Question

The population of a city increased to 2601000 in 2018 at a rate of 2 % p.a.
a) Find the population in 2016.
b)What would be it's population in 2020 ?​

Answer 1

$(2601000 \div 102) \times 100 = 2550000 \\ (2550000 \div 102) \times 100 = 2500000 \\ please \: mark \: me \: as \: branliest \: and \: follow \: me$

please mark me as brainliest

Answer 2

Step-by-step explanation:

Given :-

The population of a city increased to 2601000 in 2018 at a rate of 2 % p.a.

To find :-

Find the following :

a) Find the population in 2016 ?

b)What would be it's population in 2020 ?

Solution :-

a)

Given that

The population of a city in 2018 = 2601000

Increased percentage = 2% Per annum

Required population year = 2016

Time = 2018-2016 = 2 years

We have,

A = 2601000

R = 2%

T = 2 years

N = 2

We know that

A = P[1+(R/100)]^n

The population in 2016 =

=> 2601000 = P[1+(2/100)]²

=> 2601000 = P[1+(1/50)]²

=> 2601000 = P[(50+1)/50]²

=> 2601000 = P(51/50)²

=> 2601000 = P[(51×51)/(50×50)]

=> 2601000 = P(2601/2500)

=> P = (2601000×2500)/2601

=> P = 1000×2500

=> P = 2500000

The population in 2016 is 2500000

b)

The population of a city in 2018 = 2601000

Increased percentage = 2% Per annum

Required population year = 2020

Time = 2020-2018= 2 years

We have,

P= 2601000

R = 2%

T = 2 years

N = 2

We know that

A = P[1+(R/100)]^n

The population in 2020 =

=> A = 2601000[1+(2/100)]²

=> A =2601000 [1+(1/50)]²

=> A = 2601000[(50+1)/50]²

=> A = 2601000(51/50)²

=> A= 2601000[(51×51)/(50×50)]

=> A = 2601000(2601/2500)

=> A= (2601000×2601)/2500

=> A = 26010×2601/25

=> A = 67652010/25

=> A = 2706080.04

=> A = 2706080

The population in 2020 is 2706080

Answer:-

a) The population of a city in 2016 was 2500000

b) The population of a city in 2020 will be 2706080

Used formulae:-

→ A = P[1+(R/100)]^n

• A = amount
• P = Principle
• R = Rate of Interest
• n = number of times the compound interest is calculated