Maharashtra State BoardHSC Commerce 12th Board Exam
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The population of a town increases at a rate proportional to the population at that time. If the population increases from 40 thousands to 60 thousands in 40 years ? - Mathematics and Statistics

Sum

The population of a town increases at a rate proportional to the population at that time. If the population increases from 40 thousands to 60 thousands in 40 years, what will be the population in another 20 years?

(Given: `sqrt(3/2)= 1.2247)`

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Solution

Let ‘x’ be the population at time ‘t’

∴ `dx/dt ∝ x`

∴`dx/dt = kx`, where k is the constant of proportionality.

∴`dx/dt = k  dt`

Integrating on both sides, we get

`int dx/x = k int 1 dt`

∴ log x = kt + c  …(i)

When t = 0, x = 40000

∴ log (40000) = k(0) + c

∴ c = log (40000)

∴ log x = kt + log (40000)  …(ii) [From (i)]

When, t = 40, x = 60000

∴ log (60000) = 40k + log (40000)

∴ 40k = log (60000) – log (40000)

∴ 40k = log `(60000 /40000)`

∴ k = `1/40 log (3/2)`  …(iii)

When t = 60, we get

log x = k(60) + log (40000)   …[From (ii)]

∴ log `x =[ 1/40 log (3/2)] (60) + log(40000)` …[From (iii)]

∴ log `x = 3/2 log (3/2) + log(40000)`

=`3/2 log (3/2)^(3/2) + log(40000)`

=`3/2 log (sqrt(3/2))^3 + log(40000)`

=`log (3/2sqrt(3/2) xx  log40000)`

∴ `log x = log( (3xx1.2247)/2xx40000)`

∴ x = 73482

∴ Population in another 20 years, i.e., in 60 years will be 73482.

  Is there an error in this question or solution?
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APPEARS IN

Balbharati Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board
Chapter 8 Differential Equation and Applications
Exercise 8.6 | Q 2 | Page 170
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