Seven pieces of rope have an average (arithmetic mean) length of 68 centimeters and a median length of 84 centimeters....

1. Translate the problem requirements: We have 7 ropes with average length 68 cm and median 84 cm. The longest rope is 14 cm more than 4 times the shortest rope. We need to find the maximum possible length of the longest rope.
2. Set up the constraint equations: Use the given average and median to establish relationships between the rope lengths when arranged in order.
3. Express the longest rope in terms of the shortest: Use the given relationship to create a connection between these two extremes.
4. Maximize by strategic minimization: To maximize the longest rope, minimize the other rope lengths while respecting the median constraint.
5. Solve for the optimal configuration: Calculate the shortest rope length and then find the corresponding longest rope length.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we have in plain English. We have 7 pieces of rope that we need to arrange from shortest to longest. Think of it like lining up 7 people by height - we'll call these rope lengths \(\mathrm{r_1}, \mathrm{r_2}, \mathrm{r_3}, \mathrm{r_4}, \mathrm{r_5}, \mathrm{r_6}, \mathrm{r_7}\) where \(\mathrm{r_1}\) is the shortest and \(\mathrm{r_7}\) is the longest.

We know three key facts:

• The average length of all 7 ropes is 68 cm, which means if we add them all up and divide by 7, we get 68
• The median length is 84 cm, which means the middle rope (the 4th one when arranged in order) is exactly 84 cm
• The longest rope is 14 cm more than 4 times the shortest rope

Process Skill: TRANSLATE - Converting the problem's English into mathematical relationships

2. Set up the constraint equations

Now let's transform our understanding into mathematical relationships.

Since the average is 68 cm, the total length of all ropes must be:
Total length = \(7 \times 68 = 476\) cm

So: \(\mathrm{r_1} + \mathrm{r_2} + \mathrm{r_3} + \mathrm{r_4} + \mathrm{r_5} + \mathrm{r_6} + \mathrm{r_7} = 476\)

Since the median is 84 cm, and we have 7 ropes (an odd number), the median is simply the middle value:
\(\mathrm{r_4} = 84\) cm

This gives us: \(\mathrm{r_1} + \mathrm{r_2} + \mathrm{r_3} + 84 + \mathrm{r_5} + \mathrm{r_6} + \mathrm{r_7} = 476\)
Simplifying: \(\mathrm{r_1} + \mathrm{r_2} + \mathrm{r_3} + \mathrm{r_5} + \mathrm{r_6} + \mathrm{r_7} = 392\)

3. Express the longest rope in terms of the shortest

The problem tells us that the longest rope is 14 cm more than 4 times the shortest rope. In mathematical terms:
\(\mathrm{r_7} = 4\mathrm{r_1} + 14\)

Substituting this into our equation:
\(\mathrm{r_1} + \mathrm{r_2} + \mathrm{r_3} + \mathrm{r_5} + \mathrm{r_6} + (4\mathrm{r_1} + 14) = 392\)
Simplifying: \(5\mathrm{r_1} + \mathrm{r_2} + \mathrm{r_3} + \mathrm{r_5} + \mathrm{r_6} = 378\)
Therefore: \(\mathrm{r_2} + \mathrm{r_3} + \mathrm{r_5} + \mathrm{r_6} = 378 - 5\mathrm{r_1}\)

4. Maximize by strategic minimization

Here's the key insight: to make the longest rope as long as possible, we want to make all the other ropes as short as possible while still respecting our constraints.

Since the ropes are arranged in order, each rope must be at least as long as the previous one:

• \(\mathrm{r_2} \geq \mathrm{r_1}\) (the second shortest can't be shorter than the shortest)
• \(\mathrm{r_3} \geq \mathrm{r_2}\) (and so on...)
• \(\mathrm{r_5} \geq \mathrm{r_4} = 84\) (the 5th rope must be at least as long as the median)
• \(\mathrm{r_6} \geq \mathrm{r_5}\)

To minimize the sum \(\mathrm{r_2} + \mathrm{r_3} + \mathrm{r_5} + \mathrm{r_6}\), we should make each rope equal to its minimum possible value:
\(\mathrm{r_2} = \mathrm{r_3} = \mathrm{r_1}\) and \(\mathrm{r_5} = \mathrm{r_6} = 84\)

Substituting: \(\mathrm{r_1} + \mathrm{r_1} + 84 + 84 = 378 - 5\mathrm{r_1}\)
Simplifying: \(2\mathrm{r_1} + 168 = 378 - 5\mathrm{r_1}\)
Combining terms: \(7\mathrm{r_1} = 210\)
Therefore: \(\mathrm{r_1} = 30\)

Process Skill: APPLY CONSTRAINTS - Using the ordering constraint to minimize other rope lengths

5. Solve for the optimal configuration

Now that we know the shortest rope is 30 cm, we can find the longest rope:
\(\mathrm{r_7} = 4\mathrm{r_1} + 14 = 4(30) + 14 = 120 + 14 = 134\) cm

Let's verify our solution works:

• Rope lengths: 30, 30, 30, 84, 84, 84, 134
• Sum: \(30 + 30 + 30 + 84 + 84 + 84 + 134 = 476\) ✓
• Average: \(476 \div 7 = 68\) ✓
• Median: 84 ✓
• Longest = 4 × shortest + 14: \(134 = 4(30) + 14 = 134\) ✓

Final Answer

The maximum possible length of the longest piece of rope is 134 centimeters.

The answer is (D) 134.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding median with odd number of data points

Students often confuse how to find the median when there are 7 pieces of rope. They might think the median is the average of two middle values (like with even numbers) rather than recognizing that with 7 pieces arranged in order (\(\mathrm{r_1}, \mathrm{r_2}, \mathrm{r_3}, \mathrm{r_4}, \mathrm{r_5}, \mathrm{r_6}, \mathrm{r_7}\)), the median is simply the 4th value: \(\mathrm{r_4} = 84\).

2. Missing the ordering constraint

Students frequently overlook that the ropes must be arranged in non-decreasing order (\(\mathrm{r_1} \leq \mathrm{r_2} \leq \mathrm{r_3} \leq \mathrm{r_4} \leq \mathrm{r_5} \leq \mathrm{r_6} \leq \mathrm{r_7}\)). This constraint is crucial for the optimization step - without recognizing this, they can't minimize the other rope lengths to maximize the longest rope.

3. Incorrect optimization strategy

Students may not realize that to maximize \(\mathrm{r_7}\), they need to minimize all other rope lengths while respecting constraints. They might try random values or miss that the optimal strategy is to make \(\mathrm{r_2} = \mathrm{r_3} = \mathrm{r_1}\) and \(\mathrm{r_5} = \mathrm{r_6} = 84\) (the minimum allowed values).

Errors while executing the approach

1. Algebraic manipulation errors

When substituting \(\mathrm{r_7} = 4\mathrm{r_1} + 14\) and setting up the equation \(2\mathrm{r_1} + 168 = 378 - 5\mathrm{r_1}\), students often make errors moving terms across the equals sign. They might get \(2\mathrm{r_1} + 5\mathrm{r_1} = 378 + 168\) instead of \(7\mathrm{r_1} = 378 - 168 = 210\).

2. Arithmetic calculation mistakes

Students may correctly set up \(7\mathrm{r_1} = 210\) but then calculate \(\mathrm{r_1}\) incorrectly, or when finding \(\mathrm{r_7} = 4(30) + 14\), they might compute \(4 \times 30 = 120\) correctly but then add 14 incorrectly to get 124 instead of 134.

Errors while selecting the answer

1. Selecting an intermediate value

Students might calculate \(\mathrm{r_1} = 30\) correctly and then mistakenly select this value or another intermediate calculation result instead of the final answer \(\mathrm{r_7} = 134\). The question asks for the maximum length of the longest piece, not the shortest piece.