The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was 20, 000 in 1999 and 25000 in the year 2004, what will be the population of the

Options

A. 31250
B. 31251
C. 312510
D. 31350

Answer

b

Explanation

Let y be the population at time t , then `(dy)/(dt ) prop y.`
` rArr (dy)/(dt) = ky` ( where , k is constant )
` rArr (dy)/y = k dt `
On integrating both sides , we get
` log y = kt +C " "` …(i)
In the year 1999 , t = 0 , y = 20000
On putting these values in Eq. (i) ,we get
`log 20000 = k (0) +C`
` rArr log 20000 = C " "` ...(ii)
In the year 2004, t = 5 yr , y = 25000 ltbRgt On putting these values in Eq. (i) , we get
` log 25000 = k * 5 +C`
` rArr log 25000 = 5k + log 20000 ` [ using Eq. (ii) ]
` rArr 5k = log ((25000)/20000) = log (5/4)`
` rArr k = 1/5 log 5/4`
In the year 2009 , t= 10 yr
On putting these values of t, k and C in Eq. ( i ), we get
` log y = 10 xx 1/5 log (5/4) + log (20000)`
` rArr log y = log [ 20000 xx (5/4)^(2)]`
` [ :' log m + log n = log mn and ` ltbRgt ` n log m = log m^(n)]`
` rArr y = 20000 xx 5/4 xx5/4 rArr y = 31 250`
Hence , the population of the village in 2009 will be 31250