The population of a town increased by 16 2/3 % in 1990 and 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?
Accepted Solution
A:
First, find the population at the end of 1990 At the end of 1990, the increase would be 16 2/3 % of 30,000, or we could write it as i = [tex]16 \frac{2}{3} [/tex]% × 30,000
We need to change percent into proper fraction i = [tex]16 \frac{2}{3} [/tex]% × 30,000 i = [tex] \frac{50}{3} [/tex]% × 30,000 percent means per hundred i = [tex]\frac{50}{3} .\frac{1}{100}[/tex] × 30,000 i = [tex]\frac{50}{300}[/tex] × 30,000 i = 5,000
At the end of 1990, the population would be p = 30,000 + 5,000 p = 35,000
Second, find the population at the end of 1991 The increase of the population would be 10% of 35,000, or we could write it as i = 10% × 35,000 i = 0.1 × 35,000 i = 3,500
The population at the end of 1991 would be p = 35,000 + 3,500 p = 38,500 This is the final answer