Question

at what rate of compound interest compounded annually, will rupees 7500 become 9075 in 2 years? ​

Answer 1

QuestioN

• At what rate of compound interest compounded annually, will ₹ 7500 become ₹ 9075 in 2 years?

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${\large{\frak{\pmb{\underline{From\:the\:above\:data }}}}}$

Amount after n years when the interest is compounded annually is given by

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$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Large{\ddagger}\small{\underline{\boxed{\mathcal{\pmb{\underline{\red{ A\:=\: P\bigg(1+\bf\dfrac{R}{100}\bigg)^n}}}}}}}$

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$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\curlyeqsucc\:Principal\:(P)=\: ₹\:7500$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\curlyeqsucc\:Amount\:(A)=\: ₹\:9075$

$\sf \:\:\:\:\:\:\:\:\curlyeqsucc\: Time\:(n\:years)=\: 2\:years$

$\:\:\:\:\:\:\:\:\:\:\:\:\:──────────────────$

${\large{\mathcal{\pmb{\underline{Now, }}}}}$

$\:\:\:\:\:{\bold\bullet \:\:{ \underline{\small{ Let \:the\: required\:rate\:be\:R\%\:per\:annum.}}}}$

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$\sf \:\:\: \therefore\:\:\:\:\:\:\:\:\: ₹ 9075\:=\: ₹ 7500\bigg(1+\bf\dfrac{R}{100}\bigg)^2$

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$\sf \:\:\:\:\:\:\: \implies\: \bf\dfrac{9075}{7500}\:=\:\bigg(1+\bf\dfrac{R}{100}\bigg)^2$

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$\sf \:\:\:\:\:\:\:\:\:\:\:\:\: \twoheadrightarrow \;\bf\dfrac{121}{100}\:=\:\bigg(1+\bf\dfrac{R}{100}\bigg)^2$

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$\sf \:\:\:\:\:\:\:\:\:\: \twoheadrightarrow \:\bigg( \bf\dfrac{11}{10}\bigg)\cancel{^2}\:=\:\bigg(1+\bf\dfrac{R}{100}\bigg)\cancel{^2}$

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$\sf \:\:\:\:\:\:\:\:\:\:\:\:\: \twoheadrightarrow \:\bf\dfrac{11}{10}\:=\:1+\bf\dfrac{R}{100}$

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$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\twoheadrightarrow \:\bf\dfrac{11}{10}\:-1=\:\bf\dfrac{R}{100}$

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$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\: \twoheadrightarrow \:\bf\dfrac{1}{10}\:-1=\:\bf\dfrac{R}{100}$

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$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \twoheadrightarrow \:R\:=\: \bf\dfrac{100}{10}\:=\:\purple{10}$

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$\underline{\boxed{\sf{Hence,\:the\: required\:rate\:is\:\frak{\pink{10\%\:per\:annum.}}}}}$

Answer 2

$\begin{gathered}{\Large{\textsf{\textbf{\underline{Question\:Given:}}}}}\end{gathered}$

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• At what rate of compound interest compounded annually, will Rs 7500 become 9075 in 2 years?

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$\begin{gathered}{\Large{\textsf{\textbf{\underline{Required\:Solution:}}}}}\end{gathered}$

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$\:\:\:\:\:\bigstar\large\:\tt{Value\:Given\:To\:Us\::}$

• $\sf{Principal\:(P)\:\leadsto\:Rs.\:7500}$
• $\sf{Amount\:(A)\:\leadsto\:Rs.\:9075}$
• $\sf{Time\:(n)\:\leadsto\:2\:year}$

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$\:\:\:\:\:\bigstar\large\:\tt{Formula\:Used\:Here\::}$

• $\pink{\frak{Amount\:=\:P\:\bigg(\:1\:+\:{\dfrac{R}{100}}\:\bigg)^{n}}}$

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$\:\:\:\:\:\bigstar\large\:\tt{Putting\:Value\:in\:Formula\::}$

• Let the required rate be R% per annum.

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$\:\:\:\:\:\:\:\:{\odot}\:\sf{Rs.\:9075\:=\:Rs.\:7500\:\bigg(\:1\:+\:{\dfrac{R}{100}}\:\bigg)^{2}}$

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$\:\:\:\:\:\:\:\:\:\:\dashrightarrow\:\sf{\dfrac{9075}{7500}}\:=\:\bigg(\:1\:+\:{\dfrac{R}{100}}\:\bigg)^{2}$

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$\:\:\:\:\:\:\:\:\:\:\dashrightarrow\:\sf{\dfrac{121}{100}}\:=\:\bigg(\:1\:+\:{\dfrac{R}{100}}\:\bigg)^{2}$

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$\:\:\:\:\:\:\:\:\:\:\dashrightarrow\:\sf{\bigg(\:{\dfrac{11}{10}}\:\bigg)^{\cancel{2}}\:=\:\bigg(\:1\:+\:{\dfrac{R}{100}}\:\bigg)^{\cancel{2}}}$

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$\:\:\:\:\:\:\:\:\:\:\dashrightarrow\:\sf{\dfrac{11}{10}}\:-\:1\:=\:{\dfrac{R}{100}}$

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$\:\:\:\:\:\:\:\:\:\:\dashrightarrow\:\sf{\dfrac{1}{10}}\:-\:1\:=\:{\dfrac{R}{100}}$

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$\:\:\:\:\:\:\:\:\:\:\dashrightarrow\:\sf{R\:=\:\cancel{\dfrac{100}{10}}}$

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$\:\:\:\:\:\:\:\:\:\:\dashrightarrow\:\sf\:{\underline{\boxed{\frak{\pink{10\:\%}}}}}\:\bigstar$

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$\:\:\:\:\:\:\therefore\:{\underline{\sf{Hence,\:the\:rate\:is\:{\textsf{\textbf{10\:\% }}}}}}.$

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$\:\:\:\:\:\:\:\:━━━━━━━━━━━━━━━━━━━$

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