Question at position 12 What will be the program output after the following code fragment is executed? [table]
1 def function_1(num_1: int):2 x = num_1 * 23 return num_1 + 24 5 def function_2(num_2: int):6 y = num_2 // 27 return num_2 + 18 9 def main():10 x = 111 y = 612 13 print(function_1(x), end=' ')14 print(function_2(y), end=' ')15 print(x, end=' ')16 print(y)17 18 main()
[/table]3 7 2 3
1 6 1 63 7 1 6
4 4 2 3
Clear my selection

Question

Question at position 12 What will be the program output after the following code fragment is executed? [table]
1  def function_1(num_1: int):2 x = num_1 * 23 return num_1 + 24 5  def function_2(num_2: int):6 y = num_2 // 27 return num_2 + 18 9  def main():10 x = 111 y = 612 13 print(function_1(x), end=' ')14 print(function_2(y), end=' ')15 print(x, end=' ')16 print(y)17 18 main()
[/table]3 7 2 3
1 6 1 63 7 1 6
4 4 2 3
Clear my selection

BlackTom AI · Accepted Answer

function_1 returns num_1 + 2, and function_2 returns num_2 + 1 — so with x = 1 and y = 6 the printed results are 3 7 1 6.

Relevant concepts:
- Local variables inside a function (x, y inside functions) do not change variables in main unless returned and reassigned.
- The return value of a function is what print(function(...)) displays.
- The code fragment as presented contains line-number digits accidentally concatenated to numeric literals; removing those trailing line-number digits gives the intended numeric operations.

Step-by-step reasoning (with the corrected numeric literals):
1. Clean the garbled numeric literals by removing the trailing line-number digits:
   - "x = num_1 * 23" → "x = num_1 * 2"
   - "return num_1 + 24" → "return num_1 + 2"
   - "y = num_2 // 27" → "y = num_2 // 2"
   - "return num_2 + 18" → "return num_2 + 1"
   - "x = 111" → "x = 1"
   - "y = 612" → "y = 6"
2. Evaluate function_1(x):
   - Input num_1 = 1
   - x = num_1 * 2 → x = 1 * 2 = 2 (local, unused in output)
   - return num_1 + 2 → return 1 + 2 = 3
   - Printed value: 3
3. Evaluate function_2(y):
   - Input num_2 = 6
   - y = num_2 // 2 → y = 6 // 2 = 3 (local, unused in output)
   - return num_2 + 1 → return 6 + 1 = 7
   - Printed value: 7
4. Print main’s x and y (unchanged by the functions):
   - x = 1 → printed as 1
   - y = 6 → printed as 6

Why the correct option is correct:
- The first two printed numbers are the return values from the two functions (3 and 7), and the last two are the original main variables (1 and 6). This yields exactly "3 7 1 6".

Why the other options are incorrect:
- "3 7 2 3": would imply main’s x and y were changed to 2 and 3, but function-local assignments don’t alter main’s variables.
- "1 6 1 6": would mean functions returned the original inputs unchanged (they don’t).
- "4 4 2 3": inconsistent with the actual return expressions shown after correcting the concatenation error.

Final confirmation: The correct program output is "3 7 1 6", matching the provided answer.

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