Question

find the compounds interest on 7,500 at rate of 12% per annum for 9 months if the interest is compounded quarterly give explanation ​

Answer 1

Given :

• P= 7,500
• F= 4
• R= 12%

To Find :

• The Compounds Interest = ?

Solution :

$\dag$ Formula Used :

${\color{black}{\bigstar}} \; {\underline{\boxed{\red{\sf{A=P( + \frac{r}{n}) ^{nt} }}}}}$

Fullform Of The Formula's :-

$\begin{gathered}\boxed{\begin {array}{cc}\\ \quad \small \underline{\bf Fullform \: Of \: The \: Formula's:-}\\ \\ \sf A = Final Amount \\ \sf \: \\ \sf \: P = Initial \: Principal \: Balance \\ \sf \: \\ \sf \: R = Interest \: Rate \\ \\ \small \: \sf \: N = Number \: Of \: Times \: Interest \: Applied \: Per \: Time \\ \\ \sf \: \small \: N= Number \: Of \: Times \: Interest \: Applied \: Per \: Time \: Period \\ \\ \small \sf \: T= Number \: Of \: Time \: Period \: Elapsed\\ \\ \end {array}}\end{gathered}$

$\\ \qquad{\rule{150pt}{1pt}}$

$\dag$ Solution :

${➻ \; \; {\sf{ A = 7500(1 + \frac{ \frac{12}{4} }{100}) ^{\frac{9}{12} \times 4} }}}$

${➻ \; \; {\sf{ A = 7500(1 + \frac{3}{100}) ^{3} }}}$

${➻ \; \; {\sf{ A = 7500(1 +0.03) ^{3} }}}$

${➻ \; \; {\sf{ A = 7500(1.03) ^{3} }}}$

${➻ \; \; {\sf{ A = 8195.25 \cong8196 }}}$

${ \pmb{ \purple{ \frak{A = 8195.25 \cong8196}}}}$

$\\ \qquad{\rule{150pt}{1pt}}$

Let's Find Compound Interest:-

$\longrightarrow \sf \: C.I = A-P$

$\longrightarrow \sf \: = 8196 - 7500$

$\longrightarrow \sf \: C.I =  ₹\: 696$

$\dag$ Therefore :

❛❛ The Compound Interest Is ₹696 ❜❜

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