Question

THE NUMERATOR OF A FRACTION IS 2 LESS THAN THE DENOMINATOR . IF "1" IS ADDED TO BOTH NUMERATOR AND DENOMINATOR, THE SUM OF THE NEW AND THE ORIGINAL FRACTION IS 19/15. FIND THE ORIGINAL FRACTION?

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{denominator}=\mathrm{x} \\ \mathrm{then}\mathrm{numerator}=\mathrm{x}-2 \\ \mathrm{Now},\mathrm{original}\mathrm{fraction}=\frac{\mathrm{x}-2}{\mathrm{x}} \\ \mathrm{Now},\mathrm{according}\mathrm{to}\mathrm{question}: \\ \frac{\mathrm{x}-2+1}{\mathrm{x}+1}+\frac{\mathrm{x}-2}{\mathrm{x}}=\frac{19}{15} \\ \Rightarrow \frac{\mathrm{x}-1}{\mathrm{x}+1}+\frac{\mathrm{x}-2}{\mathrm{x}}=\frac{19}{15} \\ \Rightarrow \frac{{\mathrm{x}}^2-\mathrm{x}+{\mathrm{x}}^2-\mathrm{x}-2}{\mathrm{x}\left(\mathrm{x}+1\right)}=\frac{19}{15} \\ \Rightarrow \frac{2{\mathrm{x}}^2-2\mathrm{x}-2}{{\mathrm{x}}^2+\mathrm{x}}=\frac{19}{15} \\ \Rightarrow 30{\mathrm{x}}^2-30\mathrm{x}-30=19{\mathrm{x}}^2+19\mathrm{x} \\ \Rightarrow 11{\mathrm{x}}^2-49\mathrm{x}-30=0 \\ \Rightarrow 11{\mathrm{x}}^2-55\mathrm{x}+6\mathrm{x}-30=0 \\ \Rightarrow 11\mathrm{x}\left(\mathrm{x}-5\right)+6\left(\mathrm{x}-5\right)=0 \\ \Rightarrow \left(11\mathrm{x}+6\right)\left(\mathrm{x}-5\right)=0 \\ \Rightarrow \mathrm{x}-5=0\mathrm{or}11\mathrm{x}+6=0 \\ \Rightarrow \mathrm{x}=5\mathrm{or}\mathrm{x}=-\frac{6}{11}\left[\mathrm{rejected}\right] \\ \mathrm{So},\mathrm{original}\mathrm{fraction}=\frac{\mathrm{x}-2}{\mathrm{x}}=\frac{5-2}{5}=\frac{3}{5} \end{gathered}$$