Question

A sum of money invested at a certain rate of interest doubles itself in $8$ years . In much time will it triple itself at the same interest rate ?

Answer 1

Step 1: Find the interest rate.

Let principal $=₨\mathrm{P}$

Given:

Amount $=₨2\mathrm{P}$

Time $=8$years

Let Rate of interest $=\mathrm{R}\%$

Since, Amount $=$ Principal $+$ Simple Interest

$$\begin{gathered} \Rightarrow 2\mathrm{P}=\mathrm{P}+\mathrm{SI} \\ \Rightarrow \mathrm{SI}=2\mathrm{P}-\mathrm{P} \\ \Rightarrow \mathrm{SI}=\mathrm{P} \end{gathered}$$

We know that

Simple interest $=\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100}$

Then,

$\mathrm{P}=\frac{\mathrm{P}\times \mathrm{R}\times 8}{100}$
$$\begin{gathered} \Rightarrow \mathrm{R}=\frac{\mathrm{P}\times 100}{\mathrm{P}\times 8} \\ \Rightarrow \mathrm{R}=\frac{100}{8} \\ \Rightarrow \mathrm{R}=12.5\% \end{gathered}$$

Hence, the interest rate is $12.5\%$.

Step 2: Find the required time.

Given:

Amount $=₨3\mathrm{P}$

Rate of interest $=12.5\%$

Let Time $=\mathrm{T}$years

Since, Amount $=$ Principal $+$ Simple Interest

$$\begin{gathered} \Rightarrow 3\mathrm{P}=\mathrm{P}+\mathrm{SI} \\ \Rightarrow \mathrm{SI}=3\mathrm{P}-\mathrm{P} \\ \Rightarrow \mathrm{SI}=2\mathrm{P} \end{gathered}$$

We know that

Simple interest $=\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100}$

Then,

$2\mathrm{P}=\frac{\mathrm{P}\times 12.5\times \mathrm{T}}{100}$
$$\begin{gathered} \Rightarrow \mathrm{T}=\frac{2\mathrm{P}\times 100}{\mathrm{P}\times 12.5} \\ \Rightarrow \mathrm{T}=2\times 8 \\ \Rightarrow \mathrm{T}=16 \end{gathered}$$

Hence, the required answer is $16$ years.