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The population of a town increases by 3% every year. In the year 2000, the population was 3,000. Find the population in the year 2003.

A. 3,182
B. 3,278
C. 6,591
D. 7,515

Sagot :

To determine the population of the town in the year 2003, given that it increases by [tex]\(3\%\)[/tex] every year and the population in the year 2000 was 3,000, we will calculate the population year by year.

### Step-by-Step Solution:

1. Initial Population (Year 2000):
[tex]\[ \text{Population in 2000} = 3,000 \][/tex]

2. Population in Year 2001:
The population increases by [tex]\(3\%\)[/tex] every year. Therefore, the population at the end of 2001 will be:
[tex]\[ \text{Population in 2001} = 3,000 \times (1 + 0.03) \][/tex]
[tex]\[ \text{Population in 2001} = 3,000 \times 1.03 = 3,090 \][/tex]

3. Population in Year 2002:
Now the population in 2001 becomes the base population for 2002. Hence, we calculate:
[tex]\[ \text{Population in 2002} = 3,090 \times (1 + 0.03) \][/tex]
[tex]\[ \text{Population in 2002} = 3,090 \times 1.03 = 3,182.7 \approx 3,182 \][/tex]
(We round to the nearest whole number)

4. Population in Year 2003:
Similarly, the population in 2002 becomes the base population for 2003. Therefore:
[tex]\[ \text{Population in 2003} = 3,182 \times (1 + 0.03) \][/tex]
[tex]\[ \text{Population in 2003} = 3,182 \times 1.03 = 3,277.46 \approx 3,278 \][/tex]
(Again, rounding to the nearest whole number)

Therefore, the population of the town in the year 2003 is:

[tex]\[ \boxed{3,278} \][/tex]

None of the other options provided matches our calculated value accurately. Hence, the correct answer to the problem is [tex]\(3,278\)[/tex].