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Program to add two fractions and display their sum in reduced form

A simple program to add two fractions and display their sum in the simplest form is discussed here. Let us assume the fractions to be a/b and c/d. Now we need to add these and find out the resultant fraction.

Examples Test cases:

Input: 1 2 3 2

Output: 2 1

Input: 1 3 3 9

Output: 2 3

Input: 1 5 3 15

Output: 2 5

1) Calculate the numerator and denominator of the resultant fraction (x/y).

2) Find the gcd of x and y.

3) Divide x and y by their gcd to get the resultant fraction in the simplest form.

C

#include

int main()

{

int a, b,c,d,x,y,i,gcd;

printf(“\nEnter the numerator for 1st number : “);

scanf(“%d”,&a);

printf(“\nEnter the denominator for 1st number : “);

scanf(“%d”,&b);

printf(“\nEnter the numerator for 2nd number : “);

scanf(“%d”,&c);

printf(“\nEnter the denominator for 2nd number : “);

scanf(“%d”,&d);

x=(a*d)+(b*c); //numerator

y=b*d; //denominator

// Trick part. Reduce it to the simplest form by using gcd.

for(i=1; i <= x && i <= y; ++i)

{

if(x%i==0 && y%i==0)

gcd = i;

}

printf(“\nThe added fraction is %d/%d “,x/gcd,y/gcd);

printf(“\n”);

return 0;

}

C++

#include

using namespace std;

int main()

{

int a, b,c,d,x,y,i,gcd;

cout << “\nEnter the numerator for 1st number : “;

cin >> a;

cout << “\nEnter the denominator for 1st number : “;

cin >> b;

cout << “\nEnter the numerator for 2nd number : “;

cin >> c;

cout << “\nEnter the denominator for 2nd number : “;

cin >> d;

x=(a*d)+(b*c); //numerator

y=b*d; //denominator

// Trick part. Reduce it to the simplest form by using gcd.

for(i=1; i <= x && i <= y; ++i)

{

if(x%i==0 && y%i==0)

gcd = i;

}

cout << “\nThe added fraction is “<

cout << endl>

return 0;

}

JAVA

import java.util.*;

public class Main

{

public static void main(String[] args) {

int a, b,c,d,x,y,i,gcd = 0;

Scanner sc = new Scanner(System.in);

System.out.println(“\nEnter the numerator for 1st number : “);

a = sc.nextInt();

System.out.println(“\nEnter the denominator for 1st number : “);

b = sc.nextInt();

System.out.println(“\nEnter the numerator for 2nd number : “);

c = sc.nextInt();

System.out.println(“\nEnter the denominator for 2nd number : “);

d = sc.nextInt();

x=(a*d)+(b*c); //numerator

y=b*d; //denominator

// Trick part. Reduce it to the simplest form by using gcd.

for(i=1; i <= x && i <= y; ++i)

{

if(x%i==0 && y%i==0)

gcd = i;

}

System.out.println(“\nThe added fraction is “+ x/gcd + “/” + y/gcd);

System.out.println();

}

}

PYTHON 3

a = int(input(“Enter the numerator for 1st number : “))

b = int(input(“Enter the denominator for 1st number : “))

c = int(input(“Enter the numerator for 2nd number : “))

d = int(input(“Enter the denominator for 2nd number : “))

x=(a*d)+(b*c) #numerator

y=b*d #denominator

# Trick part. Reduce it to the simplest form by using gcd.

i = 1

while(i <= x and i <= y):

if(x % i == 0 and y % i == 0):

gcd = i

i = i + 1

print(“The added fraction is”, end = ‘ ‘)

print(int(x/gcd),end=”)

print(“/”,end=”)

print(int(y/gcd))

Output

Input- Enter the numerator for 1st number :1 Enter the denominator for 1st number:2 Enter the numerator for 2nd number :3 Enter the denominator for 2nd number:2 Output- The added fraction is 2 1