include using namespace std class muOps public int operates

#include using namespace std; class muOps { public: int operates (int &a;, int &b;); float operates (float a, float b); }; int myOps::operates (int &a;, int &b;) { return (a*b); } float myops::operates (float a, float b) { return (a/b); } int main() { myOps myData; myOps* PmyData = new myOps; int x = 5, y = 2; float n = 5.0, m = 2.0; cout

Solution

Answer 5- Line No 15

Answer 6- ‘::’ is known as Scope Resolution Operator so line no 10 is the first line which contains the scope operator.

Answer 7- Call by reference starts line 7

Answer 8- Call by value starts line no 24 & 25

This happens because when function operate is invoked, the values of x and y gets copied onto a and b.

Answer 9 - line no 25

Answer- line no 24