
Ishita invested a sum of Rs. 12,000 at 5% per annum compound interest. She received an amount of Rs. 13,230 after n years. Find the value of n.
Answer
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Hint: Here, we will the formula for the compound interest on a certain sum which is given as $CI=p{{\left( 1+\dfrac{r}{n} \right)}^{{{n}_{o}}t}}-p$. Using this formula we will find the value of n. Since, we know the principal amount, interest and the rate. So, we can easily find the value of n.
Complete step by step answer:
We know that compound interest is the addition of interest to the principal sum of a loan or deposit. It is the result of reinvesting interest, rather than paying it out, so that interest on the next period is then earned on the principal sum plus previously accumulated interest. Compound interest is standard in finances.
Compound interest is controlled with simple interest, where previously accumulated interest is not added to the principal amount of the current period, so there is no compounding.
The value of the total compound interest generated is the final value of principal minus the initial value of principal. So, we have:
$CI=p{{\left( 1+\dfrac{r}{n} \right)}^{{{n}_{o}}t}}-p.................\left( 1 \right)$
Here, P is principal amount, r is annual interest rate, ${{n}_{0}}$ is compounding frequency and t is the overall time for which interest is applied.
We have:
P = Rs. 12000
r = 5%
${{n}_{0}}$= 1
t = n
And, the final principal amount is = Rs. 13230
Therefore, interest is = 13230 – 12000 = Rs. 1230.
On substituting these values in equation (1), we get:
$\begin{align}
& 1230=12000{{\left( 1+\dfrac{5}{1} \right)}^{1}}-12000 \\
& \Rightarrow 1230+12000=12000\times {{6}^{n}} \\
& \Rightarrow 13230=12000\times {{6}^{n}} \\
& \Rightarrow {{6}^{n}}=\dfrac{13230}{12000} \\
& \Rightarrow {{6}^{n}}=1.1025 \\
\end{align}$
On taking log to both sides of this equation, we get:
$\begin{align}
& \log {{6}^{n}}=\log 1.1025 \\
& \Rightarrow n\log 6=\log 1.1025 \\
& \Rightarrow n\times 0.77=0.04 \\
& \Rightarrow n=\dfrac{0.04}{0.77}=0.05 \\
\end{align}$
Hence, the value of n is 0.05 years.
Note: Students should note here that the value of ${{n}_{0}}$ is always given by the reciprocal of the compounded time period. Since, here the time period is 1 year, so the value of ${{n}_{0}}$ is also 1 year.
Complete step by step answer:
We know that compound interest is the addition of interest to the principal sum of a loan or deposit. It is the result of reinvesting interest, rather than paying it out, so that interest on the next period is then earned on the principal sum plus previously accumulated interest. Compound interest is standard in finances.
Compound interest is controlled with simple interest, where previously accumulated interest is not added to the principal amount of the current period, so there is no compounding.
The value of the total compound interest generated is the final value of principal minus the initial value of principal. So, we have:
$CI=p{{\left( 1+\dfrac{r}{n} \right)}^{{{n}_{o}}t}}-p.................\left( 1 \right)$
Here, P is principal amount, r is annual interest rate, ${{n}_{0}}$ is compounding frequency and t is the overall time for which interest is applied.
We have:
P = Rs. 12000
r = 5%
${{n}_{0}}$= 1
t = n
And, the final principal amount is = Rs. 13230
Therefore, interest is = 13230 – 12000 = Rs. 1230.
On substituting these values in equation (1), we get:
$\begin{align}
& 1230=12000{{\left( 1+\dfrac{5}{1} \right)}^{1}}-12000 \\
& \Rightarrow 1230+12000=12000\times {{6}^{n}} \\
& \Rightarrow 13230=12000\times {{6}^{n}} \\
& \Rightarrow {{6}^{n}}=\dfrac{13230}{12000} \\
& \Rightarrow {{6}^{n}}=1.1025 \\
\end{align}$
On taking log to both sides of this equation, we get:
$\begin{align}
& \log {{6}^{n}}=\log 1.1025 \\
& \Rightarrow n\log 6=\log 1.1025 \\
& \Rightarrow n\times 0.77=0.04 \\
& \Rightarrow n=\dfrac{0.04}{0.77}=0.05 \\
\end{align}$
Hence, the value of n is 0.05 years.
Note: Students should note here that the value of ${{n}_{0}}$ is always given by the reciprocal of the compounded time period. Since, here the time period is 1 year, so the value of ${{n}_{0}}$ is also 1 year.
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