Question

TCS nqt programming questions​

Answer 1

Answer:

#include<stdio.h>

#include<conio.h>

int main()

{

int n,p,i,t=0,t1=0,p2;

int a[10];

n=5;

for(i=0;i<n;i++)

scanf("%d",&a[i]);

p=0;

p2=0;

for(i=0;i<n;i++)

{

if((a[i]>60)&&(p<=5))

{

p++;

t=p*15;

t=t;

}

else

p2++;

t1=p2*15;

t1=t1;

}

printf("%d",p);

printf("the total time(in min)=%d",(t+t1));

getch();

}

Explanation:

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