A manufacturer reduces the diameter of their spherical bearing balls by 50% to cut costs. What is the percentage decrease in volume of each ball?
87.5%
The volume of a sphere is \(V = \dfrac{4}{3}\pi r^3\), so it depends on the cube of the radius, and equally on the cube of the diameter since \(d = 2r\).
Reducing the diameter by 50% makes the new diameter half of the old one, a scaling factor of \(\dfrac{1}{2}\) on every linear dimension.
Volume scales by the cube of that linear factor: new volume \(= \left(\dfrac{1}{2}\right)^3 V = \dfrac{1}{8}V\).
So the new volume is \(\dfrac{1}{8} = 12.5\%\) of the original volume.
Percentage decrease \(= 100\% - 12.5\% = 87.5\%\).
Key concept: for similar solids, volume changes as the cube of the linear scale factor; halving a length divides volume by \(2^3 = 8\), not by 2.
Hence the volume decreases by 87.5%.
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