A fraction in reduced form is such that when it is squared and then its numerator is reduced by 33(1/3)% and denominator is reduced to 20%, its result is twice the original fraction. The sum of numerator and denominator is:

Answer:- 1

Explanation:-

Solution:

Let the fraction be x/y.
When fraction is squared its numerator is reduced by 33(1/3) and denominator is reduced by 20%.

\begin{aligned} According to question \\ {(\frac{x}{y})}^{2} \times \frac{33 (\frac{1}{3}) %}{20 %} = 2 (\frac{x}{y}) \\ Or, {(\frac{x}{y})}^{2} \times \frac{(\frac{2}{3})}{(\frac{1}{5})} = 2 (\frac{x}{y}) \\ Or, \frac{x}{y} = \frac{3}{5} \\ Sum of numerator and denominator is \\ (x + y) = 3 + 5 \\ = 8 \end{aligned}

$\eqalign{ & {\text{According}}\,{\text{to}}\,{\text{question}} \cr & {\left( {\frac{x}{y}} \right)^2} \times \frac{{33\left( {\frac{1}{3}} \right)\% }}{{20\% }} = 2\left( {\frac{x}{y}} \right) \cr & {\text{Or}},\,{\left( {\frac{x}{y}} \right)^2} \times \frac{{\left( {\frac{2}{3}} \right)}}{{\left( {\frac{1}{5}} \right)}} = 2\left( {\frac{x}{y}} \right) \cr & {\text{Or}},\,\frac{x}{y} = \frac{3}{5} \cr & {\text{Sum of numerator and denominator is}} \cr & \left( {x + y} \right) = 3 + 5 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 8 \cr}$