Question at position 14 What will be the output of the program after the following code fragment is executed? [table]
1 def function_1(num_1: int):2 x = num_1 * 23 return num_1 + 2 4 5 def function_2(num_2: int):6 y = num_2 // 27 return num_2 + 1 8 9 def main():10 x = 311 y = 412 13 print(function_1(x))14 print(function_2(y))15 print(x, y)16 17 main()
[/table] 833 4556 2555 5553 4Clear my selection

Question

Question at position 14 What will be the output of the program after the following code fragment is executed? [table]
1  def function_1(num_1: int):2 x = num_1 * 23 return num_1 + 2 4  5  def function_2(num_2: int):6 y = num_2 // 27 return num_2 + 1 8  9  def main():10 x = 311 y = 412 13 print(function_1(x))14 print(function_2(y))15 print(x, y)16 17 main()
[/table] 833 4556 2555 5553 4Clear my selection

BlackTom AI · Accepted Answer

function_1 returns num_1 + 2, so calling it with 3 yields 3 + 2 = 5. function_2 returns num_2 + 1, so calling it with 4 yields 4 + 1 = 5. The x and y assigned inside main remain 3 and 4 because assignments to local variables inside functions do not change main’s variables.

1) Relevant concepts
- Function return values: print(function_1(x)) prints the value returned by function_1, not any internal variable inside the function unless returned.
- Variable scope: variables defined inside a function are local to that function; they do not modify variables in the caller unless explicitly returned and assigned.

2) Step-by-step execution
- main sets x = 3 and y = 4.
- Call print(function_1(x)):
  - function_1 receives num_1 = 3.
  - It computes and returns num_1 + 2 = 3 + 2 = 5.
  - print prints 5.
- Call print(function_2(y)):
  - function_2 receives num_2 = 4.
  - It computes and returns num_2 + 1 = 4 + 1 = 5.
  - print prints 5.
- Call print(x, y):
  - main’s x and y are unchanged (3 and 4), so print prints "3 4".

3) Intermediate calculations shown
- function_1: 3 + 2 = 5
- function_2: 4 + 1 = 5
- main variables: x = 3, y = 4

4) Why the correct answer is correct
Because both functions return simple expressions of their arguments and do not modify main’s variables, the printed outputs are the two return values (5 and 5) followed by the original main variables (3 4).

5) Why other options are incorrect
- "8
3
3 4": first value 8 would require function_1 to produce 8 for input 3, which it does not.
- "5
5
6 2": final pair 6 2 contradicts main’s assigned x and y (3 and 4).
- "5
5
5 5": would imply main’s x and y were changed to 5, which never happens because functions only return values and do not modify main’s locals.

Conclusion: The program prints
5
5
3 4
which matches the provided correct answer.

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