Certainly! Let's tackle this problem step-by-step.
### Given:
- The population of a place in 2003 is 54,000.
- The annual growth rate is 5%.
We need to find:
1. The population in 2001.
2. The population in 2005.
### (i) Find the population in 2001
We know the population in 2003 is 54,000 and it has been increasing at a rate of 5% per annum. To find the population in 2001, we need to work backwards and see what the population was 2 years before 2003.
Let the population in 2001 be [tex]\( P_{2001} \)[/tex].
Since the population grows at a rate of 5% per annum, the population in 2003 can be expressed in terms of the population in 2001 as:
[tex]\[ 54000 = P_{2001} \times (1 + 0.05)^2 \][/tex]
To isolate [tex]\( P_{2001} \)[/tex]:
[tex]\[ P_{2001} = \frac{54000}{(1 + 0.05)^2} \][/tex]
So, we calculate:
[tex]\[ P_{2001} = \frac{54000}{1.05^2} \][/tex]
[tex]\[ P_{2001} = \frac{54000}{1.1025} \][/tex]
[tex]\[ P_{2001} \approx 48979.59183673469 \][/tex]
Therefore, the population in 2001 was approximately 48,979.59.
### (ii) Find the population in 2005
Similarly, to find the population in 2005, we can use the population in 2003 and project it forward 2 years.
Let the population in 2005 be [tex]\( P_{2005} \)[/tex].
Since the population grows at a rate of 5% per annum, the population in 2005 can be expressed in terms of the population in 2003 as:
[tex]\[ P_{2005} = 54000 \times (1 + 0.05)^2 \][/tex]
So, we calculate:
[tex]\[ P_{2005} = 54000 \times 1.05^2 \][/tex]
[tex]\[ P_{2005} = 54000 \times 1.1025 \][/tex]
[tex]\[ P_{2005} = 59535.0 \][/tex]
Therefore, the population in 2005 would be 59,535.
### Summary:
- The population in 2001 was approximately 48,979.59.
- The population in 2005 would be 59,535.
