Q.Each edge of a cube is reduced by 50%. What is the ratio of the volume of the original cube to that of the reduced cube?

Let the original length of each edge of the cube be \(x\) units. If each edge is reduced by 50%, the new edge length becomes: \(x - 50\%\text{ of }x = x - \cfrac{50x}{100} = x - \cfrac{x}{2} = \cfrac{x}{2}\) units The volume of the original cube = \(x^3\) cubic units The volume of the reduced cube = \((\cfrac{x}{2})^3 = \cfrac{x^3}{8}\) cubic units ∴ The ratio of the volumes of the original cube to the reduced cube is: \(x^3 : \cfrac{x^3}{8} = 1 : \cfrac{1}{8} = 8 : 1\)
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