Question

The sum of ages of husband and his wife is four times the sum of ages of their children. Four years ago, the ratio of sum of their ages to the sum of ages of their children was 18:1. two years hence the ratio will be 3:1. How many children do they have?

Answer 1

$$\begin{gathered} \text{Let the sum of the ages of husband and wife be}\mathit{\text{x}}\text{yrs and sum of the ages of their children be}\mathit{\text{ny}}\text{yrs,} \\ \mathrm{where}n\mathrm{is}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{children}\mathrm{and}y\mathrm{is}\mathrm{the}\mathrm{average}\mathrm{age} \\ \text{According to the question,} \\ x=4ny...\left(1\right) \\ 4\text{years ago,} \\ \text{Sum of the ages of the husband and wife}=\left(x-8\right)\text{yrs} \\ \text{Sum of the ages of their children}=\left(ny-4n\right)\text{yrs} \\ \frac{\left(x-8\right)}{\left(ny-4n\right)}=\frac{18}{1} \\ x-8=18\left(ny-4n\right) \\ \mathrm{From}\left(1\right) \\ 4ny-8=18ny-72n \\ 14ny+8=72n \\ ny=\frac{72n-8}{14}...\left(2\right) \\ \mathrm{Two}\mathrm{years}\mathrm{hence}, \\ \mathrm{Sum}\mathrm{of}\mathrm{the}\mathrm{ages}\mathrm{of}\mathrm{husband}\mathrm{and}\mathrm{wife}=\left(x+4\right)\mathrm{yrs} \\ \mathrm{Sum}\mathrm{of}\mathrm{the}\mathrm{ages}\mathrm{their}\mathrm{children}=\left(ny+2n\right)\mathrm{yrs} \\ \frac{x+4}{ny+2n}=\frac{3}{1} \\ x+4=3ny+6n \\ \mathrm{From}\left(1\right) \\ 4ny+4=3ny+6n \\ ny=6n-4....\left(3\right) \\ \mathrm{From}\left(2\right)\mathrm{and}\left(3\right) \\ \frac{72n-8}{14}=6n-4 \\ \Rightarrow 72n-8=84n-56 \\ \Rightarrow 12n=48 \\ \Rightarrow n=4 \\ \mathrm{Hence},\mathrm{they}\mathrm{have}4\mathrm{children}. \end{gathered}$$