Total students = 100
At t=0 (zero day), infected student =2
Let at t=t day infected student =x
∴ At t=t day non infected student = (100−x)
Rate of infection=dtdx
Given, dtdx∝x(100−x)
⇒∫x(100−x)dx=∫kdt
⇒1001∫x(100−x)100−x+xdx=kt+c
⇒1001∫(x1+100−x1)dx=kt+c
⇒1001[lnx−ln(100−x)]=kt+c
⇒1001ln100−xx=kt+c......(1)
Given, At, t=0,x=2
∴1001ln982=c
Putting value of c in equation (1), we get
1001ln100−xx=kt+1001ln982
⇒1001ln100−xx−1001ln982=kt
⇒1001ln2(100−x)x×98=kt
Given, At t=4,x=30
∴1001ln2(70)30×98=k×4
⇒k=4001ln21
∴1001ln2(100−x)x×98=t×4001×ln21
Now, when t = 8, then r = ?
1001ln(100−x)49x=8×4001×ln21
⇒ln(100−x)49x=2ln21
⇒100−x49x=212
⇒100−xx=4921×21
⇒100−xx=9
⇒x=900−9x
⇒10x=900
⇒x=90 
