Question

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 studetns on 8^th day will be ______.

Options

• 60

• 70

• 80

• 90

Answer 1

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 studetns on 8^th day will be 90.

Explanation:

Let the number of infected students be n.

Non-infected students be (100 – n).

`(dn)/(dt) ∝ n(100 - n) \implies (dn)/(dt)` = kn(100 – n)

`(dn)/(n(100 - n)) = k  dt \implies 1/100[((n + 100 - n)dn)/(n(100 - n))]` = k dt

`\implies 1/100 [(dn)/n + (dn)/(100 - n)]` = k dt

Take integral both side

`1/100 int (dn)/n + 1/100 int (dn)/((100 - n)) = k int dt`

`1/100  ln (n) - 1/100  ln(100 - n)` = kt + c

`1/100 ln(n/(100 - n)) = kt + c \implies ln(n/(100 - n))` = 100kt + 100c

`(n/(100 - n))` = e^100kt . c'   ...(i)

At, t = 0, n = 2,

`(2/(100 - 2)) = e^0 . c^' \implies c^' = 2/98 = 1/49`

At, t = 4, n = 30,

`(30/(100 - 30)) = e^(400k) . 1/49 \implies (30 xx 49)/70` = e^400k

400 k = ln (21) k = `1/400  ln (21)`

`(n/(100 - n)) = e^(100 xx 1/400 ln (21).t) xx 1/49` ...[From (i)]

Put t = 8 in the above equation.

`(n/(100 - n)) = (e^((1/4 ln(21)).8))/49 \implies (n/(100 - n)) = (21)^(1/4 xx 8)/49`

`(n/(100 - n)) = 441/49 \implies` n = 90.