Logical reasoning worksheet with fish length puzzle and sea creature counting problem.

Here is the step-by-step solution to the problems.

Problem 1: Who caught which fish?

Step 1: List the fish and their lengths.
First, let's look at the sizes of the fish available:
* Trout: 16 in.
* Catfish: 13 in.
* Bass: 8 in.
* Bluegill: 6 in.
* Minnow: 2 in.

Step 2: Analyze the clues for Sarah.
The clue says: "The length of Sarah's fish was not a multiple of 2."
A multiple of 2 is an even number (like 2, 4, 6, 8...). We need to find an odd number from our list.
* 16 is even.
* 13 is odd.
* 8 is even.
* 6 is even.
* 2 is even.

So, Sarah caught the Catfish (13 in.).

Step 3: Analyze the clues for Marta.
The clue says: "Marta's fish was longer than Sarah's fish."
We know Sarah's fish is 13 inches. The only fish longer than 13 inches is the Trout (16 in.).

So, Marta caught the Trout (16 in.).

Step 4: Analyze the remaining fish.
We have used the 16 in. and 13 in. fish. The remaining fish are:
* Bass (8 in.)
* Bluegill (6 in.)
* Minnow (2 in.)

The remaining people are Nancy, Yoko, and Frank.

Step 5: Analyze the clues for Nancy and Yoko.
The clue says: "Yoko's fish was 2 in. longer than Nancy's fish."
Let's check the pairs from the remaining fish to see which ones have a difference of 2 inches:
* 8 - 6 = 2 (This works!)
* 6 - 2 = 4 (No)
* 8 - 2 = 6 (No)

So, Yoko has the 8-inch fish and Nancy has the 6-inch fish.
Yoko caught the Bass (8 in.).
Nancy caught the Bluegill (6 in.).

*(Check: The clue also says "Nancy did not catch the smallest fish." The smallest fish is 2 in. Nancy caught the 6 in. fish, so this fits.)*

Step 6: Determine Frank's fish.
There is only one person left (Frank) and one fish left (Minnow, 2 in.).

So, Frank caught the Minnow (2 in.).

***

Problem 2: Kevin's Sea Creatures

Step 1: Set up the variables.
Let $s$ be the number of sand dollars.
Let $f$ be the number of starfish.
Let $u$ be the number of sea urchins.

We know the total is 50:
$$s + f + u = 50$$

Step 2: Use the clues to write equations.
Clue 1: "He has four times as many starfish as sand dollars."
$$f = 4 \times s$$

Clue 2: "He has 8 more sea urchins than sand dollars."
$$u = s + 8$$

Step 3: Substitute and Solve.
Substitute the values of $f$ and $u$ into the total equation so everything is in terms of sand dollars ($s$):

$$s + (4 \times s) + (s + 8) = 50$$

Combine the $s$ terms ($1s + 4s + 1s = 6s$):
$$6s + 8 = 50$$

Subtract 8 from both sides:
$$6s = 42$$

Divide by 6:
$$s = 7$$

So, there are 7 sand dollars.

Step 4: Calculate the other creatures.
Starfish ($f$): $4 \times 7 = 28$
Sea urchins ($u$): $7 + 8 = 15$

Step 5: Check the work.
$7 + 28 + 15 = 50$. The math is correct.

Final Answer:

1. Who caught which fish?
a. Sarah: Catfish (13 in.)
b. Nancy: Bluegill (6 in.)
c. Marta: Trout (16 in.)
d. Frank: Minnow (2 in.)
e. Yoko: Bass (8 in.)

2. How many of each type does Kevin have?
* Sand dollars: 7
* Starfish: 28
* Sea urchins: 15