If the edge of a cube is increased by 100%, the surface area of the cu

1.	If the edge of a cube is increased by 100%, the surface area of the cube is increased by how much percent?
Answer» ANSWER: The SURFACE area of the CUBE is increased by 300%. Step-by-step explanation: GIVEN : If the edge of a cube is increased by 100%. To find : The surface area of the cube is increased by how much percent? Solution : Let edge of the cube be 's'. The surface area of the original cube is $S_o=6s^2$ If the edge of a cube is increased by 100%. i.e. The increased edge is $e=s+s=2s$ The surface area of the increased edge cube is $S_n=6(2s)^2$ $S_n=6(4s^2)$ $S_n=24s^2$ The increased percentage of surface area of the cube is $P\%=\frac{S_n-S_o}{S_o}\times 100$ $P\%=\frac{24s^2-6s^2}{6s^2}\times 100$ $P\%=\frac{18s^2}{6s^2}\times 100$ $P\%=3\times 100$ $P\%=300\%$ Therefore, The surface area of the cube is increased by 300%.