Q:(f) Paper of size A0 has dimensions 1189 mm x 841 mm. Each subsequent size A(n) is defined as A(n-1) cut in half parallel to its shorter sides. Thus paper of size A1 would have dimensions 841 mm x 594 mm. Write a program to calculate and print paper sizes A0, A1, A2, … A8.
#include <stdio.h>
int main()
{
double a[9][2]={{1189,841}};
int i,j;
for(i=1;i<=8;i++)
{
for(j=0;j<=0;j++)
{
if(a[i-1][j]>a[i-1][j+1])
{
a[i][j+1]=a[i-1][j]/2;
a[i][j]=a[i-1][j+1];
}
else
{
a[i][j+1]=a[i-1][j+1]/2;
a[i][j]=a[i-1][j];
}
}
}
for(i=0;i<=8;i++)
{
for(j=0;j<=1;j++)
{
printf("%lf\t",a[i][j]);
}
printf("\n");
}
return 0;
}
Q: (e) The length and breadth of a rectangle and radius of a circle are input through the keyboard. Write a program to calculate the area and perimeter of the rectangle, and the area and circumference of the circle.
#include <stdio.h>
int main()
{
double l,b,arearec,perirec,r,areacir,circumcir;
printf("Enter Length & Breadth of Rectangle\n");
scanf("%lf%lf",&l,&b);
printf("Enter Radius of Circle\n");
scanf("%lf",&r);
arearec=l*b;
perirec=2*(l+b);
areacir=3.14*r*r;
circumcir=2*3.14*r;
printf("Area of Rectangle is %lf\n",arearec);
printf("Perimeter of Rectangle is %lf\n",perirec);
printf("Area of Circle is %lf\n",areacir);
printf("Circumference of Circle is %lf\n",circumcir);
return 0;
}
Q: (d) Temperature of a city in Fahrenheit degrees is input through the keyboard. Write a program to convert this temperature into Centigrade degrees.
#include <stdio.h>
int main()
{
double f, c;
printf("Enter temperature in Fahrenheit: ");
scanf("%lf", &f);
c=(f - 32) * 5.0 / 9.0;
printf("%lf Fahrenheit is equal to %lf Celsius\n", f, c);
return 0;
}
Q: (c) If the marks obtained by a student in five different subjects are input through the keyboard, write a program to find out the aggregate marks and percentage marks obtained by the student. Assume that the maximum marks that can be obtained by a student in each subject is 100.
/*Q: (c) If the marks obtained by a student in five different subjects are input through the keyboard,
write a program to find out the aggregate marks and percentage marks obtained by the student.
Assume that the maximum marks that can be obtained by a student in each subject is 100.*/
#include<stdio.h>
#include<conio.h>
void main()
{
double m1,m2,m3,m4,m5,tot,per;
clrscr();
printf("Enter Marks in 5 Subjects out of 100 each\n");
printf("Enter Marks in Subject 1\n");
scanf("%lf",&m1);
printf("Enter Marks in Subject 2\n");
scanf("%lf",&m2);
printf("Enter Marks in Subject 3\n");
scanf("%lf",&m3);
printf("Enter Marks in Subject 4\n");
scanf("%lf",&m4);
printf("Enter Marks in Subject 5\n");
scanf("%lf",&m5);
tot=m1+m2+m3+m4+m5;
per=tot/5;
printf("Total Marks = %lf\n",tot);
printf("Percentage = %lf",per);
getch();
}
Q: (b) The distance between two cities (in km.) is input through the keyboard. Write a program to convert and print this distance in meters, feet, inches and centimeters.
/*Q: (b) The distance between two cities (in km.) is input through the keyboard.
Write a program to convert and print this distance in meters, feet, inches and centimeters.*/
#include<stdio.h>
#include<conio.h>
void main()
{
double dkm,dm,dcm,dft,din;
clrscr();
printf("Enter Distance in KM");
scanf("%lf",&dkm);
dm=dkm*1000;
dcm=dm*100;
din=dcm*0.3937;
dft=dcm*0.0328;
printf("Distance in CM = %lf\n",dcm);
printf("Distance in Inches = %lf\n",din);
printf("Distance in Feet = %lf\n",dft);
printf("Distance in Meter = %lf",dm);
getch();
}
Q: (a) Ramesh’s basic salary is input through the keyboard. His dearness
allowance is 40% of basic salary, and house rent allowance is 20% of
basic salary. Write a program to calculate his gross salary.
Solution:
/*(a) Ramesh’s basic salary is input through the keyboard. His dearness
allowance is 40% of basic salary, and house rent allowance is 20% of
basic salary. Write a program to calculate his gross salary.*/
#include<stdio.h>
#include<conio.h>
void main()
{
double b_sal,da,hra,g_sal;
clrscr();
printf("Enter basic salary");
scanf("%lf",&b_sal);
hra=0.20*b_sal;//House Rent Allowance
da=0.40*b_sal;//Dearness Allowance
g_sal=b_sal+da+hra;//Gross Salary
printf("Gross Salary = %lf",g_sal);
getch();
}
Q: 2 Let
m=(313)4 and n=(322)4 . Find the base 4 expansion of m+n.
(1) (635)4
(2)
(32312)4
(3)
(21323)4
(4) (1301)4
Answer: (4)
Explanation:
Firstly convert m & n into base 10:
(313)4
= 3*42+1*41+3*40 = (55)10
(322)4 = 3*42+2*41+2*40 = (58)10
Now m + n = 55+58 = (113)10
Now Convert
(113)10 into base 4
Write the remainders from
bottom to top.
So, (113)10 = (1301)4 which is Correct.
UGC NET COMPUTER SCIENCE AND APPLICATIONS PAPER - 2 November 2017 Solution
Q: 1 If the time is now 4 O’clock, what will be the time after 101 hours from now ?
(1) 9 O’clock (2) 8 O’clock (3) 5 O’clock (4) 4 O’clock
Answer: (1)
The task is to find all unique triplets in an array that add up to zero. A triplet consists of three numbers, and the sum of these three numbers should be zero. For example:
Input: [-1, 0, 1, 2, -1, -4]
Output: [[-1, -1, 2], [-1, 0, 1]]
Sort the Array:
First, sort the array in ascending order. Sorting helps to efficiently find the triplets using two pointers.
Fix One Element:
Iterate through the array and fix one element. For each fixed element, find two other elements (using two pointers) such that their sum equals the negative of the fixed element.
Use Two Pointers:
After fixing one element, use two pointers:
Avoid Duplicates:
Skip duplicate elements to ensure the triplets are unique.
Add Valid Triplets:
If the sum of the triplet is zero, add it to the result list.
import java.util.*;
public class ZeroSumTriplets {
public static List<List<Integer>> findTriplets(int[] nums) {
List<List<Integer>> result = new ArrayList<>();
Arrays.sort(nums); // Step 1: Sort the array
for (int i = 0; i < nums.length - 2; i++) {
if (i > 0 && nums[i] == nums[i - 1]) {
continue; // Skip duplicates for the first element
}
int left = i + 1; // Pointer 1
int right = nums.length - 1; // Pointer 2
while (left < right) {
int sum = nums[i] + nums[left] + nums[right];
if (sum == 0) {
result.add(Arrays.asList(nums[i], nums[left], nums[right]));
// Move both pointers and skip duplicates
while (left < right && nums[left] == nums[left + 1]) left++;
while (left < right && nums[right] == nums[right - 1]) right--;
left++;
right--;
} else if (sum < 0) {
left++; // Move the left pointer to increase the sum
} else {
right--; // Move the right pointer to decrease the sum
}
}
}
return result;
}
public static void main(String[] args) {
int[] nums = {-1, 0, 1, 2, -1, -4};
List<List<Integer>> triplets = findTriplets(nums);
System.out.println("Triplets with zero sum: " + triplets);
}
}
This solution is efficient and easy to implement for finding triplets with zero sum.
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