If the numerator of the fraction is increased by 17% and its denominator is decreased by 12%, the value of fraction is$$\frac{39}{44}$$​. what is the original fraction?

Correct option is D

Given:

Numerator is increased by 17%

Denominator decreased by 12%

Value of fraction becomes = $$\frac{39}{44}$$​​

Solution:

Let the numerator be $$x$$​​

Then after 17% increase :

​$$x+17\% \ of\ x = 1.17x$$​​

Let denominator be $$y$$​​

Then after 12% decrease :

​$$y -12\% \ of \ y = 0.88y$$​​

Then fraction becomes = $$\frac{1.17x}{0.88y}$$​

Now, 

​$$\frac{1.17x}{0.88y} =\frac{39}{44}$$​

​$$\frac{x}{y} =\frac{39 \times 0.88}{44 \times 1.17} = \bf \frac{2}{3}$$​​