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The population of a town increases every year by 4%. If its present population is 50,000, then after 2 years it will be
Answer & Solution
Correct Answer:
Option
C
Here we can use the compound interest based formula,
$$\eqalign{ & {\text{Population after }}n{\text{ years}} \cr & = P \times {\left[ {1 + {\frac{r}{{100}}} } \right]^n} \cr & {\text{Population after 2 years}} \cr & = 50000 \times {\left[ {1 + {\frac{4}{{100}}} } \right]^2} \cr} $$
Population after 2 years = 54080
Alternatively,
we can use, net percentage change graphic as well,
50,000------4%↑---→ 52,000---- 4%↑---→ 54,080.
Then, population after 2 years= 54,080.
In this calculation, we need to find 1% of 50,000 first, which is easily calculated by dividing 50,000 by 100.
$$\eqalign{ & {\text{Population after }}n{\text{ years}} \cr & = P \times {\left[ {1 + {\frac{r}{{100}}} } \right]^n} \cr & {\text{Population after 2 years}} \cr & = 50000 \times {\left[ {1 + {\frac{4}{{100}}} } \right]^2} \cr} $$
Population after 2 years = 54080
Alternatively,
we can use, net percentage change graphic as well,
50,000------4%↑---→ 52,000---- 4%↑---→ 54,080.
Then, population after 2 years= 54,080.
In this calculation, we need to find 1% of 50,000 first, which is easily calculated by dividing 50,000 by 100.
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Loginso, 2000+50000 = 52000
4% of 52000 = 2080
so, 2080+52000 = 54080
Good method
2year% is 4000
50,000+4000=54000
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