Question

The population of a town is 1,21,000 if it increases at the rate of 10% per annum, what is the difference between the population 3 years hence and that of 2 year ago​

Answer 1

Answer:

Difference=(Population of 3 years- Population of 2 years ago)

Difference = (161051-98010)=63041

STEPS:

1) Population of 3 years=(121000*1.1*1.1*1.1)=161051

note : 10% increase =110/100= 1.1

for 3 years (1.1^3)

2) Population 2 years ago=(121000*0.9*0.9)=98010

note: 10% less population as it ago therefore 10 % less= 90/100= 0.9

for 2 years (0.9^2)

Answer 2

Solution :-

Population 3 years hence :-

→ Current Population = 121000

→ Rate = 10% per annum

→ Time = 3 years

So,

→ Population after 3 years = Current Population * [1 + (rate/100)]^(time)

→ Population after 3 years = 121000[1 + (10/100)]³

→ Population after 3 years = 121000[1 + (1/10)]³

→ Population after 3 years = 121000 * (11/10)³

→ Population after 3 years = 121000 * (1331/1000)

→ Population after 3 years = 161051

Population 2 years ago :-

→ Current Population = 121000

→ Rate = 10% per annum

→ Time = 2 years

So,

→ Current Population = Population 2 years ago * [1 + (rate/100)]^(time)

→ 121000 = Population 2 years ago[1 + (10/100)]²

→ 121000 = Population 2 years ago[1 + (1/10)]²

→ 121000 = Population 2 years ago * (11/10)²

→ 121000 = Population 2 years ago * (121/100)

→ Population 2 years ago = (121000 * 100)/121

→ Population 2 years ago = 100000

therefore,

→ Required difference = 161051 - 100000 = 61051 (Ans.)

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