A fraction in reduced form is such that when it is squared and then its numerator is increased by 25% and the denominator is reduced t0 80% it results in 5/8 of original fraction. The product of the numerator and denominator is :

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Let the fraction be \frac{100 x}{100 y} \\ Now according to the question, \\ {(\frac{100 x}{100 y})}^{2} = \frac{125 x^{2}}{80 y^{2}} = \frac{25 x^{2}}{16 y^{2}} \\ \frac{25 x^{2}}{16 y^{2}} = \frac{5}{8} (\frac{100 x}{100 y}) \\ (\frac{100 x}{100 y}) = \frac{2}{5} \\ hence, \\ product of numerator and denominator \\ = 2 \times 5 = 10 \end{aligned}

$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{fraction}}\,{\text{be}}\,\frac{{100x}}{{100y}} \cr & {\text{Now}}\,{\text{according}}\,{\text{to}}\,{\text{the}}\,{\text{question}}, \cr & {\left( {\frac{{100x}}{{100y}}} \right)^2} = \frac{{125{x^2}}}{{80{y^2}}} = \frac{{25{x^2}}}{{16{y^2}}} \cr & \frac{{25{x^2}}}{{16{y^2}}} = \frac{5}{8}\left( {\frac{{100x}}{{100y}}} \right) \cr & \left( {\frac{{100x}}{{100y}}} \right) = \frac{2}{5} \cr & {\text{hence}}, \cr & {\text{product of numerator and denominator}} \cr & = 2 \times 5 = 10 \cr}$