Question

Calculate the amount and compound interest on Rs 64000 for 3 years at 15/2 •\• per annum compounded annually.​

Answer 1

Solution :

Here,

• Principal (P) = Rs. 64000
• Time (n) = 3 years
• Rate (R) = $\sf {\dfrac{15}{2} }$ %

We have to find,

• Amount (A)
• Compound Interest (C.I)

$\underline{\small \sf {\maltese \; \; \; Finding \: Amount \: : \; \; \; }}$

We know that,

$\bigstar \: \boxed{\sf { {A = P \Bigg \lgroup 1 + \dfrac{R}{100} \Bigg \rgroup}^n }}$

→ $\sf {A = Rs. \: 64000 \Bigg \lgroup 1 + \dfrac{ \cfrac{15}{2} }{ 100} \Bigg \rgroup^3 }$

→ $\sf {A = {Rs. \: 64000 \Bigg \lgroup 1 + \dfrac{15}{2} \times \dfrac{1}{100} \Bigg \rgroup}^3 }$

→ $\sf {A = {Rs. \: 64000 \Bigg \lgroup 1 + \dfrac{15}{2 \times 100} \Bigg \rgroup}^3 }$

→ $\sf {A = {Rs. \: 64000 \Bigg \lgroup 1 + \dfrac{15}{200} \Bigg \rgroup}^3 }$

→ $\sf {A = {Rs. \: 64000 \Bigg \lgroup \dfrac{200 + 15}{200} \Bigg \rgroup}^3 }$

→ $\sf {A = {Rs. \: 64000 \Bigg \lgroup \dfrac{215}{200} \Bigg \rgroup}^3 }$

→ $\sf {A = {Rs. \: 64000 \Bigg \lgroup \dfrac{43}{40} \Bigg \rgroup}^3 }$

→ $\sf {A = Rs. \: 64000 \times \dfrac{43}{40} \times \dfrac{43}{40} \times \dfrac{43}{40} }$

→ $\sf {A = Rs. \: 64000 \times \dfrac{43 \times 43 \times 43 }{40 \times 40 \times 40}}$

→ $\sf {A = Rs. \: \cancel{64000} \times \dfrac{43 \times 43 \times 43 }{\cancel{64000}}}$

→ $\sf {A = Rs. \: 43 \times 43 \times 43 }$

→ $\underline{ \boxed{ \bf {Amount = Rs. \: 79507 } }}$

Also, we know that :

$\bigstar \: \boxed{\sf {CI = Amount - Principal}}$

→ C.I = Rs. ( 79507 - 64000 )

→ $\underline{ \boxed{ \bf {CI = Rs. \: 15507 } }}$

Therefore,

● $\underline{ \boxed{ \bf {Amount = Rs. \: 79507 } }}$

● $\underline{ \boxed{ \bf {CI = Rs. \: 15507 } }}$

Hence, we got the answer !

Answer 2

Provided Question :

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$\tt{\orange{Calculate \: the \: amount\:and\: Compound\:interest }} \\ \tt{\orange{on\:Rs\:64000\:for\:3\:yrs\:at\: \dfrac{15}{2} \% }} \\ \tt{\orange{per\:annum\:compounded\: annually}}$

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Required Answer :

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Given ,

• $\boxed{\sf{Principal\:(P)\:=\: Rs 64000 }}$

• $\boxed{\sf{Time\:(T)\:=\: 3 \:years }}$

• $\boxed{\sf{Rate\:(R)\:=\: \dfrac{15}{2} \% }}$

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Answer to find :

• $\boxed{\orange{\boxed{\sf{\purple{Amount\:(A) }}}}}$

• $\boxed{\orange{\boxed{\sf{\purple{ Compound\: Interest }}}}}$

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Final Answer :

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$\implies \boxed{\sf{ A \:=\: P \bigg( 1 + \dfrac{r}{100} \bigg)^{T} }}$

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$\implies \sf{ A \:=\: P \: \bigg( 1 + \dfrac{\dfrac{15}{2}}{100} \bigg)^{3} }$

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$\implies \sf{ A \:=\: P \: \bigg( 1 + \dfrac{15}{100 \times 2 } \bigg)^{3} }$

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$\implies \sf{ A \:=\: 64000 \: \bigg( 1 + \dfrac{15}{200} \bigg)^{3} }$

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$\implies \sf{ A \:=\: 64000 \: \bigg( \dfrac{1}{1} + \dfrac{15}{200} \bigg)^{3} }$

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$\implies \sf{ A \:=\: 64000 \: \bigg( \dfrac{200 + 15}{200} \bigg)^{3} }$

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$\implies \sf{ A \:=\: 64000 \: \bigg( \dfrac{\cancel{215}}{\cancel{200}} \bigg)^{3} }$

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$\implies \sf{ A \:=\: 64000 \: \bigg( \dfrac{43}{40} \bigg)^{3} }$

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$\implies \sf{ A \:=\: 64000 \times \dfrac{43}{40} \times \dfrac{43}{40} \times \dfrac{43}{40} }$

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$\implies \sf{ A \:=\: \xcancel{64000} \times \dfrac{43}{\xcancel{40}} \times \dfrac{43}{40} \times \dfrac{43}{40} }$

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$\implies \sf{ A \:=\: \xcancel{1600} \times 43 \times \dfrac{43}{\xcancel{40}} \times \dfrac{43}{40} }$

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$\implies \sf{ A \:=\: \xcancel{40} \times 43 \times 43 \times \dfrac{43}{\xcancel{40}} }$

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$\implies \sf{ A \:=\: 43 \times 43 \times 43 }$

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$\implies \sf{ A \:=\: 79507 }$

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$\boxed{\boxed{\red{\rm{\therefore A \:=\: Rs \: 79507 }}}}$

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°•° Compound Interest = A - P

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•°• C.I = $\boxed{\bf{ 79507 - 64000 }}$

ㅤ ㅤ = $\boxed{\bf{ rs \: 15507}}$

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Outcomes :

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• ${\fcolorbox{blue}{blue}{\sf{ C.I = Rs 15507 }}}$

• ${\fcolorbox{blue}{blue}{\sf{ Amount = Rs 79507 }}}$

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$\boxed{\begin{array}{| c |}\qquad\tt{:}\longrightarrow\large\textsf{ Amount = A }\\\\\\\qquad\tt{:}\longrightarrow\large\textsf{ Time = T }\\\\\\\qquad\tt{:}\longrightarrow\large\textsf{ P = Principal }\\\\\\\qquad\tt{:}\longrightarrow\large\textsf{ S.I = Simple Interest }\\\\\\\qquad\tt{:}\longrightarrow\large\textsf{ C.I = Compound Interest }\end{array}}$

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