If 1 is added to the both the numerator and the denominator of a fraction, it becomes $$\frac{1}{4}$$ . If 2 is added to both the numerator and the denominator of that fraction it becomes $$\frac{1}{3}$$ . The sum of the numerator and the denominator and the denominator of the fraction is -
Correct Answer: 8
Let the numerator and denominator x and y
According to question,
$$\eqalign{ & \frac{{x + 1}}{{y + 1}} = \frac{1}{4} \cr & 4x + 4 = y + 1 \cr & 4x - y = - 3.....(i) \cr & \frac{{x + 2}}{{y + 2}} = \frac{1}{3} \cr & 3x + 6 = y + 2 \cr & 3x - y = - 4.....(ii) \cr & {\text{Solve (i) and (ii) }} \cr & x = 1\,\,\,\& \,\,\,y = 7 \cr} $$
Sum of numerator and denominator of fraction :
= x + y
= 1 + 7
= 8
According to question,
$$\eqalign{ & \frac{{x + 1}}{{y + 1}} = \frac{1}{4} \cr & 4x + 4 = y + 1 \cr & 4x - y = - 3.....(i) \cr & \frac{{x + 2}}{{y + 2}} = \frac{1}{3} \cr & 3x + 6 = y + 2 \cr & 3x - y = - 4.....(ii) \cr & {\text{Solve (i) and (ii) }} \cr & x = 1\,\,\,\& \,\,\,y = 7 \cr} $$
Sum of numerator and denominator of fraction :
= x + y
= 1 + 7
= 8