A is the set of ten consecutive numbers, while the average of second lowest and the highest number of set A is 32. If B is the set of five consecutive odd number which lowest number is equal to second highest number of set A, then find the average of set B?

The correct option is 39.

Let's solve the problem step-by-step:

Step 1: Represent the numbers in Set A
Let Set A consist of ten consecutive numbers. We can represent these ten consecutive numbers as:
$x$, $x+1$, $x+2$, $x+3$, $x+4$, $x+5$, $x+6$, $x+7$, $x+8$, and $x+9$.
Here, $x$ is the lowest number of the set.

From this definition:
- The second lowest number of Set A is $x+1$.
- The highest number of Set A is $x+9$.
- The second highest number of Set A is $x+8$.

Step 2: Find the value of $x$
We are given that the average of the second lowest and the highest number of Set A is 32. We can set up the following equation:

$\frac{(x+1)+(x+9)}{2}=32$

Simplify the equation to solve for $x$:

$\frac{2x+10}{2}=32$

$x+5=32$

$x=27$

Step 3: Determine the second highest number of Set A
Using the value of $x=27$, the second highest number is:

$x+8=27+8=35$

Step 4: Analyze Set B and find its average
Set B consists of five consecutive odd numbers. The lowest number of Set B is equal to the second highest number of Set A, which is 35.
Therefore, the five consecutive odd numbers in Set B are:
35, 37, 39, 41, and 43.

Since the numbers in Set B are consecutive odd numbers (which form an arithmetic progression with a common difference of 2), the average of the set is simply the middle (3rd) term:
$\text{Average}=39$

Alternatively, we can calculate the average mathematically:

$\text{Average}=\frac{35+37+39+41+43}{5}$

$\text{Average}=\frac{195}{5}=39$