A fraction becomes $$\frac{1}{6}$$ when 4 is subtracted from its numerator and 1 is added to its denominator. If 2 and 1 are respectively added to its numerator and the denominator, it becomes $$\frac{1}{3}$$. Then, the LCM of the numerator and denominator of the said fraction, must be = ?
Correct Answer: 14
$$\eqalign{ & {\text{Let fraction is }}\frac{x}{y} \cr & \therefore \frac{{x - 4}}{{y + 1}} = \frac{1}{6}({\text{given}}) \cr & \Rightarrow {\text{cross multiply the equation }} \cr & \Rightarrow {\text{6}}x - 24 = y + 1 \cr & 6x - y - 25 = 0......({\text{i}}) \cr & {\text{Again,}} \cr & \frac{{x + 2}}{{y + 1}} = \frac{1}{3}({\text{given}}) \cr & \Rightarrow 3x + 6 = y + 1 \cr & 3x - y + 5 = 0.......({\text{ii}}) \cr & {\text{From equation (i) and (ii)}} \cr & {\text{6}}x - y = 25 \cr & \frac{{3x - y = - 5}}{{x\,\,\,\,\, = \,\,\,\,\,10}} \cr & \therefore y = 35 \cr & \frac{x}{y} = \frac{{10}}{{35}} = \frac{2}{7} \cr & {\text{Fraction = }}\frac{x}{y} = \frac{2}{7} \cr & {\text{Numerator = 2}} \cr & {\text{Denominator = 7}} \cr & {\text{LCM (Numerator}},{\text{Denominator}}) \cr & \Rightarrow 2 \times 7 = 14 \cr} $$