Question

The population of city is 33800, increases at rate 30 % p.a. Find population 2 years ago?


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Answer 1

Answer:

Let the polulation a year ago be x

Let the polulation a year ago be xSo, x+ 30x/100 = 338000 [30% increase means x + 30x/100]

Let the polulation a year ago be xSo, x+ 30x/100 = 338000 [30% increase means x + 30x/100]130x/100= 338000

Let the polulation a year ago be xSo, x+ 30x/100 = 338000 [30% increase means x + 30x/100]130x/100= 338000X = 338000 x 100/ 130

Let the polulation a year ago be xSo, x+ 30x/100 = 338000 [30% increase means x + 30x/100]130x/100= 338000X = 338000 x 100/ 130X= 260000

Let the polulation a year ago be xSo, x+ 30x/100 = 338000 [30% increase means x + 30x/100]130x/100= 338000X = 338000 x 100/ 130X= 260000Let y be the polulation 2 years back

Let the polulation a year ago be xSo, x+ 30x/100 = 338000 [30% increase means x + 30x/100]130x/100= 338000X = 338000 x 100/ 130X= 260000Let y be the polulation 2 years backY + 30y/100 = 260000

Let the polulation a year ago be xSo, x+ 30x/100 = 338000 [30% increase means x + 30x/100]130x/100= 338000X = 338000 x 100/ 130X= 260000Let y be the polulation 2 years backY + 30y/100 = 260000130y/100 = 260000

Let the polulation a year ago be xSo, x+ 30x/100 = 338000 [30% increase means x + 30x/100]130x/100= 338000X = 338000 x 100/ 130X= 260000Let y be the polulation 2 years backY + 30y/100 = 260000130y/100 = 260000Y = 260000 x 100/ 130

Let the polulation a year ago be xSo, x+ 30x/100 = 338000 [30% increase means x + 30x/100]130x/100= 338000X = 338000 x 100/ 130X= 260000Let y be the polulation 2 years backY + 30y/100 = 260000130y/100 = 260000Y = 260000 x 100/ 130Y= 200000

Answer 2

Answer:

Let initial population be p

As population increased by 30% per year.

So, population after 2 years =p(1+

100

30

)

2

1.69p

338000=1.69p

⟹p=2,00,000

So, population two years ago=2,00,000

Step-by-step explanation:

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