9514 1404 393
Let d represent the distance to the town. Let s represent the nominal speed of the car. The relation between time, speed, and distance is d = st.
t1 = d/s
t2 = d/(s+15)
t1 : t2 = 6 : 5 . . . increasing the speed reduces the time
Substituting for t1 and t2, we have ...
(d/s)/(d/(s+15)) = 6/5
(s +15)/s = 6/5
1 +15/s = 1 +1/5
s = 5·15 = 75 . . . . nominal speed in km/h
Decreasing the speed increases the time.
d/75 +(105/60) = d/(75-15)
d(60/75) +105 = d . . . . . . multiply by 60
105 = d/5 . . . . . . . . . . . subtract 4/5d
525 = d . . . . . . . . . . multiply by 5
The distance traveled by the car is 525 km.
4:5 or 4 over 5.
Depending on the way the problem is communicated, your first listed item is usually on the top, while your second is placed on the bottom, in the case of a fraction. In the case of a colon, the left is your first unit and the right is your second.
2.5 times taller
6+6=12 3+3=6 12+6=18 so the streetlight is 18 ft