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
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++
// C++ implementation of the approach#include <bits/stdc++.h>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:ddouble 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 codeint main(){ double a = 1, b = 2, c = 9, d = 13, x = 5; cout << sum(a, b, c, d, x); return 0;} |
Java
// Java implementation of the above approachclass 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
# 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:ddef sum(a, b, c, d, x): ans = ((x * (a + b) * (c - d)) / ((a * d) - (b * c))); return ans;# Driver codeif __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#
// C# implementation of the above approachusing 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 |
Javascript
<script>// javascript implementation of the above 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:dfunction sum(a , b, c , d, x){ var ans = (x * (a + b) * (c - d)) / ((a * d) - (b * c)); return ans;}// Driver codevar a = 1, b = 2, c = 9, d = 13, x = 5;document.write(sum(a, b, c, d, x));// This code is contributed by 29AjayKumar </script> |
Output:
12
Time Complexity: O(1)
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
