# In a Laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5,06,000.

**Solution:**

Let the initial count of the bacteria be 'P' = 506000, n = 2, R = 2.5%

The count of the bacteria after 2 hours is assumed as 'A' and calculated as follows:

A = P[1 + (R/100)]^{n}

A = 506000[1 + (25/1000)]^{2}

A = 506000[1 + (1/40)]^{2}

A = 506000 × (41/40)^{2}

A = 506000 × (41/40) × (41/40)

A = 506000 × (1681/1600)

A = 506000 × 1.050625

A = 531616(approx.)

**Video Solution:**

## In a Laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5,06,000.

### Maths NCERT Solutions Class 8 - Chapter 8 Exercise 8.3 Question 11

**Summary:**

In a Laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. The bacteria at the end of 2 hours is 531616 if the count was initially 5,06,000.