System of Equations Word Problems

Question

A motorboat takes 5 hours to travel 100 miles going upstream. The return trip takes 2 hours going downstream. What is the rate of the boat in still water and what is the rate of the current?

Answer 1

d = rt; r = d/t

Speed of boat against current is 100/5 = 20 mph

Speed of boat with current is 100/2 = 50 mph

so speed of boat in still water is (20 + 50)/2 = 35 mph (fast!!)

Then speed of current must be the difference in both directions, or 15 mph.

You could also set up a system of equations

d = rt

100 = 5(B - C) against the current; 100 = 2(B + C) with the current.

Reduces to B - C = 20 and B + C = 50.

Use elimination method to solve.

Answer 2

Distance = Rate x Time therefore Rate = Distance divided by Time

D= 100, T₁ = 5 so R₁ = 100/5 or 20 mph. This is going against the current.

T₂ = 2 so R₂ = 100/2 or 50 mph. This is traveling with the current. Since the distance is the same, you can add the equations together: 20mph = R_b - R_w We are subtracting because the current is going in the opposite direction. 50 mph= R_b + R_w

Adding the equations together yields 70 mph = 2R_b

R_b = 35 mph in still water.