|
Question 1158073: A boat can travel 39mph in still water. If it travels 235 miles with the current in the same length of time it travels 155 miles against the current, what is the speed of the current?
Found 4 solutions by josgarithmetic, ikleyn, MathTherapy, jim_thompson5910: Answer by josgarithmetic(39613) (Show Source): Answer by ikleyn(52754) (Show Source): Answer by MathTherapy(10549) (Show Source): Answer by jim_thompson5910(35256) (Show Source):
You can put this solution on YOUR website!
There are already great answers here. I'll just provide a way to set up the equation mentioned.
C = speed of current
39+C = the speed of the boat when it is going downstream (it is sped up 39 mph)
39-C = speed of boat when it is going upstream (it is slowed down by 39 mph)
Basically the boat naturally goes 39 mph, but the current will help or hinder the boat to speed up or slow down. Which is why we add or subtract that C value.
Now we use the idea that
distance = rate*time
d = r*t
Solve for t to get
t = d/r
If the boat travels 235 miles with the current (downstream), then it goes a speed of 39+C mph, and the time it takes to do so is...
t = d/r
t = 235/(39+C)
Furthermore, if the boat goes 155 miles against the current (upstream), then it goes a speed of 39-C mph, and the time it takes to do so is...
t = d/r
t = 155/(39-C)
Both time values (t) are the same in this case due to the instructions stating so. Therefore, we are able to equate the right hand sides of the equations t = 235/(39+C) and t = 155/(39-C) to get 235/(39+C) = 155/(39-C)
From here you would solve for C as directed in the steps provided by @MathTherapy
|
|
|
| |