Question

9) Find the compounded interest for six months on Rs.10,000 at an
annual rate of interest of 12% if the interest is compounded every 3
months.​

Answer 1

$\pink{ \bold{ \frak{Given}}} \begin{cases} \sf Principal \: (P) = 10000 \\ \\ \sf \: Time \: (n) = 6 \: months \\ \\ \sf Rate \: (R) = 12 \: \% \\ \\ \sf Compound \: Interest \: ( C.I) = compounded \: 3 \: months.\end{cases}$

☯$\underline{\boldsymbol{According\: to \:the\: Question\:now :}}$

$: \implies \displaystyle \sf \: C . I= P \Bigg[ \bigg(1 + \dfrac{R}{100}\bigg)^{2} - 1\Bigg]$

$: \implies \displaystyle \sf \: C . I= 10000 \Bigg[ \bigg(1 + \dfrac{3}{100}\bigg)^{2 } - 1\Bigg]$

$: \implies \displaystyle \sf \: C . I= 10000 \Bigg[ \bigg(\dfrac{100+ 3}{100}\bigg)^{2} - 1\Bigg]$

$: \implies \displaystyle \sf \: C . I= 10000 \Bigg[ \bigg(\dfrac{103}{100}\bigg)^{2} - 1\Bigg]$

$: \implies \displaystyle \sf \: C . I= 10000 \Bigg[ \bigg(\dfrac{10609}{10000}\bigg) - 1\Bigg]$

$: \implies \displaystyle \sf \: C . I= 10000 \Bigg[ \dfrac{10609}{10000} - 1\Bigg]$

$: \implies \displaystyle \sf \: C . I= 10000 \Bigg[ \dfrac{10609- 10000}{10000} \Bigg]$

$: \implies \displaystyle \sf \: C . I= 10000 \Bigg[ \dfrac{609 }{10000} \Bigg]$

$: \implies \displaystyle \sf \: C . I= 10000 \times \frac{609}{10000}$

$: \implies \underline{ \boxed{ \displaystyle \sf \: C . I=609}}$

Answer 2

$\huge\bf{\underline{\underline{\gray{GIVEN:-}}}}$

• $\bf\blue{Principle\:(P)}$ = Rs.10,000

• $\bf\blue{Time\:(T)}$ = 6 months = $\bf{\dfrac{1}{2}\:year}$

• $\bf\blue{Rate\:(R)}$ = 12% per annum

☯︎︎︎︎ Compounded every 3 months .

• $\bf\blue{n}$ = 3

$\huge\bf{\underline{\underline{\gray{TO\:FIND:-}}}}$

• Compound Interest .

$\huge\bf{\underline{\underline{\gray{SOLUTION:-}}}}$

We know that,

${\color{aqua}\bigstar}\:\bf{\red{\overbrace{\underbrace{\green{Amount\:(A)\:=\:P\:\Big(\:1\:+\:\dfrac{R}{n}\:\Big)^{n\:T}\:}}}}}$

$\\ \bf{\implies\:Amount\:(A)\:=\:10,000\times\:\Big(\:1\:+\:\dfrac{12/100}{4}\:\Big)^{3\times{\dfrac{1}{2}}}\:}$

$\\ \bf{\implies\:Amount\:(A)\:=\:10,000\times\:\Big(\:1\:+\:\dfrac{0.12}{4}\:\Big)^{3\times{\dfrac{1}{2}}}\:}$

$\\ \bf{\implies\:Amount\:(A)\:=\:10,000\times\:\Big(\:1\:+\:0.03\:\Big)^{3\times{\dfrac{1}{2}}}\:}$

$\\ \bf{\implies\:Amount\:(A)\:=\:10,000\times\:\Big(\:1.03\:\Big)^{3\times{\dfrac{1}{2}}}\:}$

$\\ \bf{\implies\:Amount\:(A)\:=\:10,000\times\:\Big(\:(1.03)^{\dfrac{1}{2}}\:\Big)^3\:}$

$\\ \bf{\implies\:Amount\:(A)\:=\:10,000\times\:\Big(\:1.01489\:\Big)^3\:}$

$\\ \\ \bf{\implies\:Amount\:(A)\:=\:10,000\times{1.045335}\:}$

$\\ \\ \bf\orange{\implies\:Amount\:(A)\:=\:Rs.10,453.35\:}$

_ _ _ _ _ _ _ _ _ _

We know that,

$\purple\bigstar\:\bf{\pink{\overbrace{\underbrace{\blue{Compound\:Interest\:(C.I)\:=\:Amount\:-\:Principle\:}}}}}$

$\\ \sf{\implies\:Compound\:Interest\:(C.I)\:=\:10,453.35\:-\:10,000\:}$

$\\\sf\gray{\implies\:Compound\:Interest\:(C.I)\:=\:Rs.453.35\:}$

$\\ \huge{\color{orange}\therefore}$ The Compound Interest is 'Rs.453.35' .