What is the output of following Python Code?

@PythonPr 1num = 75 if num <= 50: False → skip (and pass does nothing) elif num <= 75: True (75 ≤ 75) → execute print("1") The next elif is not checked because the previous condition was true.

@PythonPr In IF statements, the first True condition will execute, so 1 will output.

@PythonPr Answer: 1 75 satisfies two conditions here: › num <= 75 › num <= 80 But only "1" prints. elif always stops at the first match. The second branch never even gets checked. For beginners: one if-elif-else chain = one outcome.

@PythonPr If you think the answer is: 1 2 Sorry, you are wrong. Once the first elif satisfies the condition, it won't run the second one again

@PythonPr Output is 1 whenever the condition becomes true prints after that no checking happens.

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@PythonPr Answer: 1 Here's the logic; Variable Assignment: The variable num is assigned the integer value 75.First Condition (if num <= 50): Evaluates to False because 75 is greater than 50. The pass statement is skipped.Second Condition (elif num <= 75): Evaluates to True because 75 is

@PythonPr It prints "1" (elif chain stops at first true condition).

@PythonPr 1

@PythonPr 1

@PythonPr a=b=257;print("yes"if a is b else f"no → {hex(id(a))}≠{hex(id(b))}",[hex(x)for x in __import__("ctypes").string_at(id(a)-0x10,0x30)])

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@PythonPr 12

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@PythonPr 1

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@PythonPr Output: 1 ... .@PythonPr

@PythonPr 1

@PythonPr 1

@PythonPr 1

@PythonPr 1

@PythonPr 1.