How to solve this maths problem?

Question

1.)

starting from the year 2000,the population of a certain new town increases steadily by r% every year.given that the population in 2001 and 2004 are 6000 and 10368 respectively,find

a.) r and the population in 2000.

b.) the population of the new town in 2010

Answer 1

if we use formula

polulation in 2004 = pop(2001)(1+r)^3

10368= 6000(1+r)^3

we get r=0.2

poplution in 2000 = 6000-.2%= 5988

from above formula u can alculate (b)

Answer 2

in 2004 the population would be (3 years later than 2001)

P(2004)=P(2001)*(1+r)^3

10368/6000= (1+r^3)=1.728

1+r= (1.728)^(1/3)=1.2  so r+1=1.2 r=0.2

In 2010 after 9 years P(2010) =6000*1.2^9=30958

Answer 3

r=24.26    ,population in the year 2000=4544 ,population in the year 2010=19103

Answer 4

Formula for future population is

F = P(1 + r)^n

where:

P = present population

F = future population

r = rate of increase

n = number of years

1) You need two equations because there are two unknowns, r and P

    6000 = P(1 + r)^1 ---->1

   10368 = P (1 + r)^10 ---->2

From 1,

  P = 6000/(1 + r)^1 ---> substitute to 2

Take it from there

2) When you find r, use the above equation to find the population on 2010.

Answer 5

hi

  6000+3r=10368

r=1456

in 2000:    P = 4544

in 2010:    P =13104

Answer 6

Formula for future population is

F = P(1 + r)^n

where:

P = present population

F = future population

r = rate of increase

n = number of years

1) You need two equations because there are two unknowns, r and P

6000 = P(1 + r)^1 ---->1

10368 = P (1 + r)^10 ---->2

From 1,

P = 6000/(1 + r)^1 ---> substitute to 2

Take it from there

2) When you find r, use the above equation to find the population on 2010.