Question

The virus of a culture decreases at the rate of 5% due to a medicine. If the virus count in the culture at 10:00 am was 2.3 ×10 to the power 7 , find the virus count at 1:00 pm on the same day

Answer 1

Answer:

The virus count at 1:00 PM is $1.97\times 10^7$

Step-by-step explanation:

Assuming that the virus is decreasing at the rate of 5% per hour

If the initial number of virus are N and the rate of decreasing is r then after n hours, the virus remaining will be

$N'=N(1-\frac{r}{100})^n$

From 10 AM to 1 PM total number of hours passed = 3

Given

$N=2.3\times 10^7$

$r=5\%$

Thus,

$N'=N(1-\frac{r}{100})^n$

$N'=2.3\times 10^7(1-\frac{5}{100})^3$

$\implies N'=2.3\times 10^7(1-\frac{1}{20})^3$

$\implies N'=2.3\times 10^7(\frac{19}{20})^3$

$\implies N'=2.3\times 10^7\times \frac{6859}{8000}$

$\implies N'=1.97\times 10^7$

Hope this is helpful.

Answer 2

Answer:

Assuming that the virus is decreasing at the rate of 5% per hour

If the initial number of virus are N and the rate of decreasing is r then after n hours, the virus remaining will be

N'=N(1-\frac{r}{100})^nN

′

=N(1−

100

r

)

n

From 10 AM to 1 PM total number of hours passed = 3

Given

N=2.3\times 10^7N=2.3×10

7

r=5\%r=5%

Thus,

N'=N(1-\frac{r}{100})^nN

′

=N(1−

100

r

)

n

N'=2.3\times 10^7(1-\frac{5}{100})^3N

′

=2.3×10

7

(1−

100

5

)

3

\implies N'=2.3\times 10^7(1-\frac{1}{20})^3⟹N

′

=2.3×10

7

(1−

20

1

)

3

\implies N'=2.3\times 10^7(\frac{19}{20})^3⟹N

′

=2.3×10

7

(

20

19

)

3

\implies N'=2.3\times 10^7\times \frac{6859}{8000}⟹N

′

=2.3×10

7

×

8000

6859

\implies N'=1.97\times 10^7⟹N

′

=1.97×10