Time taken per hour for stoppage of Car

A car travels with an average speed of S km/h without any stoppage and with stoppage the speed of car reduces to an average of S1 km/h. The task is to find the time wasted per hour for the stoppage.

Examples:

Input: S = 50, S1 = 30
Output: 24 min



Input: S = 30, S1 = 10
Output: 40 min

Approach: Take the first example,

Speed of car without any stoppage = 50 kmph i.e. 50 km in 60 min.
Speed of car with stoppage = 30 kmph i.e. 30 km in 60 min.

Now, if there will be no stoppage then 30 km can be covered in 36 min.
50 km –> 60 min
30 km –> (60 / 50) * 30 = 36 min
but it takes 60 min.
So, time for stoppage per hour is 60 min – 36 min = 24 min.

This can be calculated using the below formula:

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the time taken
// per hour for stoppage
int numberOfMinutes(int S, int S1)
{
  
    int Min = 0;
  
    Min = ((S - S1) / floor(S)) * 60;
  
    return Min;
}
  
// Driver code
int main()
{
    int S = 30, S1 = 10;
  
    cout << numberOfMinutes(S, S1) << " min";
  
    return 0;
}

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Java

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// Java implementation of the approach
import java.util.*;
  
class GFG 
{
  
// Function to return the time taken
// per hour for stoppage
static int numberOfMinutes(int S, int S1)
{
    int Min = 0;
  
    Min = (int) (((S - S1) / Math.floor(S)) * 60);
  
    return Min;
}
  
// Driver code
public static void main(String[] args) 
{
    int S = 30, S1 = 10;
  
    System.out.println(numberOfMinutes(S, S1) + " min");
}
}
  
// This code is contributed by Princi Singh

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Python3

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# Python3 implementation of the approach
import math 
  
# Function to return the time taken
# per hour for stoppage
def numberOfMinutes(S, S1):
  
    Min = 0;
  
    Min = ((S - S1) / math.floor(S)) * 60;
  
    return int(Min);
  
# Driver code
if __name__ == '__main__':
    S, S1 = 30, 10;
  
    print(numberOfMinutes(S, S1), "min");
  
# This code is contributed by Rajput-Ji

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C#

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// C# implementation of the approach
using System;
  
class GFG 
{
  
// Function to return the time taken
// per hour for stoppage
static int numberOfMinutes(int S, int S1)
{
    int Min = 0;
  
    Min = (int) (((S - S1) / 
                   Math.Floor((double)S)) * 60);
  
    return Min;
}
  
// Driver code
public static void Main() 
{
    int S = 30, S1 = 10;
  
    Console.WriteLine(numberOfMinutes(S, S1) + 
                                      " min");
}
}
  
// This code is contributed 
// by Akanksha Rai 

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Output:

40 min


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