You can
put this solution on YOUR website!Let the speed of the boat in still water be s, and the speed of the current c
Since

. then on the 100-mile journey downstream, we have:

, and on the 100-mile journey upstream, we have:

,
Therefore, we have

, which when cross-multiplied gives us: 5s + 5c = 100, or s + c = 20, and when

is also cross-multiplied, we get: 25s – 25c = 100, or s – c = 4
s + c = 20---- (i)
s – c = 4 ----- (ii)
Adding equations (i) & (ii), we get: 2s = 24, or s = 12
Substituting this value for s in eq (i), we have; 12 + c = 20, or c = 8
Therefore, the speed of the boat in still water is s, or 12 mph, and the speed of the current is 8 mph
You can
put this solution on YOUR website!Let r=rate of the boat, s=speed of boat in still water, c=speed of current, t=time and d=distance
r=s + - d)
t*r=d
.
25(s-c)=100
s-c=4
s=4+c
.
5(s+c)=100
s+c=20
4+c+c=20 substitute 4+c for s.
2c=16
c=8 mph speed of current.
.
s=4+8
s=12 speed of boat in still water.
.
Ed