Question

The number of bacteria in a certain culture doubles every hour if there were 40 bacteria present in the cultire originally, the number of bacteria will be present at the end of 2nd hour and 4th hour are respectively:
(a) 40,160
(b) 40,80
(c) 80,160
(d) 160,640

Answer 1

The increase in the number of bacteria in this case follows a geometric progression.

The first term, a = 40
The common ratio, r = 2

For the first part, n = 2
For the second part, n = 4

The total number of bacteria is, therefore, given by the formula;

Nth term = ar^n - 1

For the first part;

At the end of the 2nd hour, we calculate the 3rd term

3rd Term= 40[(2^(3-1)] = 40(2²) = 40(4) = 160

For the second part;

At the end of the 4th hour, we calculate the 5th term

5th term = 40[(2^(5-1)] = 40(2^4) = 40(16) = 640

Therefore the correct answer is (d) 160, 640

Answer 2

Solution :-

The case of increasing bacteria is an example of geometric progression.

Given -

a = first term = 40 and r = common ratio = 2 (as the bacteria doubles every hour)

Xn = ar⁽ⁿ ⁻ ¹⁾

We have to find the 3rd term (number of bacteria will be present after the end of 2nd hour) and the 5th term (number of bacteria present after the end of 4th hour)

1) Bacteria after the end of 2nd hour

⇒ 40*[2⁽³ ⁻ ¹⁾]

⇒ 40*(2)²

⇒ 40*4

= 160 bacteria

2) Bacteria after the end of 4th hour

⇒ 40*[2⁽⁵ ⁻ ¹⁾]

⇒ 40*(2)⁴

⇒ 40*16

= 640 bacteria

So, after the end of 2nd hour the number of bacteria will be 160 and after the end of 4th hour the number of bacteria will be 640

So, option (d) is correct.

Answer.