The mean of five consecutive even integers is 6 less than the mean of four consecutive odd integers. If the sum of the least odd integer and the least even integer is 43, then find the difference between the second largest odd integer and the second smallest even integer.

Correct Answer: 9

Step-by-Step Explanation:

Step 1: Express the consecutive even and odd integers algebraically.

Let the five consecutive even integers be:
$e,e+2,e+4,e+6,e+8$
where $e$ is the least (smallest) even integer.

The mean (average) of these five consecutive even integers is the middle term:
$\text{Mean of even integers}=e+4$

Let the four consecutive odd integers be:
$o,o+2,o+4,o+6$
where $o$ is the least (smallest) odd integer.

The mean of these four consecutive odd integers is:
$\text{Mean of odd integers}=\frac{o+(o+2)+(o+4)+(o+6)}{4}=\frac{4o+12}{4}=o+3$

Step 2: Formulate equations based on the problem conditions.

Condition 1: The mean of the five even integers is 6 less than the mean of the four odd integers.
$e+4=(o+3)-6$
Simplifying:
$e+4=o-3$
$o-e=7$     (Equation 1)

Condition 2: The sum of the least odd integer ($o$) and the least even integer ($e$) is 43.
$o+e=43$     (Equation 2)

Step 3: Solve for the variables $o$ and $e$.

Adding Equation 1 and Equation 2:
$(o-e)+(o+e)=7+43$
$2o=50$
$o=25$

Substituting $o=25$ into Equation 2:
$25+e=43$
$e=43-25=18$

Step 4: Identify the target values and compute the difference.

The four odd integers are 25, 27, 29, and 31.
The second largest odd integer is 29.

The five even integers are 18, 20, 22, 24, and 26.
The second smallest even integer is 20.

Calculate the difference:
$\text{Difference}=29-20=9$

Thus, the correct answer is 9.