In need of help with this word problem

Question

A boat can travel 31 mph in still water. If the boat can travel 100 miles with the current in the same time it can travel 80 miles against the current, find the rate of the current r in miles per hour.

Answer 1

Let r = rate of the current
 
Speed of boat downstream (with current) = 31 + r
Speed of boat upstream (against current) = 31 - r

Remember D-I-R-T:  Distance Is Rate times Time
Distance = (rate)(time);    so time = (distance)/(rate)
 
Same time:     Time downstream = Time upstream
 100/(31+r) = 80/(31-r)
 
Cross multiplying, we have:  100(31-r) = 80(31+r)
                                         3100 - 100r = 2480 + 80r
                                             620 = 180r
                                              r ≅ 3.4 mph
 
ANSWER:  The speed of the current is approximately 3.4 mph.
 
Checking (very important):
  Is   100/(31+3.4) ≅ 80/(31-3.4)    ?
      2.91 ≅ 2.90    ?   yes

Answer 2

Let x = rate of current 

 

Speed of boat with current = x+31

Speed of boat against current = 31-x

 

Distance = (Rate)(Time), so Time = distance/rate

 

Time with current = Time against current

 

100/(x+31) = 80/(31-x)

 

Cross multiply to obtain 100(31-x) = 80(x+31)

 

3100 - 100x = 80x + 2480

 

620 = 180x

 

x = 620/180 = 62/18 = 31/9 mph

 

Answer 3

The first sentence contradicts the second: it seems unlikely that the boat could travel only 31 mph in still water, but 80 mph against the current. 

 

If you ignore the first sentence, and only look at the second, then the rate of the current is 10 mph. The rate of the boat must be 90 mph, so it travels 90 + 10 = 100 mph with the current, and 90 - 10 = 80 mph against the current.