Questions

Answer D is correct because a sunk cost is a project expense that will eventualy need to be paid. Answer A is incorrect because a fixed cost remains constant, regardless of any change in a company's activity. Answer B is incorrect because direct cost can be directly linked to producing specific goods or services. Answer C is incorrect because a variable cost changes in proportion to a change in the company's activity or business.

Okhla Bird Sanctuary is located in Uttar Pradesh.

Okhla Bird Sanctuary is a bird sanctuary at the Okhla barrage over Yamuna River. It is situated in Noida, Gautam Buddh Nagar district, on Delhi-Uttar Pradesh state border and known as a haven for over 300 bird species.

IUPAC stands for International Union of Pure and Applied Chemistry.

A chemical nomenclature is a set of rules to generate systematic names for chemical compounds. The nomenclature used most frequently worldwide is the one created and developed by the International Union of Pure and Applied Chemistry (IUPAC).

#include "string.h"

#include "stdlib.h"

struct stud

{

       int rollno;

       int marks;

       char name[30];

};

int sort_m (struct stud *, struct stud *);

int sort_name (struct stud *, struct stud *);

int sort_marks (struct stud *, struct stud *);

main()

{

static struct stud ss[] = {

                                            { 15, 96, "Akshay" },

                                            { 2, 97, "Madhuri" },

                                            { 8, 85, "Aishvarya" },

                                            { 10, 80, "Sushmita" }

                                   };

int x,w;

clrscr();

w = sizeof (struct stud);

printf ('\nIn order of roll numbers:");

qsort (ss, 4, w, sort_rn);

for(x=0; x<4;x++)

     printf ("\n%d%s%d", ss[x].rollno, ss[x].name,ss[x].marks);

printf("\n\nIn order of names:");

qsort(ss, 4, sort_name);

for (x=0; x<4;x++)

      printf("\n%d%s%d",ss[x].rollno, ss[x].name,ss[x].marks);

printf("\n\nIn order of marks:");

qsort(ss,4,w,sort_marks);

for (x=0;x<4;x++)

      printf ("\n%d%s%d",ss[x].rollno,ss[x].name,ss[x].marks);

}

int sort_rn (struct stud *t1, struct stud *t2)

{

     return (t1->rollno-t2->rollno);

}

int sort_name (struct stud *t1, struct stud *t2)

{

     return (strcmp(t1->name,t2->name));

}

int sort_marks (struct stud *t1, struct stud *t2)

{

     return (t2->marks-t1->marks);

}