If x = 80%, by what percent is x larger than x^2?
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27 Jan 2024, 19:35
If \(x=80\)%, then \(x=\frac{80}{100} = \frac{4}{5}\)
\(x=\frac{4}{5}\)
\(x^2=\frac{16}{25}\)
First, let us find out how much is \(x\) larger than \(x^2\) in absolute terms
\(x-x^2 = \frac{4}{5}-\frac{16}{25} = \frac{20-16}{25} = \frac{4}{25}\)
In terms of percentage, \(x\) is larger than \(x^2\) by
\(\frac{x-x^2}{x^2} \times 100 = \dfrac{\frac{4}{25}}{\frac{16}{25}} \times 100 = \dfrac{1}{4} = 25\)%
The correct answer is D.