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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?

Respuesta :

gustanika
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
shristiparmar1221

This question is based on the percentage. Therefore, the population  at the end of 1991  is 38,500.

Given:

The population of a town increased by  [tex]16\dfrac{2}{3} \%[/tex] in 1990 and 10% in 1991.

Now, we have to find the population at the end of 1990.

At the end of 1990, the increase  i would be  [tex]16\dfrac{2}{3} \%[/tex]  of 30,000, or it can be written as

[tex]i =16\dfrac{2}{3} \% \times 30000\\\\i = \dfrac{50}{3 \times 100} \times 30000\\\\i = \dfrac{30000}{6} \\\\i = 5000[/tex]

At the end of 1990, the population p would be,

p = 30,000 + 5,000

p = 35,000

Now, calculate the population at the end of 1991.

The increase of the population would be 10% of 35,000, or it can be written 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

Therefore, the population  at the end of 1991  is 38,500.

For more details, please prefer this link:

https://brainly.com/question/13703753

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