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.
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