Question 200208
Let the speed of the boat in still water be s, and the speed of the current c
Since {{{D = (S/T)}}}. then on the 100-mile journey downstream, we have:

{{{100/(s + c) = 5}}}, and on the 100-mile journey upstream, we have:
{{{100/(s - c) = 25}}},

Therefore, we have {{{100/(s + c) = 5}}}, which when cross-multiplied gives us:     5s + 5c = 100, or s + c = 20, and when {{{100/(s - c) = 25}}} 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