Answer:
myInt=40
myFloat=4.8
Explanation:
First look at the function definition .It has two arguments intVal is passed by reference while floatVal is passed by value.So the changes done on the myInt variable will be reflected on the original argument because when a variable is passed by reference the the changes are reflected on the original argument but when a variable is passed by value the function created a duplicate copy of it and work on them so changes are not reflected on the original argument.So myInt will get doubled while myFloat will remain the same.
Answer:
A or B would work.
If the developer tell you how to, which depends on what you ask then A would work
But Following links could give several smaller bits of info and resources required which makes B able to work
