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

• Last Updated : 09 Sep, 2022

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,
• 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

Below is the implementation of the above approach:

## C++

 `// C++ program to find percentage increase` `// in the volume of the cube` `// if a side of cube is increased` `// by a given percentage`   `#include ` `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;` `}`

## Java

 `// 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..`

## Python3

 `# 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 `

## C#

 `// 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..`

## Javascript

 ``

Output:

`percentage increase in the volume of the cube is 33.1%`

Time Complexity: O(1) because pow function will take constant time

Auxiliary Space: O(1), since no extra space has been taken.

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