A boat can go 20 mi against a current in the same time that

A boat can go 20 mi against a current in the same time that it can go 60 mi with the current. The current is 4 mph. Find the speed of the boat in still water.

Solution

Let the speed of the boat be x mph
speed of the current = 4mph
so speed against the current = (x-4)mph
Distance traveled against the current = 20 miles
So time taken = distance/speed
= 20/(x-4)
Now speed with the current = (x+4)mph
Distance traveled with the current = 60 miles
So time taken = distance/speed
= 60/(x+4)
As the time taken is the same,
20/(x-4) = 60/(x+4)
Multiplying both the sides by (x-4)(x+4),
20(x+4) = 60(x-4)
==> 20x + 80 = 60x - 240
==> 240 + 80 = 60x - 20x [ collecting the like terms to one side]
==> 320 = 40x
==> 320/40 = x
==> 8 = x
Thus the speed of the boat in still water = 8mph
Best of luck!!!