WGU C960 – Discrete Mathematics II Objective Assessment Review NEWEST EXAM COMPLETE 50 QUESTIONS AND CORRECT DETAILED ANSWERS (VERIFIED ANSWERS) ALREADY GRADED A+ BRAND NEW!!

WGU C960 – Discrete Mathematics II Objective
Assessment Review NEWEST EXAM 2026-2027
COMPLETE 50 QUESTIONS AND CORRECT
DETAILED ANSWERS (VERIFIED ANSWERS)
ALREADY GRADED A+ BRAND NEW!!
Pre-Assessment: Algorithms - Algorithm Structures:

Question 1:
Consider the following pseudocode that merges two lists of numbers into one:

Merge0(List1, List2)

Set OUTlist to empty

While List1 is not empty OR List2 is not empty

If one list is empty and the other is not,

Remove the first number from the non-empty list and add it to OUTlist

If both lists are non-empty,
Remove the first number from List1 and add it to OUTlist

Remove the first number from List2 and add it to OUTlist

Return OUTlist



If ListA is [1, 3, 5] and ListB is [2, 4, 6] then what is the result of Merge0 (ListA, Merge0 (ListB,
ListA))? - CORRECT ANSWER: [ 1, 2, 3, 1, 5, 4, 3, 6, 5 ]

Step 1:

To solve this, we need to understand how the Merge0 function works. It merges two lists by
taking the first element of each list and adding it to the output list in ascending order. If one list is
empty and the other is not, it removes the first element from the non-empty list and adds it to the
output list.

Using the given example, we have:
ListA = [1, 3, 5]

,ListB = [2, 4, 6]

Now, let's analyze the innermost function Merge0(ListB, ListA):

1. OUTlist = empty

2. OUTlist = [2]
3. OUTlist = [2, 1]

4. OUTlist = [2, 1, 4]

5. OUTlist = [2, 1, 4, 3]

6. OUTlist = [2, 1, 4, 3, 6]

7. OUTlist = [2, 1, 4, 3, 6, 5]

8. ListA is now empty, so we return OUTlist = [2, 1, 4, 3, 6, 5]

Step 2:

Now, let's use the result of Merge0(ListB, ListA) as the second argument for the outer Merge0
function. So we have:

List1 = ListA = [1, 3, 5]
List2 = Merge0(ListB, ListA) = [2, 1, 4, 3, 6, 5]

Using the Merge0 function with these lists, we have:

1. OUTlist = empty

2. OUTlist = [1]

3. OUTlist = [1, 2]

4. OUTlist = [1, 2, 3]

5. OUTlist = [1, 2, 3, 1]
6. OUTlist = [1, 2, 3, 1, 5]

7. OUTlist = [1, 2, 3, 1, 5, 4]

8. OUTlist = [1, 2, 3, 1, 5, 4, 3]

9. OUTlist = [1, 2, 3, 1, 5, 4, 3, 6]

10. OUTlist = [1, 2, 3, 1, 5, 4, 3, 6, 5]
11. return OUTlist = [1, 2, 3, 1, 5, 4, 3, 6, 5]

,Therefore, the result of Merge0(ListA, Merge0(ListB, ListA)) is [1, 2, 3, 1, 5, 4, 3, 6, 5].



Pre-Assessment: Algorithms - Algorithm Structures:

Question 2:


Given this pseudocode:

S = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}

x=2

While(x<11):

For i in S:

If 0 ≡ i mod x and i ≠ x:

delete i from S
end-If

end-For

x=x+1

end-While



What is S at the end of this code? - CORRECT ANSWER: {2, 3, 5, 7, 11, 13, 17, 19}


This code is nothing but finding the prime numbers from the given set.

Start from x = 2

1) When x = 2, we have to delete that elements from S which satisfy 0 == i mod 2 and i ≠ 2 i.e
multiples of 2 except 2. So we remove {4, 6, 8, 10, 12, 14, 16, 18, 20}

x = x+1

2) When x = 3, we have to delete that elements from S which satisfy 0 == i mod 3 and i ≠ 3 i.e
multiples of 3 except 3. So we remove {6, 9, 12, 15, 18}

x = x+1

, 3) When x = 4, we have to delete that elements from S which satisfy 0 == i mod 4 and i ≠ 4 i.e
multiples of 4 except 4. So we remove {8, 12, 16, 20}

x = x+1

4) When x = 5, we have to delete that elements from S which satisfy 0 == i mod 5 and i ≠ 5 i.e
multiples of 5 except 5. So we remove {10, 15, 20}

x = x+1

5) When x = 6, we have to delete that elements from S which satisfy 0 == i mod 6 and i ≠ 6 i.e
multiples of 6 except 6. So we remove {12, 18}

x = x+1

6) When x = 7, we have to delete that elements from S which satisfy 0 == i mod 7 and i ≠ 7 i.e
multiples of 7 except 7. So we remove {14}

x = x+1

7) When x = 8, we have to delete that elements from S which satisfy 0 == i mod 8 and i ≠ 8 i.e
multiples of 8 except 8. So we remove {16}

x = x+1

8) When x = 9, we have to delete that elements from S which satisfy 0 == i mod 9 and i ≠ 9 i.e
multiples of 9 except 9. So we remove {18}

x = x+1

9) When x = 10, we have to delete that elements from S which satisfy 0 == i mod 10 and i ≠ 10
i.e multiples of 10 except 10. So we remove {20}

x = x+1
Now x = 11, we break the while loop and the program is terminated.

So, after deleting the above elements, we are left with

{2, 3, 5, 7, 11, 13, 17, 19}



Pre-Assessment: Algorithms - Algorithm Structures:

Question 3:


Given the pseudocode fragment: