The population of a town increased by \( 16 \frac{2}{3} \% \) in 1990 and by \( 10 \% \) in 1991 . If the population of the town at the beginning of 1990 was 30,000 , what was it at the end of 1991 ?

To find the population at the end of 1991, we first calculate the population after the increase in 1990. Beginning of 1990 population: 30,000 Increase in 1990: \[ 16 \frac{2}{3} \% = \frac{50}{3} \% = \frac{50}{3} \times \frac{1}{100} = \frac{50}{300} = \frac{1}{6} \] So, the increase is: \[ \text{Increase in 1990} = 30,000 \times \frac{1}{6} = 5,000 \] Thus, the population at the end of 1990 is: \[ 30,000 + 5,000 = 35,000 \] Now, for the year 1991, we apply the \(10\%\) increase: \[ 10 \% = 0.10 \] Thus, the increase is: \[ \text{Increase in 1991} = 35,000 \times 0.10 = 3,500 \] Therefore, the population at the end of 1991 is: \[ 35,000 + 3,500 = 38,500 \] The population of the town at the end of 1991 is **38,500**.