1.

Chinmay invested ₹ 60,000 at an interest rate of 12% per annum compounded half yearly. What amount would he get after 1 and half year?​

Answer»

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  • Principal = ₹ 60,000
  • Rate (r) = 12% (Compounded half yearly)
  • Time (t) = 1 & ½ YEARS (3 parts)

We have to adjust Rate and time According to the half yearly compounded as:-

\circ \ {\pmb{\underline{\boxed{\sf{ Amount = P \left( 1 + \dfrac{r}{100} \right)^ n }}}}} \\

\\ \bigstar \ {\pmb{\underline{\sf{ According \ to \ Question: }}}} \\ \\ \\ \colon\implies{\sf{ 60000 \left( 1 + \dfrac{6}{100} \right)^ 3 }} \\ \\ \\ \colon\implies{\sf{ 60000 \left( \cancel{ \dfrac{106}{100} } \right)^ 3 }} \\ \\ \\ \colon\implies{\sf{ 60000 \left( \dfrac{53}{50} \right)^ 3 }} \\ \\ \\ \colon\implies{\sf{ 60000 \times \dfrac{53}{50} \times \dfrac{53}{50} \times \dfrac{53}{50} }} \\ \\ \\ \colon\implies{\sf{ 60 \cancel{000} \times \dfrac{53}{5 \cancel{0} } \times \dfrac{53}{5 \cancel{0} } \times \dfrac{53}{5 \cancel{0} } }} \\ \\ \\ \colon\implies{\sf{ 60 \times \dfrac{53}{5} \times \dfrac{53}{5} \times \dfrac{53}{5} }} \\ \\ \\ \colon\implies{\sf{ \cancel{ \dfrac{8932620}{125} } }} \\ \\ \\ \colon\implies{\underline{\boxed{\sf\green{ Rs. \ 71460.96 _{(Amount)} }}}} \\

HENCE,

\\ {\pmb{\underline{\sf\green{ The \ Amount \ of \ the \ Investment \ is \ Rs. \ 71460.96. }}}}


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