Question 1200814
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Starting table
<table border = "1" cellpadding = "5"><tr><td></td><td>Distance (miles)</td><td>Rate (mph)</td><td>Time (hours)</td></tr><tr><td>With Current</td><td></td><td></td><td></td></tr><tr><td>Against Current</td><td></td><td></td><td></td></tr></table>


Fill in the given distance (150 miles) and the mentioned time values in their correct corresponding slots.
<table border = "1" cellpadding = "5"><tr><td></td><td>Distance (miles)</td><td>Rate (mph)</td><td>Time (hours)</td></tr><tr><td>With Current</td><td>150</td><td></td><td>3</td></tr><tr><td>Against Current</td><td>150</td><td></td><td>5</td></tr></table>


Then use the formula
rate = distance/time
to get the following
<table border = "1" cellpadding = "5"><tr><td></td><td>Distance (miles)</td><td>Rate (mph)</td><td>Time (hours)</td></tr><tr><td>With Current</td><td>150</td><td>50</td><td>3 </td></tr><tr><td>Against Current</td><td>150</td><td>30</td><td>5</td></tr></table>
Eg: 
rate = distance/time = 150/3 = 50 for the "with current" row.



b = speed of boat in still water
c = speed of the current
b+c = 50 ... downstream, i.e. with current
b-c = 30 ... upstream, i.e. against current


Add the equations straight down to eliminate variable c.
2b = 80
b = 80/2
b = 40
notice that 40 mph is the midpoint of 30 mph and 50 mph


Let's determine c based on b = 40
b+c = 50
40+c = 50
c = 50-40
c = 10
Or
b-c = 30
40-c = 30
-c = 30-40
-c = -10
c = 10
Technically you only would need to solve one of those to find c, but it doesn't hurt to have more practice solving both.
It helps to verify the answer is correct.


Answers:
speed of the boat in still water = <font color=red>40 mph</font>
speed of the current = <font color=red>10 mph</font>
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