Question

The population of a certain town was 50000. In a year,male population was increased by 5% and female population was increased by 3%. Now the population became 52020. Then what was the number of males and females in the previous year??

Answer 1

Dear Student,
Here is the solution of your asked query:
$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{male}\mathrm{population}\mathrm{was}\mathrm{x}\mathrm{and}\mathrm{number}\mathrm{of}\mathrm{female}\mathrm{population}\mathrm{was}\mathrm{y}. \\ \mathrm{So},\mathrm{total}\mathrm{number}\mathrm{of}\mathrm{population}=\mathrm{x}+\mathrm{y}=50000 \\ \mathrm{x}+\mathrm{y}=50000...\left(\mathrm{i}\right) \\ \mathrm{When}\mathrm{male}\mathrm{population}\mathrm{was}\mathrm{increased}\mathrm{by}5\%\mathrm{and}\mathrm{female}\mathrm{population}\mathrm{was}\mathrm{increased}\mathrm{by}3\%,\mathrm{we}\mathrm{have}; \\ \left(\mathrm{x}+\mathrm{x}\times \frac{5}{100}\right)+\left(\mathrm{y}+\mathrm{y}\times \frac{3}{100}\right)=52020 \\ \Rightarrow \left(\mathrm{x}+\frac{\mathrm{x}}{20}\right)+\left(\mathrm{y}+\frac{3\mathrm{y}}{100}\right)=52020 \\ \Rightarrow \frac{21\mathrm{x}}{20}+\frac{103\mathrm{y}}{100}=52020 \\ \Rightarrow 105\mathrm{x}+103\mathrm{y}=5202000...\left(\mathrm{ii}\right) \\ \mathrm{Multiplying}\left(\mathrm{i}\right)\mathrm{by}105\mathrm{and}\mathrm{subtracting}\left(\mathrm{ii}\right)\mathrm{from}\left(\mathrm{i}\right)\mathrm{we}\mathrm{get}; \\ 105\mathrm{x}+105\mathrm{y}-105\mathrm{x}-103\mathrm{y}=5250000-5202000 \\ \Rightarrow 2\mathrm{y}=48000 \\ \Rightarrow \mathrm{y}=24000 \\ \mathrm{Then},\mathrm{from}\left(\mathrm{i}\right)\mathrm{we}\mathrm{get}; \\ \mathrm{x}=50000-24000=26000 \\ \mathrm{Therefore},\mathrm{number}\mathrm{of}\mathrm{male}\mathrm{population}\mathrm{was}26000\mathrm{and}\mathrm{number}\mathrm{of}\mathrm{female}\mathrm{population}\mathrm{was}24000 \end{gathered}$$
Regards