# 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,
• 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.. `

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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