
The difference between the compound interest compounded annually and the simple interest on a certain sum $2$ years at $6\%$ per annum is $Rs.18.$ Find the sum.
Answer
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Hint: We know that the simple interest can be found by using the formula given by $S.I=\dfrac{P\times T\times R}{100}.$ We also know that the compound interest compounded annually can be found by using the formula $C.I=P{{\left( 1+\dfrac{R}{100} \right)}^{T}}-P.$
Complete step by step solution:
Let us consider the given problem for finding the sum when the difference of compound interest compounded annually and the simple interest on certain sum $2$ years at $6\%$ per annum is $Rs.18.$
Now, we can write the difference between compound interest and simple interest is equal to $18.$ That is, we write $C.I-S.I=18.$
We know that the compound interest compounded annually can be found by using the formula $C.I=P{{\left( 1+\dfrac{R}{100} \right)}^{T}}-P$ where $P$ is the principal amount, $R$ is the rate of interest and $T$ is the time duration in years.
Similarly, we can find the simple interest by using the formula $S.I=\dfrac{P\times R\times T}{100}$ where $P$ is the principal amount, $R$ is the rate of interest and $T$ is the time duration in years.
Therefore, we can write $P{{\left( 1+\dfrac{R}{100} \right)}^{T}}-P-\dfrac{P\times R\times T}{100}=18.$
In our case, $R=6\%$ and $T=2years.$ We need to find the value of $P.$
Let us substitute the values to get $P{{\left( 1+\dfrac{6}{100} \right)}^{2}}-P-\dfrac{P\times 6\times 2}{100}=18.$
Now from this, we will get $P{{\left( \dfrac{53}{50} \right)}^{2}}-P-\dfrac{3P}{25}=18.$
And, we will get $P{{\left( \dfrac{53}{50} \right)}^{2}}-\left( P+\dfrac{3P}{25} \right)=18.$
This will give us $P{{\left( \dfrac{53}{50} \right)}^{2}}-\left( \dfrac{28P}{25} \right)=18.$
Now, we will get ${{\left( 1.06 \right)}^{2}}P-1.12P=18.$
After the calculations, we will get $1.1236P-1.12P=18.$
So, we will get $0.0036P=18.$
We will get $P=\dfrac{18}{0.0036}=5000.$
Hence the principal amount or the sum is $5000.$
Note: We know that the simple interest is based on the principal amount of a loan or deposit whereas the compound interest is based on the principal amount and the interest that accumulates on it in every period.
Complete step by step solution:
Let us consider the given problem for finding the sum when the difference of compound interest compounded annually and the simple interest on certain sum $2$ years at $6\%$ per annum is $Rs.18.$
Now, we can write the difference between compound interest and simple interest is equal to $18.$ That is, we write $C.I-S.I=18.$
We know that the compound interest compounded annually can be found by using the formula $C.I=P{{\left( 1+\dfrac{R}{100} \right)}^{T}}-P$ where $P$ is the principal amount, $R$ is the rate of interest and $T$ is the time duration in years.
Similarly, we can find the simple interest by using the formula $S.I=\dfrac{P\times R\times T}{100}$ where $P$ is the principal amount, $R$ is the rate of interest and $T$ is the time duration in years.
Therefore, we can write $P{{\left( 1+\dfrac{R}{100} \right)}^{T}}-P-\dfrac{P\times R\times T}{100}=18.$
In our case, $R=6\%$ and $T=2years.$ We need to find the value of $P.$
Let us substitute the values to get $P{{\left( 1+\dfrac{6}{100} \right)}^{2}}-P-\dfrac{P\times 6\times 2}{100}=18.$
Now from this, we will get $P{{\left( \dfrac{53}{50} \right)}^{2}}-P-\dfrac{3P}{25}=18.$
And, we will get $P{{\left( \dfrac{53}{50} \right)}^{2}}-\left( P+\dfrac{3P}{25} \right)=18.$
This will give us $P{{\left( \dfrac{53}{50} \right)}^{2}}-\left( \dfrac{28P}{25} \right)=18.$
Now, we will get ${{\left( 1.06 \right)}^{2}}P-1.12P=18.$
After the calculations, we will get $1.1236P-1.12P=18.$
So, we will get $0.0036P=18.$
We will get $P=\dfrac{18}{0.0036}=5000.$
Hence the principal amount or the sum is $5000.$
Note: We know that the simple interest is based on the principal amount of a loan or deposit whereas the compound interest is based on the principal amount and the interest that accumulates on it in every period.
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