two years ago the population of a town was 10000 during first year it

1.	two years ago the population of a town was 10000 during first year it incresed at the rate of 5% per nannum and during the second year it increased at the rate of 6% per annum what is its present population
Answer» Given : Population of TOWN (2 years AGO) : 10000 RATE of incresion in first year : 5% Rate of incresion in second year : 6% To find : The PRESENT population of the town. Solution : First, we'll find the population of the town ONE year ago. Population of the town (1 year ago) : $\sf \dashrightarrow 10000 + (5\% \: of \: 10000)$ $\sf \dashrightarrow 10000 + \bigg( \dfrac{5}{100} \times 10000 \bigg)$ $\sf \dashrightarrow 10000 + \bigg( \dfrac{1}{20} \times 10000 \bigg)$ $\sf \dashrightarrow 10000 + \bigg( \dfrac{1 \times 10000}{20} \bigg)$ $\sf \dashrightarrow 10000 + \bigg( \dfrac{10000}{20} \bigg)$ $\sf \dashrightarrow 10000 + 500$ $\sf \dashrightarrow 10500$ Now, we can find the present population of the town. Present population of the town : $\sf \dashrightarrow 10500 + (6\% \: of \: 10500)$ $\sf \dashrightarrow 10500 + \bigg( \dfrac{6}{100} \times 10500 \bigg)$ $\sf \dashrightarrow 10500 + \bigg( \dfrac{3}{20} \times 10500 \bigg)$ $\sf \dashrightarrow 10500 + \bigg( \dfrac{3 \times 10500}{20} \bigg)$ $\sf \dashrightarrow 10500 + \bigg( \dfrac{31500}{20} \bigg)$ $\sf \dashrightarrow 10500 + 1575$ $\sf \dashrightarrow 12075$ Hence, the present population of the town is 12075.