Answer :
Let's tackle the problem step-by-step with detailed calculations and explanations.
1. The population of a place increased to 54,000 in 2003 at a rate of 5% per annum.
(i) Find the population in 2001.
(ii) What would be its population in 2005?
### Part (i): Finding the Population in 2001
The population of the place in 2003 is 54,000, and the annual growth rate is 5%. We need to determine the population two years before 2003, which is in 2001.
To find the population in 2001, we can use the formula for population growth:
[tex]\[ P_{2003} = P_{2001} \times (1 + r)^t \][/tex]
where:
- [tex]\( P_{2003} \)[/tex] is the population in 2003,
- [tex]\( P_{2001} \)[/tex] is the population in 2001,
- [tex]\( r \)[/tex] is the annual growth rate,
- [tex]\( t \)[/tex] is the number of years.
We need to rearrange the formula to solve for [tex]\( P_{2001} \)[/tex]:
[tex]\[ P_{2001} = \frac{P_{2003}}{(1 + r)^t} \][/tex]
Given:
- [tex]\( P_{2003} = 54,000 \)[/tex]
- [tex]\( r = 0.05 \)[/tex]
- [tex]\( t = 2 \)[/tex] (since 2001 is 2 years before 2003)
Substituting in the values:
[tex]\[ P_{2001} = \frac{54,000}{(1 + 0.05)^2} \][/tex]
[tex]\[ P_{2001} = \frac{54,000}{(1.05)^2} \][/tex]
[tex]\[ P_{2001} \approx \frac{54,000}{1.1025} \][/tex]
[tex]\[ P_{2001} \approx 48,979.59 \][/tex]
So, the population in 2001 was approximately 48,979.59.
### Part (ii): Finding the Population in 2005
Next, we need to find the population two years after 2003, which is in 2005. We will use the same growth formula but now to find the population in 2005:
[tex]\[ P_{2005} = P_{2003} \times (1 + r)^t \][/tex]
Given:
- [tex]\( P_{2003} = 54,000 \)[/tex]
- [tex]\( r = 0.05 \)[/tex]
- [tex]\( t = 2 \)[/tex] (since 2005 is 2 years after 2003)
Substituting in the values:
[tex]\[ P_{2005} = 54,000 \times (1 + 0.05)^2 \][/tex]
[tex]\[ P_{2005} = 54,000 \times (1.05)^2 \][/tex]
[tex]\[ P_{2005} = 54,000 \times 1.1025 \][/tex]
[tex]\[ P_{2005} = 59,535 \][/tex]
So, the population in 2005 would be 59,535.
### Summary
- The population in 2001 was approximately 48,979.59.
- The population in 2005 would be approximately 59,535.
1. The population of a place increased to 54,000 in 2003 at a rate of 5% per annum.
(i) Find the population in 2001.
(ii) What would be its population in 2005?
### Part (i): Finding the Population in 2001
The population of the place in 2003 is 54,000, and the annual growth rate is 5%. We need to determine the population two years before 2003, which is in 2001.
To find the population in 2001, we can use the formula for population growth:
[tex]\[ P_{2003} = P_{2001} \times (1 + r)^t \][/tex]
where:
- [tex]\( P_{2003} \)[/tex] is the population in 2003,
- [tex]\( P_{2001} \)[/tex] is the population in 2001,
- [tex]\( r \)[/tex] is the annual growth rate,
- [tex]\( t \)[/tex] is the number of years.
We need to rearrange the formula to solve for [tex]\( P_{2001} \)[/tex]:
[tex]\[ P_{2001} = \frac{P_{2003}}{(1 + r)^t} \][/tex]
Given:
- [tex]\( P_{2003} = 54,000 \)[/tex]
- [tex]\( r = 0.05 \)[/tex]
- [tex]\( t = 2 \)[/tex] (since 2001 is 2 years before 2003)
Substituting in the values:
[tex]\[ P_{2001} = \frac{54,000}{(1 + 0.05)^2} \][/tex]
[tex]\[ P_{2001} = \frac{54,000}{(1.05)^2} \][/tex]
[tex]\[ P_{2001} \approx \frac{54,000}{1.1025} \][/tex]
[tex]\[ P_{2001} \approx 48,979.59 \][/tex]
So, the population in 2001 was approximately 48,979.59.
### Part (ii): Finding the Population in 2005
Next, we need to find the population two years after 2003, which is in 2005. We will use the same growth formula but now to find the population in 2005:
[tex]\[ P_{2005} = P_{2003} \times (1 + r)^t \][/tex]
Given:
- [tex]\( P_{2003} = 54,000 \)[/tex]
- [tex]\( r = 0.05 \)[/tex]
- [tex]\( t = 2 \)[/tex] (since 2005 is 2 years after 2003)
Substituting in the values:
[tex]\[ P_{2005} = 54,000 \times (1 + 0.05)^2 \][/tex]
[tex]\[ P_{2005} = 54,000 \times (1.05)^2 \][/tex]
[tex]\[ P_{2005} = 54,000 \times 1.1025 \][/tex]
[tex]\[ P_{2005} = 59,535 \][/tex]
So, the population in 2005 would be 59,535.
### Summary
- The population in 2001 was approximately 48,979.59.
- The population in 2005 would be approximately 59,535.
