Percentage increase in volume of the cube if a side of cube is increased by a given percentage


Given here is a cube, whose one side is increased by a given percentage. The task is to find percentage increase in the volume of the cube.

Examples:

Input: x = 10
Output: 33.1%

Input: x = 50
Output: 237.5%

Approach



  • In a cube, all sides are equal, so,
    length = breadth = height
  • let side of the cube = a
  • given percentage increase = x%
  • so, volume before increase = a^3
  • after increase, new side = a + ax/100
  • so, new volume = (a + ax/100)^3 = a^3 + (ax/100)^3 + 3a^3x/100 + 3a^3x^2/10000
  • increase in volume = new volume – old volume = (a^3 + (ax/100)^3 + 3a^3x/100 + 3a^3x^2/10000) – a^3 = (ax/100)^3 + 3a^3x/100 + 3a^3x^2/10000
  • so, percentage increase in volume = (((ax/100)^3 + 3a^3x/100 + 3a^3x^2/10000)/a^3) * 100 = ((x/100)^3 + 3x/100 + 3x^2/10000) * 100 = x^3/10000 + 3x + 3x^2/100

   <strong>Percentage increase in the volume = (given percentage)^3/10000 + 3*given percentage + 3*(given percentage)^2/100</strong>

Below is the implementation of the above approach:

C++

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// C++ program to find percentage increase
// in the volume of the cube
// if a side of cube is increased
// by a given percentage
  
#include <bits/stdc++.h>
using namespace std;
  
void newvol(double x)
{
    cout << "percentage increase "
         << "in the volume of the cube is "
         << pow(x, 3) / 10000 + 3 * x
                + (3 * pow(x, 2)) / 100
         << "%" << endl;
}
  
// Driver code
int main()
{
    double x = 10;
    newvol(x);
    return 0;
}

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Java

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// Java program to find percentage increase
// in the volume of the cube
// if a side of cube is increased
// by a given percentage
import java.io.*;
  
class GFG 
{
  
static void newvol(double x)
{
    System.out.print( "percentage increase "
    +"in the volume of the cube is "
        + (Math.pow(x, 3) / 10000 + 3 * x
                + (3 * Math.pow(x, 2)) / 100) );
                System.out.print("%");
}
  
// Driver code
public static void main (String[] args)
{
    double x = 10;
    newvol(x);
}
}
  
// This code is contributed by anuj_67..

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Python3

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# Python program to find percentage increase
# in the volume of the cube
# if a side of cube is increased
# by a given percentage
  
def newvol(x):
  
    print("percentage increase"
            "in the volume of the cube is ",
            ((x**(3)) / 10000 + 3 * x
                + (3 * (x**(2))) / 100),"%");
  
x = 10;
newvol(x);
  
# This code is contributed by PrinciRaj1992 

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C#

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// C# program to find percentage increase
// in the volume of the cube
// if a side of cube is increased
// by a given percentage
using System;
  
class GFG 
{
  
static void newvol(double x)
{
    Console.Write( "percentage increase "
    +"in the volume of the cube is "
        + (Math.Pow(x, 3) / 10000 + 3 * x
                + (3 * Math.Pow(x, 2)) / 100) );
                Console.Write("%");
}
  
// Driver code
public static void Main ()
{
    double x = 10;
    newvol(x);
}
}
  
// This code is contributed by anuj_67..

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Output:

percentage increase in the volume of the cube is 33.1%


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Improved By : vt_m, princiraj1992



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