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 $\dfrac{2}{21}$. Find the fraction.

Let the numerator of fraction be x and denominator be y.

The fraction is $\dfrac{x}{y}$.

Given,

Sum of numerator and denominator of the fraction is 10.

⇒ x + y = 10

⇒ y = 10 - x     ………(1)

Given,

If 1 is subtracted from both the numerator and denominator, the fraction is decreased by $\dfrac{2}{21}$.

$\Rightarrow \dfrac{x}{y} - \dfrac{x - 1}{y - 1} = \dfrac{2}{21} \\[1em] \Rightarrow \dfrac{x(y - 1) - y(x - 1)}{y(y - 1)} = \dfrac{2}{21} \\[1em] \Rightarrow \dfrac{xy - x - xy + y }{y(y - 1)} = \dfrac{2}{21} \\[1em] \Rightarrow \dfrac{y - x}{y(y - 1)} = \dfrac{2}{21} \text{ ……(2)}$

Substituting the value of y from equation (1) in equation (2),

$\Rightarrow \dfrac{y - x}{y(y - 1)} = \dfrac{2}{21} \\[1em] \Rightarrow \dfrac{(10 - x) - x}{(10 - x)(10 - x - 1)} = \dfrac{2}{21} \\[1em] \Rightarrow \dfrac{10 - 2x}{(10 - x)(9 - x)} = \dfrac{2}{21} \\[1em] \Rightarrow \dfrac{10 - 2x}{90 - 10x - 9x + x^2} = \dfrac{2}{21} \\[1em] \Rightarrow \dfrac{10 - 2x}{90 - 19x + x^2} = \dfrac{2}{21} \\[1em] \Rightarrow 21(10 - 2x)= 2(90 - 19x + x^2) \\[1em] \Rightarrow 210 - 42x = 180 - 38x + 2x^2 \\[1em] \Rightarrow 2x^2 - 38x + 180 - 210 + 42x = 0 \\[1em] \Rightarrow 2x^2 + 4x - 30 = 0 \\[1em] \Rightarrow 2(x^2 + 2x - 15) = 0 \\[1em] \Rightarrow x^2 + 2x - 15 = 0 \\[1em] \Rightarrow x^2 + 5x - 3x - 15 = 0 \\[1em] \Rightarrow x(x + 5) - 3(x + 5) = 0 \\[1em] \Rightarrow (x - 3)(x + 5) = 0 \\[1em] \Rightarrow (x - 3) = 0 \text{ or }(x + 5) = 0 \text{ [Using zero - product rule] } \\[1em] \Rightarrow x = 3 \text{ or } x = -5 \\[1em]$

Since fraction is positive.

Thus, x = 3.

Substituting value of x in equation (1), we get :

⇒ y = 10 - 3

⇒ y = 7.

The fraction is $\dfrac{3}{7}$.

Hence, the required fraction = $\dfrac{3}{7}$.