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

A. 8

B. 13

C. 17

D. 15

E. None of these

Solution(By Examveda Team)

Let the fraction be $$\frac{{\text{x}}}{{\text{y}}}$$
When fraction is squared its numerator is reduced by $$33\frac{1}{3}$$ and denominator is reduced by 20%
\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}

1. pls give detail solution . not able to understand

2. if we read it properly it says reduced by 33 1/3 percent(i.e,1-1/3= 2/3) and 20 percent (1-1/5=4/5).

3. It should be 6/5 and sum is 11

4. 33 (1/3)=1/3 not 2/3

5. 6. x/y=6/5
7. 8. 