When 8 is added to the numerator of a fraction and 12 is added to its denominator, the fraction becomes $$\frac{1}{2}$$. When 2 is subtracted from its numerator and denominator, the fraction becomes $$\frac{1}{8}$$. Find the original fraction.

Let's assume the original fraction is $$\frac{P}{Q}$$.

When 8 is added to the numerator of a fraction and 12 is added to its denominator, the fraction becomes $$\frac{1}{2}$$.

$$\frac{P+8}{Q+12} = \frac{1}{2}$$

2P+16 = Q+12

2P-Q = 12-16

2P-Q = -4    Eq.(i)
When 2 is subtracted from its numerator and denominator, the fraction becomes $$\frac{1}{8}$$.

$$\frac{P-2}{Q-2} = \frac{1}{8}$$

8P-16 = Q-2
8P-Q = 16-2

8P-Q = 14    Eq.(ii)

Eq.(ii)-Eq.(i) ==> 8P-Q-(2P-Q) = 14-(-4)

8P-Q-2P+Q = 14+4

6P = 18

P = 3

Put the value of 'P' in Eq.(i).

$$2\times3-Q = -4$$

6-Q = -4

Q = 6+4

Q = 10

Original fraction = $$\frac{P}{Q}\ =\ \frac{3}{10}$$