# The population of a town increases every year by 4%. If its present population is 50,000, then after 2 years it will be

A. 53,900

B. 54,000

C. 54,080

D. 54,900

### Solution(By Examveda Team)

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.

1. Fayezur good method

2. 4% of 50000 = 2000
so, 2000+50000 = 52000

4% of 52000 = 2080
so, 2080+52000 = 54080

3. Morning, ଶୁଭ ସକାଳ
Good method

4. 1 year increase is= 4/100*50000= 2000. now add 2000 and 50000 will be 52000. again 4/100*52000= 2080. now add 2080 with 52000 will 54080

5. he did not mention whether it is simple or compound interest

6. I think this solution is correct.

7. Mian,u r wrong.

8. 9. 