# Word Problems About Fishing and Fishermen

TR Smith is a product designer and former teacher who uses math in her work every day.

Fishing is one of the oldest human activities, and solving riddles and math puzzles was a common pastime before people had TVs and smart phones, so it's no wonder that math word problems about fishing and fishermen abound. Here are five word problems and brain teasers about fishing from middle school level to high school level.

## Fish Weights

Luisa caught a fish that was too big to fit on her scale, so she asked her grandson Carlos to cut it into four pieces -- head, front middle, back middle, and tail -- and weigh them individually. Because Carlos liked to annoy his grandmother, he cut and weighed the pieces and reported the following weights to her, thinking she would not be able to deduce the weight of the whole fish:

• The front middle, back middle, and tail together weigh 18 lbs.
• The head, front middle, and back middle together weigh 19 lbs.
• The head and tail together weigh 9 lbs.

Luisa surprised Carlos by thanking him and telling him the total weight of the fish. How much did the fish weigh and how did she figure it out?

Solution: Considering all of the statements about weight Carlos gave his grandmother, each section of the fish is mentioned exactly twice. That means if you added all the weights Carlos reported, the sum would count the weight of each section twice. In other words, the sum of his statements is two times the total weight of the fish. Luisa figured out the weight by computing 18 + 19 + 9 = 46, and then computing 46/2 = 23 pounds.

## The Fishmonger and the Fibbing Fishermen

Four fisherman discuss the day's catch with each other, but only in pairs. Alice and Carly are enemies who don't talk to each other, as are Barb and Dawn. The conversations go like this:

• First, Alice and Barb compare the number of fish they caught.
• Next, Barb and Carly compare the number of fish they caught. Barb tells Carly, "You caught one more fish than Alice."
• Next, Carly and Dawn compare the number of fish they caught. Carly tells Dawn, "You caught two more than Barb."
• Next, Dawn and Alice compare their catches. Dawn tells Alice, "You caught three more than Carly."
• Finally, Barb and Alice reconvene. Alice tells Barb, "You caught one more than Dawn."

After they chat in pairs, all four go individually to sell their fish to the fishmonger. The last one to sell her fish asks the fishmonger how many he bought from the four fishermen that day. He tells her the amounts, remarking what a coincidence it was that each sold him one fewer than the previous fisherman. After hearing this she realizes that her friend lied to her, but she doesn't feel too bad because she also lied, and because her enemy didn't catch the most fish.

Assuming the fishmonger and two of the fisherman told the truth, and two fishermen lied, who are the liars? Who was the last person to visit the fishmonger?

Solution: First, the fishmonger's statement implies that if the first fisherman sold X fish, then the next sold X-1, the next sold X-2, and the last sold X-3. The value of X is irrelevant, so for the sake of argument we can pretend the first sold 10, the second sold 9, the third sold 8, and the last sold 7. Moreover, the last seller was both lied to and a liar. Then there are four possibilities for who told lies and who was the last to visit the fishmonger

• Alice and Barb lied, and Barb visited last.
• Alice and Dawn lied, and Alice visited last.
• Carly and Barb lied, and Carly visited last.
• Carly and Dawn lied, and Dawn visited last.

Now let's analyze each case:

• If the first case is true, then Carly and Dawn must be telling the truth, and Alice caught three more than Carly. That would mean that Alice caught 10 and Carly caught 7, making Carly the last person to visit the fishmonger, but this is a contradiction. Therefore, we can rule out this possibility.
• If the second case is true, then Carly and Barb are telling the truth. But there is no way to order the three women such that Alice is last, Carly is third, and Dawn is two places ahead of Barb. So this possibility is also ruled out.
• If the third case is true, then Alice and Dawn are telling the truth. That would mean the order of fisherman from first to last is ABDC. But in this case, Carly's enemy Alice catches the most fish, which contradicts the statement of the problem.
• If the fourth case is true, then Alice and Barb are telling the truth. That would mean the order is either CABD or BDCA. Since Dawn needs to be last in this scenario, it has to be CABD.

Therefore, the two liars are Carly and Dawn. The order in which the four fishermen visited the fishmonger is Carly, Alice, Barb, and Dawn.

## Three Ice fishermen

William and his brother Walter go ice fishing on a circular lake that is 160 meters in diameter. William puts his shack 60 meters due east of the lake's center; Walter puts his shack 60 meters due west of the lake's center. Assuming the fish in the lake are evenly distributed and that a fish will go to the closest hole, William and Walter agree each of their holes will be the closer to exactly half the fish in the lake.

As soon as they set up their ice fishing shacks, their younger sister Wendy shows up. She asks if she can join them and set up her shack in a position such that each sibling is the closest hole to exactly one third of the fish.

Wendy says it's possible for her to set up her shack on the lake so that each of them has control of a third of the lake, without either of her brothers having to move. Walter says it's possible, but only if one of the brothers moves his shack. William says it is not possible unless both brothers move their shacks. Who is right?

Solution: Wendy is right, there exists a spot on the lake where she can place her shack that would allow each sibling to control exactly one third of the fish, without either of her brothers having to move.

In the diagram below, look at the positions of the brothers' shacks (green dots) with respect to the center of the lake. Somewhere along the pink line is where Wendy must place her ice fishing hut if her brothers' areas of control are to remain equal.

Suppose Wendy places her shack X meters due north of the center of the lake. If X = 0, i.e., she places her shack in the center of the lake, then she will control 46.6% of the lake and her brothers will each control 26.7%. If X = 60, i.e., 60 meters due north of the center, then by symmetry she will control 25% of the lake and her brothers will each control 37.5%. Wendy's areas of control are shaded in pink in the diagram below.

The proportion of the lake under Wendy's control is a continuous function of X, therefore, there is some value of X between 0 and 60 such that the placement results in Wendy controlling exactly 1/3, and her brothers each controlling 1/3 as well.

## Fishing for Salmon in a Stocked Pond

Talia likes to fish in an artificial pond that is currently stocked with 120 fish. Her favorite fish is salmon, but only 24 of the fish in the pond are salmon. The owner of the pond says Talia can catch at most 8 fish, but she must keep everything she catches and not release any back into the lake. What is the probability she gets at least 2 salmon among her 8 fish?

Solution: For this problem you need to apply the combination formula. The total number of different sets of 8 fish is (120 C 8) = 840,261,910,995. The number of sets that have no salmon among them is (96 C 8) = 132,601,016,340. The number of sets that have exactly 1 salmon and 7 non-salmon is (96 C 7)*(24 C 1) = 286,060,619,520. Therefore, the probability that Talia catches 0 or 1 salmon is

(132,601,016,340 + 286,060,619,520) / 840,261,910,995
≈ 0.498251, or 49.8%

The probability of her catching 2 or more salmon is ≈ 1 - 0.498251 = 0.501749, or 50.2%. In plain terms, she has about a 50-50 chance of getting at least 2 salmon in her catch.

## Dopey Fisherman

Three fisherman agreed that at the end of the day they would split their total catch evenly among the three of them. So at the end of a long day of fishing they dumped all their fish in the cooler and took a break. The first fisherman went to the cooler, counted all the fish, divided the total by 3, and took his third.

The second fisherman went to the cooler not realizing that the first had already taken his third. He counted the fish, divided the total by 3, and discovered there was one left over. So he took his third and left the extra fish in the cooler

The third fisherman went to the cooler not realizing his two friends had already been there before to collect their thirds. He counted the fish, divided the total by 3, and discovered there was one left over. So he took his third and left the extra fish in the cooler.

The first fisherman came back, looked in the cooler, and seeing that the number left in the cooler was the same as the amount he took, he concluded that one of his friends hadn't yet taken his share.

How many fish did they catch in total?

Solution: They caught a total of 15 fish. The first fisherman looked in the cooler, saw 15, took 5, and left 10 behind. The second looked in the cooler, saw 10, took 3, and left 7 behind. The third looked in the cooler, saw 7, took 2, and left 5 behind. The first looked in the cooler again and saw 5, the exact number he took earlier.

Fish photos courtesy public domain, diagrams created by author.

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• Clay 4 months ago

How do you solve this one: 2 mackerel cost the same as 3 herring. 1 mackerel and 1 herring cost the same as 2 cod. 2 cod, 3 mackerel, and 4 herring cost the same as 7 salmon. With this information, is it possible to order the fish from most expensive to least expensive?

I got salmon, mackerel, cod, herring. Is that correct?

• Author

TR Smith 4 months ago

Hi Clay, thanks for the question. There are a lot of ways to solve problems like this. One straight forward approach is to tackle one statement at a time. Since the actual cost of each type of fish doesn't matter, start by pretending that 1 mackerel costs \$1.

The first statement tells you that 1 herring costs (2 x \$1)/3 or about \$0.67. (Don't worry about the round off error.)

The second statement tells you that 1 cod costs about (\$1+ \$0.67)/2, or about \$0.83.

The third statement tells you that 1 salmon costs about (2 x \$0.83 + 3 x \$1 + 4 x \$0.67)/7, or about \$1.05.