Program to find the head start in a race

Given the head start that A gives to B and C in a 100-meters race. The task is to find the head-start that B can give to C in the same race.

Examples:

Input: B = 10 meters, C = 28 meters
Output: 20 meters
B can give C a start of 20 meters.

Input: B = 20 meters, C = 50 meters
Output: 62 meters
B can give C a start of 62 meters.

Approach:

Total meters in a race = 100 meters.
A is ahead of B by 10 meters. When A completed it’s 100 meters B completed it’s 90 meters.
Similarly, A is ahead of C by 28 meters. When A completed it’s 100 meters C completed it’s 72 meters.
Now, When B completed it’s 90 meters C completed it’s 72 meters.
So when B completed it’s 100 meters C completed it’s 80 meters.
–> (( C * 100) / B)
–> (( 72 * 100) / 90) i.e 80 meters
So B can give C a start of 20 meters


Below is the implementation of the above approach:

C++

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// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the B start to C
int Race(int B, int C)
{
    int result = 0;
  
    // When B completed it's 100 meter
    // then Completed meters by C is
    result = ((C * 100) / B);
  
    return 100 - result;
}
  
// Driver Code.
int main()
{
    int B = 10, C = 28;
  
    // When A completed it's 100 meter
    // Then completed meters of B and C is
    B = 100 - B;
    C = 100 - C;
  
    cout << Race(B, C) << " meters";
  
return 0;
}

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Java

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// Java implementation of above approach
public class GFG 
{
      
// Function to find the B start to C
static int Race(int B, int C)
{
    int result = 0;
  
    // When B completed it's 100 meter
    // then Completed meters by C is
    result = ((C * 100) / B);
  
    return 100 - result;
}
  
// Driver Code
public static void main(String[] args) 
{
    int B = 10;
    int C = 28;
      
    // When A completed it's 100 meter
    // Then completed meters of B and C is
    B = 100 - B;
    C = 100 - C;
      
    System.out.println(Race(B, C) + " meters");
}
}
  
// This code is contributed 
// by ChitraNayal

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Python 3

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# Python 3 implementation 
# of above approach
  
# Function to find the 
# B start to C
def Race(B, C):
  
    result = 0;
  
    # When B completed it's 100 meter
    # then Completed meters by C is
    result = ((C * 100) // B)
  
    return 100 - result
  
# Driver Code
if __name__ == "__main__":
    B = 10
    C = 28
      
    # When A completed it's 100 meter
    # Then completed meters of B and C is
    B = 100 - B;
    C = 100 - C;
      
    print(str(Race(B, C)) + " meters")
  
# This code is contributed 
# by ChitraNayal

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C#

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// C# implementation of above approach
using System;
class GFG 
{
  
// Function to find the B start to C
static int Race(int B, int C)
{
    int result = 0;
  
    // When B completed it's 100 meter
    // then Completed meters by C is
    result = ((C * 100) / B);
  
    return 100 - result;
}
  
// Driver Code
public static void Main() 
{
    int B = 10;
    int C = 28;
      
    // When A completed it's 100 meter
    // Then completed meters of B and C is
    B = 100 - B;
    C = 100 - C;
      
    Console.Write(Race(B, C) + " meters");
}
}
  
// This code is contributed 
// by ChitraNayal

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PHP

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<?php 
// PHP implementation of above approach
  
// Function to find the B start to C
function Race($B, $C)
{
    $result = 0;
  
    // When B completed it's 100 meter
    // then Completed meters by C is
    $result = (($C * 100) / $B);
  
    return 100 - $result;
}
  
// Driver Code
$B = 10;
$C = 28;
  
// When A completed it's 100 meter
// Then completed meters of B and C is
$B = 100 - $B;
$C = 100 - $C;
  
echo Race($B, $C) . " meters";
  
// This code is contributed
// by ChitraNayal
?>

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Output:

20 meters


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