SOLUTION: When Sheila asked Margie to deliever camp supplies across the lake, they agreed to meet on the water. Shelia's boat can travel at 10km/h, while Margie's boat can do 12km/h. If the

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: When Sheila asked Margie to deliever camp supplies across the lake, they agreed to meet on the water. Shelia's boat can travel at 10km/h, while Margie's boat can do 12km/h. If the       Log On


   

Question 52810: When Sheila asked Margie to deliever camp supplies across the lake, they agreed to meet on the water. Shelia's boat can travel at 10km/h, while Margie's boat can do 12km/h. If the lake is 6.6 km wide and Sheila and Margie leave at the same time, how much time will it be before they meet?
Answer by funmath(2933) About Me  (Show Source):
You can put this solution on YOUR website!
Distance = rate*time or d=rt
They travel the same amount of time: t (in hours)
Sheila travels: 10t (km)
Margie travels: 12t (km)
Altogether they travel: 6.6 (km)
Our problem to solve is:
10t+12t=6.6
(10+12)t=6.6
22t=6.6
22t/22=6.6/22
t=.3 (hours)
Most of us don't know how long that really is, so convert it to minutes:
t=.3(60)min
t=18 min