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Find the amount and the compound interest of Rs. 10000 in 3 years, if the rate of interest for the successive years is 10%, 15% and 20%.

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Last updated date: 29th Feb 2024
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IVSAT 2024
Answer
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Hint: We solve this problem by taking each year and applying the interest for the successive year to find the amount. The formula for amount having the principal value P, time period T and rate of interest R is given as
\[\text{amount}=P+\dfrac{P\times T\times R}{100}\]
After finding the amount after 3 years we calculate the compound interest by using the formula
\[C.I=\text{amount}-\text{principal}\]

Complete step-by-step answer:
We are given that the principal amount is Rs. 10000
Let us assume that the principal amount ass
\[\Rightarrow P=10,000\]
Let us assume the first year
We are given that the interest for first one year is 10%
Let us assume that the interest as
\[\Rightarrow {{R}_{1}}=10\%\]
Let us assume that the time period as
\[\Rightarrow {{T}_{1}}=1year\]
We know that the formula for amount having the principal value P, time period T and rate of interest R is given as
\[\text{amount}=P+\dfrac{P\times T\times R}{100}\]
Let us assume that the amount after first year as \[{{A}_{1}}\]
By using the above formula to first year we get
\[\Rightarrow {{A}_{1}}=P+\dfrac{P\times {{T}_{1}}\times {{R}_{1}}}{100}\]
By substituting the required values in above equation we get
\[\begin{align}
  & \Rightarrow {{A}_{1}}=10,000+\dfrac{10,000\times 1\times 10}{100} \\
 & \Rightarrow {{A}_{1}}=10,000+1,000 \\
 & \Rightarrow {{A}_{1}}=11,000 \\
\end{align}\]
We know that the amount after first year will be the principal for the second year.
We are given that the rate of interest for second year as 15%
Let us assume that the interest as
\[\Rightarrow {{R}_{2}}=15\%\]
Let us assume that the time period as
\[\Rightarrow {{T}_{2}}=1year\]
Let us assume that the amount after first year as \[{{A}_{2}}\]
By using the above formula to first year we get
\[\Rightarrow {{A}_{2}}={{A}_{1}}+\dfrac{{{A}_{1}}\times {{T}_{2}}\times {{R}_{2}}}{100}\]
By substituting the required values in above equation we get
\[\begin{align}
  & \Rightarrow {{A}_{2}}=11,000+\dfrac{11,000\times 1\times 15}{100} \\
 & \Rightarrow {{A}_{2}}=11,000+1,650 \\
 & \Rightarrow {{A}_{2}}=12,650 \\
\end{align}\]
We know that the amount after second year will be the principal for the third year.
We are given that the rate of interest for third year as 20%
Let us assume that the interest as
\[\Rightarrow {{R}_{3}}=15\%\]
Let us assume that the time period as
\[\Rightarrow {{T}_{3}}=1year\]
Let us assume that the amount after first year as \[{{A}_{3}}\]
By using the above formula to first year we get
\[\Rightarrow {{A}_{3}}={{A}_{2}}+\dfrac{{{A}_{2}}\times {{T}_{3}}\times {{R}_{3}}}{100}\]
By substituting the required values in above equation we get
\[\begin{align}
  & \Rightarrow {{A}_{3}}=12,650+\dfrac{12,650\times 1\times 20}{100} \\
 & \Rightarrow {{A}_{3}}=12,650+2,530 \\
 & \Rightarrow {{A}_{3}}=15,180 \\
\end{align}\]
Therefore the final amount after the three years is Rs. 15180.
We know that the formula for compound interest is given as
\[C.I=\text{amount}-\text{principal}\]
By using the above formula after the three years we get
\[\begin{align}
  & \Rightarrow C.I=15,180-10,000 \\
 & \Rightarrow C.I=5,180 \\
\end{align}\]
Therefore the compound interest is Rs. 5180.

Note: We have a direct formula for the amount.
If the rates of interest for the three successive years are \[{{R}_{1}},{{R}_{2}},{{R}_{3}}\] having the principal amount as P then the amount I given as
\[\Rightarrow A=P\left( 1+\dfrac{{{R}_{1}}}{100} \right)\left( 1+\dfrac{{{R}_{2}}}{100} \right)\left( 1+\dfrac{{{R}_{3}}}{100} \right)\]
By substituting the required values in above equation we get
\[\begin{align}
  & \Rightarrow A=10,000\left( 1+\dfrac{10}{100} \right)\left( 1+\dfrac{15}{100} \right)\left( 1+\dfrac{20}{100} \right) \\
 & \Rightarrow A=10,000\times \dfrac{11}{10}\times \dfrac{23}{20}\times \dfrac{6}{5} \\
 & \Rightarrow A=15,180 \\
\end{align}\]
Therefore the final amount after the three years is Rs. 15180.

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