Problema Solution
Susan started her bicycle ride at 1:30 in the afternoon. After riding a certain distances, she got a flat tire and had to walk home. She arrived home at 4:30. If Susan averaged 21km/hr on her bicycle and 6km/hr walking, how far did she ride before getting the flat tire?
Answer provided by our tutors
let
the total time of the travel is 4:30 - 1:30 = 3 hours
t = the time that Susan was riding her bicycle (before she got the flat tire)
3 - t = the time that Susan was walking
v1 = 21 km/hr the Susan's average bicycle speed
v2 = 6 km/hr Susan's average walking speed
d = the distance in one direction (we are assuming that Susan rode her bicycle to the point where she got the flat tire and then she walked the same distance back home)
since speed = distance/time follows distance = speed*time
d = v1*t
21t = d
d = v2*(3 - t)
6(3 - t) = d
18 - t = d
by solving the system of equations:
21t = d
18 - t = d
we find:
t = 9/11 hr
click here to see the step by step solution:
v1*t = 21(9/11) =17.18 km
Susan rode her bicycle for about 17.18 kilometers.