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
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 <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
// 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
# 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#
// 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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