Correct Answer: A

Explaination:

Choice A is the correct answer.

The code does not compile because the constructor of class A is declared as private. This creates a problem when the subclass constructor makes an implicit super() call to the parent class constructor at the time B is instantiated.

Since the code does not compile, all the other choices are incorrect. If the constructor of A had not been private, the output would have been 7.

Correct Answer: B & C

Explaination:

Choices B and C are the correct answers.

The List interface is implemented by the LinkedList class. So List<Number> is the super type of LinkedList<Number>, which accepts only objects of type Number. LinkedList<Integer> declares a list that accepts only Integer objects. Assigning a LinkedList raw type to it gives warnings of unsafe operations, however, it does compile and run without errors.

Choice A is incorrect because generic parameters cannot be primitives.

Choice D is incorrect because the right hand side of the assignment uses the wrong syntax.

Reference:
http://java.sun.com/docs/books/tutorial/extra/generics/index.html