Find two numbers in the ratio of 8 : 7 such that when each is decreased by $12\dfrac{1}{2}$, they are in ratio 11 : 9.

Since, numbers are in ratio 8 : 7, let the required numbers be 8x and 7x.

According to question,

$\dfrac{8x - 12\dfrac{1}{2}}{7x - 12\dfrac{1}{2}} = \dfrac{11}{9} \\[1em] \Rightarrow \dfrac{8x - \dfrac{25}{2}}{7x - \dfrac{25}{2}} = \dfrac{11}{9} \\[1em] \Rightarrow \dfrac{\dfrac{16x - 25}{2}}{\dfrac{14x - 25}{2}} = \dfrac{11}{9} \\[1em] \Rightarrow \dfrac{16x - 25}{14x - 25} = \dfrac{11}{9} \\[1em] \Rightarrow 9(16x - 25) = 11(14x - 25) \\[1em] \Rightarrow 144x - 225 = 154x - 275 \\[1em] \Rightarrow 144x - 154x = -275 + 225 \\[1em] \Rightarrow -10x = -50 \\[1em] \Rightarrow x = 5.$

∴ x = 5, 8x = 40, 7x = 35.

Hence, the required numbers are 40, 35.