# Sum of two numbers if the original ratio and new ratio obtained by adding a given number to each number is given

Given a ratio a : b of two unknown numbers. When both the numbers are incremented by a given integer x, the ratio becomes c : d. The task is to find the sum of the two numbers.

Examples:

Input: a = 2, b = 3, c = 8, d = 9, x = 6
Output: 5
Original numbers are 2 and 3
Original ratio = 2:3
After adding 6, ratio becomes 8:9
2 + 3 = 5

Input: a = 1, b = 2, c = 9, d = 13, x = 5
Output: 12

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Let the sum of the numbers be S. Then, the numbers can be (a * S)/(a + b) and (b * S)/(a + b).

Now, as given:

```(((a * S) / (a + b)) + x) / (((b * S) / (a + b)) + x) = c / d
or ((a * S) + x * (a + b)) / ((b * S) + x * (a + b)) = c / d
So, S = (x * (a + b) * (c - d)) / (ad - bc)
```

Below is the implementation of the above approach:

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the sum of numbers  ` `// which are in the ration a:b and after  ` `// adding x to both the numbers  ` `// the new ratio becomes c:d ` `double` `sum(``double` `a, ``double` `b, ``double` `c, ``double` `d, ``double` `x) ` `{ ` `    ``double` `ans = (x * (a + b) * (c - d)) / ((a * d) - (b * c)); ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``double` `a = 1, b = 2, c = 9, d = 13, x = 5; ` `     `  `    ``cout << sum(a, b, c, d, x); ` `     `  `    ``return` `0; ` `} `

 `// Java implementation of the above approach ` `class` `GFG ` `{ ` `     `  `    ``// Function to return the sum of numbers  ` `    ``// which are in the ration a:b and after  ` `    ``// adding x to both the numbers  ` `    ``// the new ratio becomes c:d ` `    ``static` `double` `sum(``double` `a, ``double` `b,  ` `                      ``double` `c, ``double` `d,  ` `                      ``double` `x) ` `    ``{ ` `        ``double` `ans = (x * (a + b) * (c - d)) /  ` `                         ``((a * d) - (b * c)); ` `        ``return` `ans; ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `main (String[] args) ` `    ``{ ` `        ``double` `a = ``1``, b = ``2``, c = ``9``, d = ``13``, x = ``5``; ` `         `  `        ``System.out.println(sum(a, b, c, d, x)); ` `    ``} ` `} ` ` `  `// This code is contributed by ihritik `

 `# Python3 implementation of the approach ` ` `  `# Function to return the sum of numbers  ` `# which are in the ration a:b and after  ` `# adding x to both the numbers  ` `# the new ratio becomes c:d ` `def` `sum``(a, b, c, d, x): ` ` `  `    ``ans ``=` `((x ``*` `(a ``+` `b) ``*` `(c ``-` `d)) ``/`  `               ``((a ``*` `d) ``-` `(b ``*` `c))); ` `    ``return` `ans; ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` ` `  `    ``a, b, c, d, x ``=` `1``, ``2``, ``9``, ``13``, ``5``; ` `     `  `    ``print``(``sum``(a, b, c, d, x)); ` `     `  `# This code is contributed by PrinciRaj1992  `

 `// C# implementation of the above approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `    ``// Function to return the sum of numbers  ` `    ``// which are in the ration a:b and after  ` `    ``// adding x to both the numbers  ` `    ``// the new ratio becomes c:d ` `    ``static` `double` `sum(``double` `a, ``double` `b,  ` `                      ``double` `c, ``double` `d,  ` `                      ``double` `x) ` `    ``{ ` `        ``double` `ans = (x * (a + b) * (c - d)) /  ` `                         ``((a * d) - (b * c)); ` `        ``return` `ans; ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `Main () ` `    ``{ ` `        ``double` `a = 1, b = 2,  ` `               ``c = 9, d = 13, x = 5; ` `         `  `        ``Console.WriteLine(sum(a, b, c, d, x)); ` `    ``} ` `} ` ` `  `// This code is contributed by ihritik `

Output:
```12
```

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Improved By : ihritik, princiraj1992

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