The population of a town increased by 10% in 1 year and 15% on secound year find the netchange in %
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We can proceed with a layman method by finding the population every year by taking into account the corresponding percent changes every year..
OR
We can directly go for the compound interest and depreciation formula.
Initially, the population of the town is P = 100000
For the next two years, the population increased by 10% each year.
The population after 2 years, P' = P(1+10100)2 =100000(1110)2 =121000P1101002 10000011102 121000
In the next two years, the population decreased by 1o% each year from P' to P''.
P'' =P(1−10100)2 =121000(910)2 =98010P1101002 1210009102 98010
Therefore, the population of the town after 4 years is 98,010.
I hope it helps!
OR
We can directly go for the compound interest and depreciation formula.
Initially, the population of the town is P = 100000
For the next two years, the population increased by 10% each year.
The population after 2 years, P' = P(1+10100)2 =100000(1110)2 =121000P1101002 10000011102 121000
In the next two years, the population decreased by 1o% each year from P' to P''.
P'' =P(1−10100)2 =121000(910)2 =98010P1101002 1210009102 98010
Therefore, the population of the town after 4 years is 98,010.
I hope it helps!
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