A piece of tape 14 inches long was cut into two pieces. If the length of the shorter piece was...

1. Translate the problem requirements: A 14-inch tape is cut into two pieces where the shorter piece is \(\frac{2}{3}\) the length of the longer piece. We need to find the length of the shorter piece.
2. Set up the relationship using variables: Define the longer piece length and express the shorter piece in terms of it to avoid working with fractions directly.
3. Apply the constraint equation: Use the fact that both pieces must add up to the total tape length to create a solvable equation.
4. Solve and identify the answer: Calculate the shorter piece length and match it to the given answer choices.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we have in everyday terms. We have a piece of tape that's 14 inches long, and someone cut it into exactly two pieces. The problem tells us that the shorter piece has a length that is \(\frac{2}{3}\) of whatever the longer piece's length is.

Think of it this way: if the longer piece were 6 inches, then the shorter piece would be \(\frac{2}{3} \times 6 = 4\) inches. But we don't know the actual lengths yet - we just know this relationship exists between them.

Our goal is to find how long the shorter piece actually is.

Process Skill: TRANSLATE - Converting the fraction relationship into a clear mathematical understanding

2. Set up the relationship using variables

To avoid working with messy fractions, let's call the length of the longer piece 'L' inches.

Since the shorter piece is \(\frac{2}{3}\) the length of the longer piece, we can say:
Shorter piece = \(\frac{2}{3} \times \mathrm{L}\)

This gives us a clean way to express both pieces:

• Longer piece: L inches
• Shorter piece: \(\frac{2}{3}\mathrm{L}\) inches

Notice how this setup makes the relationship crystal clear and avoids having fractions in multiple places.

3. Apply the constraint equation

Here's the key insight: both pieces must add up to the original 14-inch tape length. This gives us our constraint.

In plain English: Longer piece + Shorter piece = Total length

Substituting our expressions:
\(\mathrm{L} + \frac{2}{3}\mathrm{L} = 14\)

To solve this, we can factor out L:
\(\mathrm{L}(1 + \frac{2}{3}) = 14\)
\(\mathrm{L}(\frac{3}{3} + \frac{2}{3}) = 14\)
\(\mathrm{L}(\frac{5}{3}) = 14\)

Therefore: \(\mathrm{L} = 14 \times \frac{3}{5} = \frac{42}{5} = 8.4\) inches

Process Skill: APPLY CONSTRAINTS - Using the total length requirement to create a solvable equation

4. Solve and identify the answer

Now that we know the longer piece is 8.4 inches, we can find the shorter piece:

Shorter piece = \(\frac{2}{3} \times 8.4\)
Shorter piece = \((2 \times 8.4) \div 3\)
Shorter piece = \(16.8 \div 3\)
Shorter piece = 5.6 inches

Let's verify: \(8.4 + 5.6 = 14.0\) ✓
Also verify the ratio: \(5.6 \div 8.4 = \frac{2}{3}\) ✓

Final Answer

The shorter piece is 5.6 inches long, which matches answer choice C.

Common Faltering Points

Errors while devising the approach

Students often assume that since \(\frac{2}{3} < 1\), the piece described as "\(\frac{2}{3}\) the length of the other" must automatically be the shorter piece. However, they fail to verify this assumption. If they mistakenly think the piece that is \(\frac{2}{3}\) of the other is actually the longer piece, they would set up the relationship incorrectly from the start.

The problem states "the shorter piece was \(\frac{2}{3}\) the length of the longer piece," but students might misread this as "the longer piece was \(\frac{2}{3}\) the length of the shorter piece." This leads to setting up the equation as Longer piece = \(\frac{2}{3} \times\) Shorter piece instead of Shorter piece = \(\frac{2}{3} \times\) Longer piece.

Errors while executing the approach

When combining \(\mathrm{L} + \frac{2}{3}\mathrm{L} = 14\), students often make mistakes converting to a common denominator. They might incorrectly write this as \(\frac{6}{3}\mathrm{L}\) instead of \(\frac{5}{3}\mathrm{L}\), or make errors when multiplying \(14 \times \frac{3}{5}\) to solve for L.

After finding L = 8.4 inches (the longer piece), students might forget that the question asks for the shorter piece and mistakenly select an answer choice close to 8.4. They stop at finding the longer piece without completing the final step of calculating \(\frac{2}{3} \times 8.4 = 5.6\).

Errors while selecting the answer

Even if students correctly calculate both pieces (longer = 8.4 inches, shorter = 5.6 inches), they might select an answer choice that's close to 8.4 because they lose track of what the question is actually asking for. Since the question asks for the shorter piece, they should select 5.6, but they might be drawn to a distractor answer designed to catch this error.