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

  • 153,900
  • 254,000
  • 354,080
  • 454,900
Answer:- 1
Explanation:-

Solution:
Here we can use the compound interest based formula,Population after n years=P×[1+(r100)]nPopulation after 2 years=50000×[1+(4100)]2Population 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.

Post your Comments

Your comments will be displayed only after manual approval.


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. ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "
Solution:
Here we can use the compound interest based formula,Population after n years=P×[1+(r100)]nPopulation after 2 years=50000×[1+(4100)]2Population 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. ", "dateCreated": "7/24/2019 10:09:12 AM" } }
Test
Classes
E-Book