For a number, greater than one, the difference between itself and its reciprocal is 20% of the sum of itself and its reciprocal. By how much percentage (nearest to an integer) is the square of the number less than its cube?

Let's assume the original number is 'y'.

For a number, greater than one, the difference between itself and its reciprocal is 20% of the sum of itself and its reciprocal.

$$y-\frac{1}{y}\ =\ 20\%\ \left(y+\frac{1}{y}\right)$$

$$y-\frac{1}{y}\ =\ \frac{1}{5}\ \left(y+\frac{1}{y}\right)$$

$$\frac{y^2-1}{y}\ =\ \frac{1}{5}\ \left(\frac{y^2+1}{y}\right)$$

$$5y^2-5\ =\ y^2+1$$

$$4y^2\ =\ 6$$

$$y^2\ =\ \frac{3}{2}$$

percentage less = $$\frac{\left(y^3\ -\ y^2\right)\ \times\ 100}{y^3}$$

= $$\frac{\left(\left(\frac{3}{2}\right)^{\frac{3}{2}}\ -\ \frac{3}{2}\right)\ \times\ 100}{\left(\frac{3}{2}\right)^{\frac{3}{2}}}$$

= $$\frac{\left(\frac{3\sqrt{\ 3}}{2\sqrt{\ 2}}\ -\ \frac{3}{2}\right)\ \times\ 100}{\frac{3\sqrt{\ 3}}{2\sqrt{\ 2}}}$$

= $$\frac{\left(\frac{3\sqrt{\ 3}}{2\sqrt{\ 2}}\ -\ \frac{3\sqrt{\ 2}}{2\sqrt{\ 2}}\right)\ \times\ 100}{\frac{3\sqrt{\ 3}}{2\sqrt{\ 2}}}$$

= $$\frac{\left(\sqrt{\ 3}\ -\ \sqrt{\ 2}\right)\ \times\ 100}{\sqrt{\ 3}}$$

= $$\frac{\left(1.7\ -\ 1.4\right)\ \times\ 100}{1.7}$$

= $$\frac{\left(0.3\right)\ \times\ 100}{1.7}$$

= 17.64 %

= 18% (approx.)