The population of a town was 48600. It increa | vedclass

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(D) Initial population = $48600$.Population after the first year (increase of $25\%$): $48600 	imes (1 + \frac{25}{100}) = 48600 	imes \frac{125}{10

Vedclass · Accepted Answer

$55890$

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(D) Initial population = $48600$.Population after the first year (increase of $25\%$): $48600 	imes (1 + \frac{25}{100}) = 48600 	imes \frac{125}{100} = 48600 	imes 1.25 = 60750$.Population after the second year (decrease of $8\%$): $60750 	imes (1 - \frac{8}{100}) = 60750 	imes \frac{92}{100} = 60750 	imes 0.92 = 55890$.Therefore,the population at the end of $2$ years is $55890$.

The population of a town was $48600$. It increased by $25\%$ in the first year and decreased by $8\%$ in the second year. What will be the population of the town at the end of $2$ years?

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