PCEP Practice Question: Data Types, Variables, Basic I/O and Operators
A program contains this code: print(1 and 2 or 3). What is the output?
⚠ Common exam trap
Python Institute often tests the misconception that `and`/`or` always return `True` or `False`, leading candidates to pick `True` (option C) instead of the actual operand value `2`.
Answer choices
Why each option matters
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
✓
2
In Python, the `and` and `or` operators short-circuit and return the last evaluated operand, not a boolean. `1 and 2` evaluates to `2` because both are truthy and `and` returns the last truthy value. Then `2 or 3` short-circuits because `2` is truthy, so `or` returns `2` without evaluating `3`. Thus, the output is `2`.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✗
3
Why it's wrong here
Python's `and`/`or` return operands, not Booleans: `1 and 2` yields 2, and `2 or 3` short-circuits to 2, so the output is 2. Choosing 3 assumes `or` evaluates its right operand, which only happens when the left is falsy — tempting because `or` is often misread as always returning the final value.
- ✗
1
Why it's wrong here
and returns its second operand when the first is truthy, so 1 and 2 evaluates to 2, and 2 or 3 then returns 2. It is tempting because 1 is truthy and appears first, but short-circuit evaluation only returns the first operand when it is falsy.
- ✗
True
Why it's wrong here
Python's and/or return operands, not booleans, so 1 and 2 yields 2, then 2 or 3 yields 2; True never appears. It is tempting because and/or are logical operators, and True would be correct if the expression were wrapped in bool() or compared with a boolean context.
- ✓
2
Why this is correct
Python's `and` binds tighter than `or`, so `1 and 2` evaluates first. Since 1 is truthy, `and` returns its second operand, 2. That truthy result short-circuits the `or`, so 3 is never evaluated, satisfying the stem's precedence and short-circuit constraints.
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