List S consists of 10 consecutive odd integers, and list T consists of 5 consecutive even integers. If the least...

1. Translate the problem requirements: We need to understand that 'consecutive odd integers' means integers like 1, 3, 5, 7, etc., and 'consecutive even integers' means integers like 2, 4, 6, 8, etc. The problem asks for the difference between the arithmetic means of these two lists, given that the smallest number in list S is 7 more than the smallest number in list T.
2. Express the lists using the relationship given: Set up the two lists based on the constraint that the least integer in S is 7 more than the least integer in T, allowing us to express both lists in terms of one variable.
3. Apply the key insight about averages of consecutive sequences: Recognize that for any sequence of consecutive integers (odd or even), the average equals the middle value, which simplifies our calculation significantly.
4. Calculate the difference between averages: Use the relationship between the starting points of the two sequences to find how much greater one average is than the other.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we're dealing with in everyday terms:

List S: Contains 10 consecutive odd integers. Think of odd integers as numbers like 1, 3, 5, 7, 9, 11... They're spaced 2 apart and skip all the even numbers.

List T: Contains 5 consecutive even integers. Think of even integers as numbers like 2, 4, 6, 8, 10... They're also spaced 2 apart but include only even numbers.

The key relationship: The smallest number in list S is exactly 7 more than the smallest number in list T.

What we need to find: How much bigger is the average of list S compared to the average of list T?

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

2. Express the lists using the relationship given

Let's use concrete numbers to make this crystal clear. Since we know the relationship between the starting points, let's call the smallest integer in list T as 'a'.

List T starts with: a
So List T = \(\{\mathrm{a}, \mathrm{a}+2, \mathrm{a}+4, \mathrm{a}+6, \mathrm{a}+8\}\)

List S starts with: a + 7 (since it's 7 more than T's smallest)
So List S = \(\{\mathrm{a}+7, \mathrm{a}+9, \mathrm{a}+11, \mathrm{a}+13, \mathrm{a}+15, \mathrm{a}+17, \mathrm{a}+19, \mathrm{a}+21, \mathrm{a}+23, \mathrm{a}+25\}\)

Notice how each list follows the pattern of adding 2 to get the next number, maintaining the consecutive odd/even pattern.

3. Apply the key insight about averages of consecutive sequences

Here's a powerful insight that makes this problem much simpler: For any list of consecutive integers (whether odd or even), the average is always the middle value.

Let me show you why this works:

• In any evenly spaced sequence, the numbers are symmetric around the middle
• The average of symmetric values always equals the middle value

For List T (5 numbers):
The middle number is the 3rd number = \(\mathrm{a} + 4\)
So average of T = \(\mathrm{a} + 4\)

For List S (10 numbers):
With 10 numbers, we take the average of the 5th and 6th numbers:
5th number = \(\mathrm{a} + 15\)
6th number = \(\mathrm{a} + 17\)
Middle = \((\mathrm{a} + 15 + \mathrm{a} + 17) ÷ 2 = (2\mathrm{a} + 32) ÷ 2 = \mathrm{a} + 16\)
So average of S = \(\mathrm{a} + 16\)

Process Skill: SIMPLIFY - Using the middle-value property instead of calculating all terms

4. Calculate the difference between averages

Now we can find how much greater the average of S is than the average of T:

Difference = Average of S - Average of T
Difference = \((\mathrm{a} + 16) - (\mathrm{a} + 4)\)
Difference = \(\mathrm{a} + 16 - \mathrm{a} - 4\)
Difference = 12

Notice how the variable 'a' cancels out completely! This makes sense because the difference between averages depends only on how far apart the starting points are, not on what the actual starting values are.

Final Answer

The average of the integers in S is 12 greater than the average of the integers in T.

The answer is (D) 12.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding "consecutive odd/even integers" spacing: Students often think consecutive odd integers are spaced 1 apart (like 7, 8, 9) instead of 2 apart (like 7, 9, 11). This fundamental misunderstanding leads to incorrectly writing the sequences as {a+7, a+8, a+9...} instead of {a+7, a+9, a+11...}.

2. Misinterpreting the constraint "7 more than": Students may incorrectly assume that EVERY integer in S is 7 more than EVERY integer in T, rather than understanding that only the smallest integer in S is 7 more than the smallest integer in T. This leads to trying to compare corresponding terms instead of focusing on the starting relationship.

3. Attempting to calculate without using average properties: Instead of recognizing that the average of consecutive integers equals the middle value, students often plan to add all terms and divide, making the problem unnecessarily complex and time-consuming.

Errors while executing the approach

1. Incorrect identification of middle values: For list S with 10 terms, students often pick the 5th term (a+15) as the middle instead of correctly finding the average of the 5th and 6th terms. For list T with 5 terms, they might pick the wrong middle term due to counting errors.

2. Arithmetic errors in sequence generation: When writing out the sequences, students frequently make addition errors, especially in longer sequences. For example, writing List S as {a+7, a+9, a+11, a+13, a+14...} instead of maintaining the consistent +2 pattern.

3. Variable manipulation errors: When calculating (a+16) - (a+4), students sometimes forget to distribute the negative sign correctly, getting a+16-a+4 = 20 instead of a+16-a-4 = 12.

Errors while selecting the answer

1. Confusing the direction of comparison: Students may calculate the correct difference of 12 but then select it as "how much greater T is than S" instead of "how much greater S is than T," potentially leading them to look for -12 among the choices or second-guess their work.

2. Using the constraint value as the answer: Seeing that the smallest integer in S is "7 more" than the smallest in T, students might hastily select choice (B) 7, confusing the given constraint with the requested difference between averages.

Alternate Solutions

Smart Numbers Approach

This problem can be effectively solved using smart numbers by choosing a convenient starting value for list T.

Step 1: Choose a smart number for the least integer in T

Let's set the least integer in T = 2 (choosing an even number since T consists of consecutive even integers)

Step 2: Set up both lists using our smart number

List T: Since the least integer is 2 and we have 5 consecutive even integers:
\(\mathrm{T} = \{2, 4, 6, 8, 10\}\)

List S: Since the least integer in S is 7 more than the least integer in T:
Least integer in S = 2 + 7 = 9
With 10 consecutive odd integers starting from 9:
\(\mathrm{S} = \{9, 11, 13, 15, 17, 19, 21, 23, 25, 27\}\)

Step 3: Calculate the average of each list

Average of T:
For consecutive integers, the average equals the middle value.
The middle value of {2, 4, 6, 8, 10} is 6
Average of T = 6

Average of S:
For 10 consecutive odd integers, the average is the mean of the 5th and 6th terms.
The 5th term is 17 and the 6th term is 19
Average of S = \((17 + 19)/2 = 18\)

Step 4: Find the difference

Difference = Average of S - Average of T = 18 - 6 = 12

Answer: (D) 12

Note: This smart number approach works because the relationship between the starting points of the two lists is fixed (difference of 7), and the difference in averages will be the same regardless of which specific even number we choose as our starting point for T.