Answer
414.9k+ views
Hint: We will first discuss about the simple interest. Simple interest is the method to calculate the interest for money for a time period. Now the simple interest for a principal amount of $P$ at an interest rate of $R$ for a time period of $T$ years is
$SI=\dfrac{PRT}{100}$
So, from the above equation, the value of the principle amount is given by
$P=\dfrac{SI\times 100}{RT}$
Complete step-by-step solution:
Given that,
Simple interest $\left( SI \right)=4,320$
Rate of Interest $\left( R \right)=4$
Time period $\left( T \right)=6$years
From the formula of Simple Interest $SI=\dfrac{PRT}{100}$we have principle amount as
$\begin{align}
& P=\dfrac{SI\times 100}{RT} \\
& =\dfrac{4320\times 100}{4\times 6} \\
& =18000
\end{align}$
Hence the principle amount is Rs.$18000$
Note: In the problem they mentioned simple interest. If they mentioned compounded interest then we had to use the formula $A=P{{\left( 1+\dfrac{R}{n} \right)}^{nt}}$ where \[A\] is the Total amount, $P$ is the principle amount, $R$ is the interest rate in Decimals, $n$ is the number of times that the interest is compounded yearly, $t$ is the time period in years. So, assume the principle amount $P$ as $x$ and from given data $R=\dfrac{4}{100}=0.04$
$n$ is the number of times interest compounded yearly i.e. $n=1$ and time period is $t=6$years. Now the total amount is
$\begin{align}
& A=P{{\left( 1+\dfrac{R}{n} \right)}^{nt}} \\
& =x{{\left( 1+\dfrac{0.04}{1} \right)}^{1\left( 6 \right)}} \\
& =x{{\left( 1.004 \right)}^{6}} \\
& =1.0242x
\end{align}$
Now the compounded interest is given by
$\begin{align}
& CI=A-p \\
& =1.0242x-x \\
& =x\left( 1.0242-1 \right) \\
& =0.0242x
\end{align}$
But from the given data we have compounded interest as $CI=4,320$, then
$\begin{align}
& 0.0242x=4320 \\
& x=\dfrac{4320}{0.0242} \\
& =178512
\end{align}$
So, the principle amount if the interest is compounded annually is Rs.$1,78,512$
$SI=\dfrac{PRT}{100}$
So, from the above equation, the value of the principle amount is given by
$P=\dfrac{SI\times 100}{RT}$
Complete step-by-step solution:
Given that,
Simple interest $\left( SI \right)=4,320$
Rate of Interest $\left( R \right)=4$
Time period $\left( T \right)=6$years
From the formula of Simple Interest $SI=\dfrac{PRT}{100}$we have principle amount as
$\begin{align}
& P=\dfrac{SI\times 100}{RT} \\
& =\dfrac{4320\times 100}{4\times 6} \\
& =18000
\end{align}$
Hence the principle amount is Rs.$18000$
Note: In the problem they mentioned simple interest. If they mentioned compounded interest then we had to use the formula $A=P{{\left( 1+\dfrac{R}{n} \right)}^{nt}}$ where \[A\] is the Total amount, $P$ is the principle amount, $R$ is the interest rate in Decimals, $n$ is the number of times that the interest is compounded yearly, $t$ is the time period in years. So, assume the principle amount $P$ as $x$ and from given data $R=\dfrac{4}{100}=0.04$
$n$ is the number of times interest compounded yearly i.e. $n=1$ and time period is $t=6$years. Now the total amount is
$\begin{align}
& A=P{{\left( 1+\dfrac{R}{n} \right)}^{nt}} \\
& =x{{\left( 1+\dfrac{0.04}{1} \right)}^{1\left( 6 \right)}} \\
& =x{{\left( 1.004 \right)}^{6}} \\
& =1.0242x
\end{align}$
Now the compounded interest is given by
$\begin{align}
& CI=A-p \\
& =1.0242x-x \\
& =x\left( 1.0242-1 \right) \\
& =0.0242x
\end{align}$
But from the given data we have compounded interest as $CI=4,320$, then
$\begin{align}
& 0.0242x=4320 \\
& x=\dfrac{4320}{0.0242} \\
& =178512
\end{align}$
So, the principle amount if the interest is compounded annually is Rs.$1,78,512$
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)