Question

Suppose that an outbreak of cholera follows severe flooding in an isolated town of 2500 people. Initially (Day 0), 30 people are infected. Every day after, 18% of those still healthy fall ill. By what day will at least 87% of the population be infected?

Answer 1

Answer:

On 11th day ( approx )

Step-by-step explanation:

Since, if a population change with a constant rate,

Then final population,

$A=P(1-\frac{r}{100})^t$

Where,

P = initial population,

r = rate of change per period,

t = number of periods,

Given,

Initial population = 2500,

Out of which, infected population = 30,

So, healthy people, initially = 2500 - 30 = 2470

Every day after, 18% of those still healthy fall ill.

So, after x days the number of healthy people,

$P=2470(1-\frac{18}{100})^x=2470(1-0.18)^x=2470(0.82)^x----(1)$

Now, if 87% of 2470 are ill,

Then reaming healthy population = (100 - 87)% of 2470

= 13% of 2470

$=\frac{13\times 2470}{100}$

$=\frac{32110}{100}$

= 321.10

If P = 321.10,

From equation (1),

$321.10 = 2470(0.82)^x$

$\frac{321.10}{2470}=0.82^x$

$\log(\frac{321.1}{2470}) =x \log(0.82)$

$\implies x = \frac{\log(\frac{321.1}{2470})}{\log(0.82)}=10.2807315584$

Hence, by 11 day at least 87% of the population be infected.