Maharashtra State BoardHSC Commerce 12th Board Exam
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The rate of growth of population is proportional to the number present. If the population doubled in the last 25 years and the present population is 1 lac, when will the city have population 4,00,000? - Mathematics and Statistics

Sum

The rate of growth of population is proportional to the number present. If the population doubled in the last 25 years and the present population is 1 lac, when will the city have population 4,00,000?

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Solution

Let ‘x’ be the population at time ‘t’ years.

∴ `dx/dt prop x`

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

∴ `dx/x = kdt`

Integrating on both sides, we get

`int dx/x = k int dt`

∴ log x = kt + c  …(i)

When t = 0, x = 50000

∴ log (50000) = k (0) + c

∴ c = log (50000)

∴ log x = kt + log (50000) …(ii)[From (i)]

When t = 25, x = 100000, we have

log (100000) = 25k + log (50000)

∴ log 2 = 25k

∴ `k = 1/25 log 2` …(iii)

When x = 400000, we get

`log (400000) = [ 1/25 log (2)] t + log (50000)` …[From (ii) and (iii)]

∴ `log [ 400000/50000] = t/25 log2`

∴ `log 8 = t/25 log2`

∴ `3log 2 = t/25 log2`

∴ `t/25 = 3`

∴ t = 75 years.

Thus, the population will be 4,00,000 after 75 – 25 = 50 years from present date.

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APPEARS IN

Balbharati Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board
Chapter 8 Differential Equation and Applications
Miscellaneous Exercise 8 | Q 4.09 | Page 173
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