Question

The sum of the numerator and denominator of a certain fraction is 10. If 1 is subtracted from both the numerator and denominator, the fraction is decreased by $$\frac{2}{21}$$. Find the fraction.

Answer 1

Let the numerator of the fraction be $$x$$. 

Then, the denominator is $$(10 - x)$$. 

The original fraction is $$\frac{x}{10 - x}$$. 

Subtracting 1 from both numerator and denominator gives $$\frac{x - 1}{(10 - x) - 1} = \frac{x - 1}{9 - x}$$. 

According to the problem: $$\frac{x}{10 - x} - \frac{x - 1}{9 - x} = \frac{2}{21}$$ 

$$\frac{x(9 - x) - (x - 1)(10 - x)}{(10 - x)(9 - x)} = \frac{2}{21}$$

$$\frac{(9x - x^2) - (11x - x^2 - 10)}{x^2 - 19x + 90} = \frac{2}{21}$$

$$\frac{10 - 2x}{x^2 - 19x + 90} = \frac{2}{21}$$

$$\frac{2(5 - x)}{x^2 - 19x + 90} = \frac{2}{21}$$ 

$$\frac{5 - x}{x^2 - 19x + 90} = \frac{1}{21}$$

$$21(5 - x) = x^2 - 19x + 90$$

$$105 - 21x = x^2 - 19x + 90$$

$$x^2 + 2x - 15 = 0$$ 

Factoring: $$(x + 5)(x - 3) = 0$$ 

$$x = -5 \quad \text{or} \quad x = 3$$ 

Since the numerator is a positive integer, $$x = 3$$. 

Denominator $$= 10 - 3 = 7$$. 

Hence, the required fraction is $$\frac{3}{7}$$.