1Z0-811 Java Basics and Syntax Practice Question
A company requires a method that accepts an integer and returns true if the integer is even, otherwise false. Which implementation best follows Java conventions?
⚠ Common exam trap
Oracle often tests the distinction between returning a value versus printing it, and the requirement that a method returning a boolean must have a `boolean` return type, not `int` or `void`; the trap here is that candidates may choose Option C because it 'works' syntactically, overlooking the conventional naming and unnecessarily verbose code.
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
✓
public boolean isEven(int num) { return num % 2 == 0; }
It defines a method with the appropriate return type `boolean`, uses a clear and conventional name `isEven`, and returns the result of the expression `num % 2 == 0` directly. This follows Java naming conventions (camelCase with a verb for boolean methods) and leverages the fact that `%` yields the remainder, which is compared to zero to produce a boolean result.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✗
public void isEven(int num) { System.out.println(num%2==0); }
Why it's wrong here
Prints result but doesn't return it.
- ✗
public int isEven(int num) { return num % 2; }
Why it's wrong here
Returns int, not boolean.
- ✗
public boolean evenCheck(int num) { if(num%2==0) return true; else return false; }
Why it's wrong here
Redundant if-else; use direct return.
- ✓
public boolean isEven(int num) { return num % 2 == 0; }
Why this is correct
Correct: boolean return, descriptive name, straightforward logic.
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