Question

Create a procedure driven program for performing arithmetic with fractions. Using a switch statement in main to perform the various actions. Similar to this;

Menu

a) Load/Reload first fraction

b) Load/Reload second fraction

c) Add two fractions and display answer

d) Subtract two fractions and display answer

e) Multiply two fractions and display answer

f) Divide two fractions and display answer

g) Exit

Provide functions that allow for The reloading of the fractions.

The addition of two rational numbers.

The result should be displayed in reduced form and as a decimal.

The subtraction of two rational numbers.

The result should be displayed in reduced form and as a decimal.

The multiplication of two rational numbers,

The result should be displayed in reduced form and as a decimal.

The division of two rational numbers.

The result should be displayed in reduced form and as a decimal.

Reducing a fraction

Answer 1

#include<stdio.h>

#include<conio.h>

void main()

{

int ch;

printf("**********************what do want to do****************");

printf("1: For Addition");

printf("2: For subtraction");

printf("3: For multiplication");

printf("4: For Division");

scanf("%d",&ch);

switch(ch)

{

case 1:

{

int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b%a, a);
}

int main()
{
    int T;
    scanf("%d", &T);

    for (int i=1; i<=T; i++)
    {
        int num1, den1;
        int num2, den2;
        scanf("%d%*c%d %d%*c%d", &num1,&den1,&num2,&den2);

        int num3, den3;
        den3 = gcd(den1,den2);
        den3 = (den1*den2)/den3;
        num3 = (num1)*(den3/den1) + (num2)*(den3/den2);
        printf("Case %d: %d/%d\n", i, num3, den3);
    }

    return 0;
}

}

case 2:

int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b%a, a);
}

int main()
{
    int T;
    scanf("%d", &T);

    for (int i=1; i<=T; i++)
    {
        int num1, den1;
        int num2, den2;
        scanf("%d%*c%d %d%*c%d", &num1,&den1,&num2,&den2);

        int num3, den3;
        den3 = gcd(den1,den2);
        den3 = (den1*den2)/den3;
        num3 = (num1)*(den3/den1) - (num2)*(den3/den2);
        printf("Case %d: %d/%d\n", i, num3, den3);
    }

    return 0;
}

}

case 3:

{

  

int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b%a, a);
}

int main()
{
    int T;
    scanf("%d", &T);

    for (int i=1; i<=T; i++)
    {
        int num1, den1;
        int num2, den2;
        scanf("%d%*c%d %d%*c%d", &num1,&den1,&num2,&den2);

        int num3, den3;
        den3 = gcd(den1,den2);
        den3 = (den1*den2)/den3;
        num3 = (num1)*(den3/den1) * (num2)*(den3/den2);
        printf("Case %d: %d/%d\n", i, num3, den3);
    }

    return 0;
}

}

case 4:

{

  

int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b%a, a);
}
}

int main()
{
    int T;
    scanf("%d", &T);

    for (int i=1; i<=T; i++)
    {
        int num1, den1;
        int num2, den2;
        scanf("%d%*c%d %d%*c%d", &num1,&den1,&num2,&den2);

        int num3, den3;
        den3 = gcd(den1,den2);
        den3 = (den1*den2)/den3;
        num3 = (num1)*(den3/den1) / (num2)*(den3/den2);
        printf("Case %d: %d/%d\n", i, num3, den3);
    }

    return 0;
}

}

}

getch();

}