# Ex.8.3 Q11 Comparing Quantities Solutions - NCERT Maths Class 8

## Question

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 \(\rm{}\,5, 06,000\).

## Text Solution

**What is known?**

Original Count, Time Period and Rate of Increase

**What is unknown?**

The total count after \(2\) hours

**Reasoning:**

\(A = P \left( 1 + \frac{r}{100} \right)^{\rm{n}}\)

\(P= \rm{}\,5,06,000\)

\(N = 2 \) hours

\(R= 2.5\% \) hour \(=\begin{align}\frac{{25}}{{10}} \end{align}\)hours

**Steps:**

\[\begin{align}A &= P\left( {{1 + }\frac{{r}}{{{100}}}} \right)^{n} \\ &= 506000\left( {{1 + }\frac{{{25}}}{{{1000}}}} \right)^{2} \\ &= 506000\left( {{1 + }\frac{{1}}{{{40}}}} \right)^{2} \\ &= 506000\left( {\frac{{41}}{{40}}} \right)^2 \\ &= 506000 \times \frac{{41}}{{40}} \times \frac{{41}}{{40}} \\ &= 506000 \times \frac{{1681}}{{1600}} \\ &= {506000} \times 1.050625 \\ &= 531616 \\ \end{align}\]

The total count of bacteria after \(2\) hours \( =\rm{}\, 531616\)