With reference to a number greater than one, the difference between itself and its reciprocal is 25% of the sum of itself and its reciprocal. By how much percentage (correct one decimal place) is the fourth power of the number greater than its square ?

Let the number be x, then reciprocal = $$\frac{1}{x}$$

According to the question,

$$x-\frac{1}{x}$$= 25% of $$(x+\frac{1}{x})$$

$$x-\frac{1}{x}$$=  $$\frac{1}{4}(x+\frac{1}{x})$$

$$4x-\frac{4}{x}$$=$$(x+\frac{1}{x})$$

$$3x=\frac{5}{x}$$

$$x^{2}=\frac{5}{3}$$

$$x = \sqrt{\frac{5}{3}}$$

Now,

$$x^{4}=(x^{2)^{2}}=(\frac{5}{3})^{2}=\frac{25}{9}$$

$$x^{2}=\frac{5}{3}$$

$$(x^{2)^{2}}-x^{2}$$=$$\frac{10}{9}$$

Required % = $$\frac{10}{9}\times\frac{3}{5}\times100$$=66.67%