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++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the time taken// per hour for stoppageint numberOfMinutes(int S, int S1){ int Min = 0; Min = ((S - S1) / floor(S)) * 60; return Min;}// Driver codeint main(){ int S = 30, S1 = 10; cout << numberOfMinutes(S, S1) << " min"; return 0;} |
Java
// Java implementation of the approachimport java.util.*;class GFG {// Function to return the time taken// per hour for stoppagestatic int numberOfMinutes(int S, int S1){ int Min = 0; Min = (int) (((S - S1) / Math.floor(S)) * 60); return Min;}// Driver codepublic static void main(String[] args) { int S = 30, S1 = 10; System.out.println(numberOfMinutes(S, S1) + " min");}}// This code is contributed by Princi Singh |
Python3
# Python3 implementation of the approachimport math # Function to return the time taken# per hour for stoppagedef numberOfMinutes(S, S1): Min = 0; Min = ((S - S1) / math.floor(S)) * 60; return int(Min);# Driver codeif __name__ == '__main__': S, S1 = 30, 10; print(numberOfMinutes(S, S1), "min");# This code is contributed by Rajput-Ji |
C#
// C# implementation of the approachusing System;class GFG {// Function to return the time taken// per hour for stoppagestatic int numberOfMinutes(int S, int S1){ int Min = 0; Min = (int) (((S - S1) / Math.Floor((double)S)) * 60); return Min;}// Driver codepublic static void Main() { int S = 30, S1 = 10; Console.WriteLine(numberOfMinutes(S, S1) + " min");}}// This code is contributed // by Akanksha Rai |
Javascript
<script>// JavaScript implementation of the approach // Function to return the time taken // per hour for stoppage function numberOfMinutes(S, S1) { let Min = 0; Min = ((S - S1) / Math.floor(S)) * 60; return Min; } // Driver code let S = 30, S1 = 10; document.write(numberOfMinutes(S, S1) + " min"); // This code is contributed by Surbhi Tyagi.</script> |
40 min
Time Complexity: O(1), since there is only basic arithmetic that happens in constant time.
Auxiliary Space: O(1), since no extra space has been taken.
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