A hostel has 100 students. On a certain day (consider it day zero) it was found that two students are infected with some virus. Assume that the rate at which the virus spreads is directly proportional to the product of the number of infected students and the number of non-infected students. If the number of infected students on 4^th day is 30, then number of infected students on 8^th day will be __________.

<p>Total students = 100</p> <p>At t = 0 (zero day), infected student = 2</p> <p>Let at t = t day infected student = x</p> <p>$\therefore$ At t = t day non infected student = (100 $-$ x)</p> <p>Rate of infection $= {{dx} \over {dt}}$</p> <p>Given, ${{dx} \over {dt}} \propto x(100 - x)$</p> <p>$$ \Rightarrow \int\limits_{}^{} {{{dx} \over {x(100 - x)}} = \int\limits_{}^{} {k\,dt} } $$</p> <p>$$ \Rightarrow {1 \over {100}}\int\limits_{}^{} {{{100 - x + x} \over {x(100 - x)}}dx = k\,t + c} $$</p> <p>$$ \Rightarrow {1 \over {100}}\int\limits_{}^{} {\left( {{1 \over x} + {1 \over {100 - x}}} \right)dx = k\,t + c} $$</p> <p>$\Rightarrow {1 \over {100}}\left[ {\ln x - \ln (100 - x)} \right] = k\,t + c$</p> <p>$\Rightarrow {1 \over {100}}\ln {x \over {100 - x}} = k\,t + c$ ...... (1)</p> <p>Given, At, t = 0, x = 2</p> <p>$\therefore$ ${1 \over {100}}\ln {2 \over {98}} = c$</p> <p>Putting value of c in equation (1), we get</p> <p>${1 \over {100}}\ln {x \over {100 - x}} = kt + {1 \over {100}}\ln {2 \over {98}}$</p> <p>$$ \Rightarrow {1 \over {100}}\ln {x \over {100 - x}} - {1 \over {100}}\ln {2 \over {98}} = kt$$</p> <p>$\Rightarrow {1 \over {100}}\ln {{x \times 98} \over {2(100 - x)}} = kt$</p> <p>Given, At t = 4, x = 30</p> <p>$\therefore$ ${1 \over {100}}\ln {{30 \times 98} \over {2(70)}} = k \times 4$</p> <p>$\Rightarrow k = {1 \over {400}}\ln 21$</p> <p>$\therefore$ $${1 \over {100}}\ln {{x \times 98} \over {2(100 - x)}} = t \times {1 \over {400}} \times \ln 21$$</p> <p>Now, when t = 8, then r = ?</p> <p>$${1 \over {100}}\ln {{49x} \over {(100 - x)}} = 8 \times {1 \over {400}} \times \ln 21$$</p> <p>$\Rightarrow \ln {{49x} \over {(100 - x)}} = 2\ln 21$</p> <p>$\Rightarrow {{49x} \over {100 - x}} = {21^2}$</p> <p>$\Rightarrow {x \over {100 - x}} = {{21 \times 21} \over {49}}$</p> <p>$\Rightarrow {x \over {100 - x}} = 9$</p> <p>$\Rightarrow x = 900 - 9x$</p> <p>$\Rightarrow 10x = 900$</p> <p>$\Rightarrow x = 90$</p>