and the distance she jogged 8-x. lets the time she walked t and the time she jogged 1-t(1hour altogether). walking rate is 5 and jogging rate is 9
:
x=5t.........eq 1
8-x=9(1-t)...eq 2
:
take x's value from eq 1 and plug it into eq 2
:
8-5t=9(1-t)
:
8-5t=9-9t
:
4t=1
: hour she walked
:
we are after distance she jogged so lets find x ...x=5(1/4)=5/4=1.25..distance she walked
:
distance she jogged is 8-x...km
(