The ratio of the present ages of A and B is 8: 9. After 9 years, this ratio will become 19: 21. C is 3 years younger to B. What is the present age (in years) of C?

Let the present age of A and B is x and y.

Given that the ratio of the age of A and B, $$\dfrac{x}{y}=\dfrac{8}{9}$$

$$x=\dfrac{8y}{9}-------------(i)$$

After 9 years, age of A=x+9 and age of B=y+9,

The ratio in the age after 9 years,

$$\dfrac{x+9}{y+9}=\dfrac{19}{21}-------------(ii)$$

From equation (i) and (ii)

$$\Rightarrow \dfrac{ \dfrac{8y}{9} +9}{y+9}=\dfrac{19}{21}$$

$$\Rightarrow \dfrac{ 8y+81}{9(y+9)}=\dfrac{19}{21}$$

$$\Rightarrow \dfrac{ 8y+81}{9y+81}=\dfrac{19}{21}$$

$$\Rightarrow 21 \times 8y+21\times 81=19\times9y+19\times 81$$

$$\Rightarrow 168y+21\times 81=171y+19\times 81$$

$$\Rightarrow 171y-168y=21\times y -19\times 81$$

$$\Rightarrow 3y=2\times 81$$

$$\Rightarrow y=54$$years

Hence the age of C $$=y-3=51$$Years.