Question

the population of a town is 10000 and was 8000 two years ago. If it grows at the same rate, what will it be 2yr hence ?​

Answer 1

Answer:

There Are Two Methods. Pick One Method You Think It Is Easy

Step-by-step explanation:

First Method:-

The population of a town is 10000 now and was 8000 two years ago.If it grows at the same rate ,

The population of a town is 10000 now and was 8000 two years ago.If it grows at the same rate ,population after 2 years = 8000*(10000/8000)*(10000/8000) = 12500

Second Method:-

Let x be the rate at which population grows

Let x be the rate at which population grows=> before one year , population = 8000 + x% of 8000 = (8000 + 80x) = y

Let x be the rate at which population grows=> before one year , population = 8000 + x% of 8000 = (8000 + 80x) = yPresent population = y + x % of y

Let x be the rate at which population grows=> before one year , population = 8000 + x% of 8000 = (8000 + 80x) = yPresent population = y + x % of y=> 10000= 8000 + (80x) + 0.8x^2 + 80x

Let x be the rate at which population grows=> before one year , population = 8000 + x% of 8000 = (8000 + 80x) = yPresent population = y + x % of y=> 10000= 8000 + (80x) + 0.8x^2 + 80xsolving for x, x = 11.8

Let x be the rate at which population grows=> before one year , population = 8000 + x% of 8000 = (8000 + 80x) = yPresent population = y + x % of y=> 10000= 8000 + (80x) + 0.8x^2 + 80xsolving for x, x = 11.8So after two years, population will be

Let x be the rate at which population grows=> before one year , population = 8000 + x% of 8000 = (8000 + 80x) = yPresent population = y + x % of y=> 10000= 8000 + (80x) + 0.8x^2 + 80xsolving for x, x = 11.8So after two years, population will be10000 + 1180 + 11.8(10000+1180)/100 = 12500

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Answer 2

hi bro just do like this

10000 - 8000 = 2000

=> 10000 + 2000 + 2000 = 14000

this means after 2 yrs it will be 14000