Question

$\huge\bold{\textbf{\textsf{{\color{cyan}{Question}}}}}$
Mr Vig has a Recurring Deposit Account in a bank for 5 years at 9% p.a. simple interest. If he gets 53082 at the time of maturity, find the monthly ínstalment.​

Answer 1

$\large \rm {\underbrace{\underline{Elucidation:-}}}$

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$\sf \red {\underline{\underline{Provided\: that:}}}$

$\tt {No\: of\: months(n)=5 ×12=60months}$

Rate of iηterest(r)=9%

$\tt {Maturity\: value(M.V)=53082}$

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$\sf \blue {\underline{\underline{To\: determine:}}}$

$\tt {Monthly\: iηstalment(P)=?}$

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$\sf \pink {\underline{\underline{We\: Know:}}}$

$\tt {\fbox{\underline{Maturity\: value=Total\: sum\: deposited+iηterest\: on\: it}}}$

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$\to \tt{Total\: sum\: deposited=Monthly\: iηstalment×No\: of\: months}$

$\implies \tt {\fbox{Total\: sum\: deposited=p×n}}$

$\implies \tt {Total\: sum\: deposited=p×60}$

$\colon \implies \tt \green{Total\: sum\: deposited=60p}$

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$\to \tt {Simple\: iηterest (I)=p×\frac{n(n+1)}{2×12}×\frac{r}{100}}$

➻Substituting the given values,

$\to \tt {I=p×\frac{60(60+1)}{2×12}×\frac{9}{100}}$

$\to \tt {I=p×\frac{\cancel{60}^{5}(61)}{2×{\cancel{12}}}×\frac{9}{100}}$

$\to \tt {I=p×\frac{5×61}{2}×\frac{9}{100}}$

$\to \tt {I=p×\frac{305}{2}×\frac{9}{100}}$

$\to \tt {I=p×\frac{305×9}{2×100}}$

$\to \tt {I=p×\frac{2745}{200}}$

$\to \tt \green {\fbox{\underline{Simple\:Iηterest=13.725p}}}$

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➻Now,

$\tt {Maturity\: value=Total\: sum\: deposited+iηterest\: on\: it}$

$\to \tt {53082=60p+13.725p}$

$\to \tt {53082=73.725p}$

$\to \tt {p=\frac{53082}{73.725}}$

$\implies \tt \green {\fbox{\underline{p=720}}}$

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$\sf \purple {\underline{\underline{Therefore,}}}$

➻Monthly iηstallment by Mr.vig is ₹720

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$\sf \orange {\underline{\underline{NOTE:-}}}$

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