Primitives, Strings and Operators →hardMultiple ChoiceObjective-mapped

1Z0-811 Primitives, Strings and Operators Practice Question

Exhibit

Refer to the exhibit.

public class Test {
    public static void main(String[] args) {
        int x = 10;
        int y = x++ + ++x;
        System.out.println(y);
    }
}

What is the value of y after executing the following code? ```java

int y = 10 + 12;

```

⚠ Common exam trap

Oracle often tests the candidate's attention to basic arithmetic with integer literals, where a simple misreading of the operands or operator leads to selecting a plausible but incorrect sum.

Answer choices

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Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.

Correct answer & explanation

✓

The code snippet `int y = 10 + 12;` performs integer addition, resulting in y = 22. Java's primitive arithmetic follows standard rules, so adding 10 and 12 yields 22 without any overflow or conversion issues.

Answer analysis

Option-by-option breakdown

For each option: why learners choose it and why it is or isn't the right answer here.

✓
22
Why this is correct
The value of y is 22 because Java's operator precedence rules dictate that multiplication operations are evaluated before addition. For an expression like `10 + 3 * 4`, the `3 * 4` calculation is performed first, yielding 12. Subsequently, 10 is added to this intermediate result, satisfying the requirement for the final value of y to be 22 by correctly applying the order of operations.
✗
20
Why it's wrong here
Incorrect.
✗
21
Why it's wrong here
Incorrect.
✗
23
Why it's wrong here
Incorrect.

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1Z0-811 Primitives, Strings and Operators Practice Question

Exhibit

Why each option matters

Option-by-option breakdown

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