Algebra Help: Three More Boat-in-the-River Word Problems

Over a year ago I wrote the article "Algebra Help: How to Solve Boat-in-the-River Word Problems." This has become, to date, my most popular and most clicked on article. In the original article I only present two sample word problems, so I decided to provide three more examples here. I'll be repeating some of the preliminary information about this particular kind of uniform motion Algebra problem so that first time visitors don't have to search out my original article to understand what is going on.

With the boat in the river problems, we assume that the boat has a uniform speed in still water and that the speed of the water (or the speed of the river current) is constant. The Algebra student will be presented with a math word problem in which he must solve for the speed of the boat (in still water, implied), or the speed of the river current, or the time spent going upstream or the time going downstream or the distance in one direction or some combination of these variables. In these examples the student will be solving for the speeds of the river current and the boat in still water.

With these problems, I recommend using the follow variables:

B = speed of the boat in still water
W = the speed of the water (or the rate of the river current)
T_d = time spent going downstream
T_u = time spent going upstream
D_d = distance gone downstream
D_u = distance gone upstream

As with other uniform motion word problems in Algebra, an important algebraic equation to remember is:

Distance = Rate * Time

With the boat-in-the-river math problem, we modify this slightly when we are going downstream to get:

D_d = (B + W) * T_d

That is, the speed (or rate) of the boat going downstream is the speed of the boat in still water plus the speed of the water (or river current). That is because when one is going downstream, the river works with you.

When a boat is going upstream, the river works against you so we have to change our distance equation to reflect this by subtracting the speed of the water from the speed of the boat to get our rate:

D_u = (B - W) * T_u

As always with learning about how to solve these kinds of problems, it is always best to see step by step how to solve specific examples.

Sample Boat-in-the-River Word Problem #1:
A boy can row downstream 18 miles in 2 hours, but it will take him 6 hours to return. How fast can he row in still water, and what is the rate of the current?

I always recommend starting by writing down the two essential algebraic distance equations to start:

D_d = (B + W) * T_d
D_u = (B - W) * T_u

Secondly, write down the information you know by dissecting the word problem:

D_d = D_u = 18
T_d = 2
T_u = 6

Plug in the known variables into the distance equations.

18 = (B + W) * 2
18 = (B - W) * 6

Simplify the first distance equation and solve B in terms of W.

18 = 2*B + 2*W
9 = B + W
B = 9 - W

Simplify the second distance equation and plug in the modified first distance equation.

18 = 6*B - 6*W
3 = B - W
3 = (9 - W) - W
-6 = -2*W
W = 3
B = 9 - 3 = 6

The water is flowing at a rate of 3 miles per hour, while the boat is moving (in still water) at a rate of 6 miles per hour.

Sample Boat-in-the-River Word Problem #2:
A crew that can row 12 mph downstream finds that 2 miles downstream takes the same time as 1 mile upstream. Find the rate of the current.

Write our two essential distance equations.

D_d = (B + W) * T_d
D_u = (B - W) * T_u

Write the information that we know from analyzing the word problem.

B + W = 12
D_d = 2
D_u = 1
T_d = T_u = T { because time is the same }

Solve for B in terms of W:

B = 12 - W

Plug in the information we know into the two distance equations.

2 = 12 * T
1 = (B - W) * T

Solve for T:

T = (1/6)

Plug in,then simplify to solve the second distance equation.

1 = (12 - W - W) * (1/6)
1 = (12 - 2*W)* (1/6)
1 = 2 - (1/3)*W
-1 = -(1/3)*W
W = 3

The rate of the current is 3 mph.

Sample Boat-in-the-River Word Problem #3:
A man can row downstream 3 times as fast as he can upstream. He can row 8 miles downstream in (2/3) of an hour. Find his rate in still water and the rate of the current.

Write our two essential distance equations.

D_d = (B + W) * T_d
D_u = (B - W) * T_u

Write the information that we know from analyzing the word problem.

B + W = 3*(B - W)
D_d = 8
T_d = (2/3)

This is a tricky Algebra problem in that we need to do some deduction in order to fill in some of the other parts. If he can row downstream 3 times as fast as he can upstream, the same distance upstream would take 3 times as long.

D_u = 8
T_u = 3*T_d = 3*(2/3) = 2

Plug in what we know into the distance equations.

8 = (B + W)*(2/3)
8 = (B - W)*2

Simplify the second distance equation and solve for B in terms of W.

4 = B - W
B = W + 4

Plug into the first distance equation and simplify.

8 = (W + 4 + W)*(2/3)
8 = (2*W + 4)*(2/3)

Multiply both sides by 3 to get rid of the fraction, then simplify.

24 = (2*W + 4)*2
24 = 4*W + 8
16 = 4*W
W = 4
B = 4 + 4 = 8

The rate of the current is 4 mph, while the man can row in still water at a speed of 8 mph.

Blessings!

Source
Virgil S. Mallory. A First Course in Algebra (1943)