Given that person A takes a days to do a certain piece of work while person B takes b days to do the same work. If A and B started the work together and n days before the completion of work, A leaves the work. Find the total number of days taken to complete work.
Examples:
Input: a = 10, b = 20, n = 5
Output: 10Input: a = 5, b = 15, n = 3
Output: 6
Approach: Let D be the total number of days to complete the work. A and B work together for D – n days. So,
(D – n) * (1/a + 1/b) + (n) * (1/b) = 1
D * (1/a + 1/b) – (n/a) – (n/b) + (n/b) = 1
D * (1/a + 1/b) = (n + a) / a
D = b * (n + a) / (a + b)
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 // number of days required int numberOfDays(int a, int b, int n) { int Days = b * (n + a) / (a + b); return Days; } // Driver code int main() { int a = 10, b = 20, n = 5; cout << numberOfDays(a, b, n); return 0; } |
Java
// Java implementation of the approach import java.io.*; class GFG { // Function to return the // number of days required static int numberOfDays(int a, int b, int n) { int Days = b * (n + a) / (a + b); return Days; } // Driver code public static void main (String[] args) { int a = 10, b = 20, n = 5; System.out.println (numberOfDays(a, b, n)); } } // This code is contributed by jit_t |
Python3
# Python implementation of the approach # Function to return the # number of days required def numberOfDays(a, b, n) : Days = b * (n + a) // (a + b) return Days # Driver code a = 10b = 20n = 5 print(numberOfDays(a, b, n)) # This code is contributed by ihritik |
C#
// C# implementation of the approach using System; class GFG { // Function to return the // number of days required static int numberOfDays(int a, int b, int n) { int Days = b * (n + a) / (a + b); return Days; } // Driver code public static void Main () { int a = 10, b = 20, n = 5; Console.WriteLine(numberOfDays(a, b, n)); } } // This code is contributed by AnkitRai01 |
10
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