Java checked exception propagation: at Point X on line 5, what must be added so test() compiles when runTest() declares throws TestException? public class ExceptionTest {\n class TestException extends Exception {}\n public void runTest() throws TestException {}\n public void test() /* Point X */ {\n runTest();\n }\n}

Introduction / Context:
This question focuses on Java's checked exception rules. A method that calls another method which declares a checked exception must either handle it with a try/catch or declare it in its own throws clause with the same type or a supertype.

Given Data / Assumptions:

• runTest() declares throws TestException, which is a checked exception.
• test() calls runTest() directly.
• No try/catch surrounds the call in test().

Concept / Approach:
For checked exceptions, the compiler enforces one of two options: catch the exception locally, or declare it. Declaring throws Exception on test() is legal because Exception is a supertype of TestException. A try/catch could work too, but the prompt asks what to add “at Point X” in the method signature, not inside the body.

Step-by-Step Solution:
Identify the thrown type: TestException extends Exception.To compile, test() must either catch or declare.Adding throws Exception to the signature satisfies the rule because it covers TestException.Therefore, the correct addition is a throws clause with Exception (or TestException).

Verification / Alternative check:
Replace with throws TestException and recompile; it also succeeds. Wrapping in try { runTest(); } catch(TestException e){} would also compile.

Why Other Options Are Wrong:
Option A: Compiler error persists without handling/declaring.
Option C: A catch cannot be added at the signature point; it belongs inside the method body.
Option D: Declaring RuntimeException does not satisfy checked-exception requirements.
Option E: final is unrelated.

Common Pitfalls:
Thinking any throws clause will do; it must be the same checked type or a supertype.

Final Answer:
throws Exception