If the average (arithmetic mean) of the five numbers 14, 20, 27, n + 2, text{ and } 2n -...

1. Translate the problem requirements: We need to find the value of n when the arithmetic mean (average) of five specific numbers equals 21. The five numbers are: 14, 20, 27, \(\mathrm{(n + 2)}\), and \(\mathrm{(2n - 3)}\).
2. Apply the fundamental average relationship: Use the core principle that average equals sum divided by count, so if average is 21 and we have 5 numbers, their total sum must be \(21 \times 5 = 105\).
3. Set up the sum equation: Add all five given expressions and set this sum equal to 105, creating a single equation with one unknown variable n.
4. Solve for n using algebraic manipulation: Combine like terms and isolate n to find its value, then verify this matches one of the given answer choices.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding exactly what we're looking for. We have five numbers, and we need to find the value of n when their average equals 21.

The five numbers are:

• 14 (a regular number)
• 20 (another regular number)
• 27 (another regular number)
• \(\mathrm{n + 2}\) (a number that depends on n)
• \(\mathrm{2n - 3}\) (another number that depends on n)

We're told that when we add all these up and divide by 5, we get 21. Our job is to figure out what value of n makes this true.

Process Skill: TRANSLATE - Converting the word problem into clear mathematical understanding

2. Apply the fundamental average relationship

Here's the key insight about averages that makes this problem manageable: if you know the average and you know how many numbers you have, you can find the total sum.

Think about it this way: if 5 test scores average to 21 points each, then the total points earned across all 5 tests must be \(21 \times 5 = 105\) points.

This works because: \(\text{Average} = \text{Total Sum} \div \text{Number of Items}\)
So: \(\text{Total Sum} = \text{Average} \times \text{Number of Items}\)

In our case: \(\text{Total Sum} = 21 \times 5 = 105\)

This means our five numbers must add up to exactly 105.

3. Set up the sum equation

Now we can write an equation. Since we know the five numbers must sum to 105, we have:

\(14 + 20 + 27 + (\mathrm{n + 2}) + (\mathrm{2n - 3}) = 105\)

Let's first add up the regular numbers: \(14 + 20 + 27 = 61\)

So our equation becomes:

\(61 + (\mathrm{n + 2}) + (\mathrm{2n - 3}) = 105\)

Now let's simplify the terms with n. We can remove the parentheses:

\(61 + \mathrm{n} + 2 + \mathrm{2n} - 3 = 105\)

Combining like terms:

• Regular numbers: \(61 + 2 - 3 = 60\)
• Terms with n: \(\mathrm{n} + \mathrm{2n} = \mathrm{3n}\)

This gives us: \(60 + \mathrm{3n} = 105\)

4. Solve for n using algebraic manipulation

Now we solve for n step by step:

\(60 + \mathrm{3n} = 105\)

Subtract 60 from both sides:

\(\mathrm{3n} = 105 - 60\)
\(\mathrm{3n} = 45\)

Divide both sides by 3:

\(\mathrm{n} = 45 \div 3 = 15\)

Let's verify this works by substituting \(\mathrm{n = 15}\) back into our original five numbers:

• 14
• 20
• 27
• \(\mathrm{n + 2} = 15 + 2 = 17\)
• \(\mathrm{2n - 3} = \mathrm{2(15)} - 3 = 30 - 3 = 27\)

Sum: \(14 + 20 + 27 + 17 + 27 = 105\) ✓
Average: \(105 \div 5 = 21\) ✓

Final Answer

\(\mathrm{n = 15}\), which corresponds to answer choice A.

The solution confirms that when \(\mathrm{n = 15}\), the five numbers (14, 20, 27, 17, and 27) have an average of exactly 21.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the average formula relationship

Students often try to set up the average formula as: \((14 + 20 + 27 + \mathrm{n} + 2 + \mathrm{2n} - 3) \div 5 = 21\), then attempt to solve this fraction equation directly. While this approach can work, it's more error-prone than recognizing that Average × Number of Items = Total Sum, which gives us the cleaner equation: \(14 + 20 + 27 + (\mathrm{n + 2}) + (\mathrm{2n - 3}) = 105\).

2. Incorrectly interpreting the expressions with n

Students may misread "\(\mathrm{n + 2}\)" and "\(\mathrm{2n - 3}\)" as separate terms to be handled individually, rather than recognizing these as algebraic expressions that will be combined during the solving process. This can lead to setting up incorrect equations or getting confused about how to handle the variable terms.

Errors while executing the approach

1. Sign errors when combining like terms

When simplifying \(61 + (\mathrm{n + 2}) + (\mathrm{2n - 3})\), students frequently make mistakes with the signs. A common error is writing \(61 + \mathrm{n} + 2 + \mathrm{2n} + 3 = 105\) instead of \(61 + \mathrm{n} + 2 + \mathrm{2n} - 3 = 105\), missing the negative sign in front of the 3. This leads to \(66 + \mathrm{3n} = 105\) instead of the correct \(60 + \mathrm{3n} = 105\).

2. Arithmetic mistakes in basic calculations

Students often make computational errors when adding the constant terms (\(14 + 20 + 27 = 61\)) or when performing the final division (\(45 \div 3 = 15\)). These seemingly simple calculations are frequent sources of incorrect answers, especially under time pressure.

3. Coefficient combination errors

When combining the n terms, students sometimes incorrectly add the coefficients, writing \(\mathrm{n} + \mathrm{2n} = \mathrm{2n}\) instead of \(\mathrm{n} + \mathrm{2n} = \mathrm{3n}\). This fundamental algebra mistake leads to solving \(60 + \mathrm{2n} = 105\), which gives \(\mathrm{n = 22.5}\) instead of the correct \(\mathrm{n = 15}\).

Errors while selecting the answer

1. Failing to verify the solution

Students may arrive at the correct value \(\mathrm{n = 15}\) but fail to substitute back into the original expressions to confirm their answer. Without verification, they might second-guess themselves and change to a different answer choice, especially if they made and caught an error earlier in their work.