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
Answer: Option C
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.
This Question Belongs to Arithmetic Ability >> Percentage
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Comments ( 12 )
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Wait no answer should be 54000 because they are asking for after 2year population that's mean 4% in first year and 4% in 2 nd year and the 4% of 50000 is 2000 then after including 2 year population 50000+2000+2000 = 54000
Good method 54080 confirm answer
Answer should be 52000.
Fayezur good method
4% of 50000 = 2000
so, 2000+50000 = 52000
4% of 52000 = 2080
so, 2080+52000 = 54080
Morning, ଶୁଭ ସକାଳ
Good method
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
he did not mention whether it is simple or compound interest
I think this solution is correct.
Mian,u r wrong.
wrong answer
2year% is 4000
50,000+4000=54000
how its possible as in second there is 54000 answer and in first there is 54080
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