AP CSA Unit 2 Day 1: Boolean Algebra Equivalence
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Unit 2, Selection & Iteration • Cycle 2
Day 1 Advanced Practice • Harder Difficulty
Focus: De Morgan's Law
Hard
Boolean Algebra (Equivalence)
Advanced Practice Question
Format: Boolean Algebra (Equivalence)
Which expression is logically equivalent to `!(a < 5 || b == 0)`?
Which expression is logically equivalent to `!(a < 5 || b == 0)`?
Difficulty: Hard |
Topic: Boolean Algebra (Equivalence) |
Cycle: 2 (Advanced)
Why This Answer?
Apply De Morgan's Law: `!(P || Q)` becomes `!P && !Q`. Here, `!(a < 5)` is `a >= 5`, and `!(b == 0)` is `b != 0`.
Common Mistake
Watch Out!
Switching `||` to `&&` but forgetting to negate BOTH parts (or flipping the inequality the wrong way).
AP Exam Strategy
When you negate a compound condition, negate each part AND flip `||` ↔ `&&`.
Master This Topic
This Cycle 2 HARD question tests de morgan's law. Review Unit 2 concepts to build mastery of selection and iteration.
- Understanding de morgan's law
- Tracing code execution accurately
- Avoiding common pitfalls
Ready for More Challenges?
Cycle 2 questions prepare you for the hardest AP CSA exam questions.
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