Two boat landings are 1.0 km apart on the same bank of a stream that flows at 1.0 km/h. a motorboat makes the round trip between the two landings in 50 minutes. what is the speed of the boat relative to the water?

Refer to the figure shown below.

Let V = speed of the boat relative to the water

Given:
u = 1 km/h the speed of flowing water.

When traveling downstream from A to B, the actual speed of the boat is
V₁ = V + u = V + 1 km/h
When traveling upstream from B to A, the actual speed of the boat is
V₂ = V - u =V - 1 km/h

Because the distance Ab is 1 km, the time taken for the round trip is
t = (1 km)/(V+1 km/h) + (1 km)/(v-1 km/h)
  = (V-1 + V+1)/(V² - 1)
  = (2V)/(V² - 1)

The time for the round trip is 50 min = 5/6 h.
Therefore
(2V)/(V² - 1) = 5/6
5(V² - 1) = 12V
5V² - 12V - 5 = 0

Solve with the quadratic formula.
V = (1/10)*[12 +/- √(144 + 100)] = 2.762 or -0.362 km/h

Ignore negative speed, so that
V = 2.762 k/h

The speed of the boat relative to the water is 2.76 km/h (nearest hundredth)