Question

2. Abhimanyu invested 100000 13% p.a compounded annually. Find the following . i.The amount standing to his credit the end of the second year . ii.The interest for the third year​

Answer 1

$\large \; {\underline{\underline{\maltese \; {\pmb{\red{\sf{ Given \; :- }}}}}}}$

• Sum Invested = 100000
• Rate of Interest = 13 %
• Compounded = Annually

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$\large \; {\underline{\underline{\maltese \; {\pmb{\blue{\sf{ To \: Find \; :- }}}}}}}$

• Amount at the end of 2nd year = ?
• Interest for the 3rd year = ?

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$\large \; {\underline{\underline{\maltese \; {\pmb{\orange{\sf{ Solution \; :- }}}}}}}$

${\color{orange}{❒}}$ Formula Used :

• Amount :

$\large{\pink{\dashrightarrow}} \; {\underline{\boxed{\color{darkblue}{\sf{ A = P \bigg\lgroup 1 + \dfrac{R}{100} \bigg\rgroup ^n }}}}}$

• Compound Interest :

$\large{\pink{\dashrightarrow}} \; {\underline{\boxed{\color{darkblue}{\sf{ C.I = Amount - Principal }}}}}$

Where :

• A = Amount
• R = Rate
• P = Principal
• n = Time
• C.I = Compound Interest

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${\color{orange}{❒}}$ Calculating the Amount at the end of 2nd year :

${\longmapsto{\qquad{\sf{ A = P \bigg\lgroup 1 + \dfrac{R}{100} \bigg\rgroup ^n }}}} \\ \\ \ {\longmapsto{\qquad{\sf{ A = 100000 \bigg\lgroup 1 + \dfrac{13}{100} \bigg\rgroup ^2 }}}} \\ \\ \ {\longmapsto{\qquad{\sf{ A = 100000 \bigg\lgroup 1 + \cancel\dfrac{13}{100} \bigg\rgroup ^2 }}}} \\ \\ \ {\longmapsto{\qquad{\sf{ A = 100000 \bigg\lgroup 1 + 0.13 \bigg\rgroup ^2 }}}} \\ \\ \ {\longmapsto{\qquad{\sf{ A = 100000 \bigg\lgroup 1.13 \bigg\rgroup ^2 }}}} \\ \\ \ {\longmapsto{\qquad{\sf{ A = 10000 \times 1.13 \times 1.13 }}}} \\ \\ \ {\longmapsto{\qquad{\sf{ A = 100000 \times 1.2769 }}}} \\ \\ \ {\qquad{\textsf{ Amount at the end of 2nd year = {\green{\sf{ ₹ \; 127690 }}}}}}$

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${\color{orange}{❒}}$ Calculating the Interest for the 3rd year :

${\longmapsto{\qquad{\sf{ A = P \bigg\lgroup 1 + \dfrac{R}{100} \bigg\rgroup ^n }}}} \\ \\ \ {\longmapsto{\qquad{\sf{ A = 100000 \bigg\lgroup 1 + \dfrac{13}{100} \bigg\rgroup ^3 }}}} \\ \\ \ {\longmapsto{\qquad{\sf{ A = 100000 \bigg\lgroup 1 + \cancel\dfrac{13}{100} \bigg\rgroup ^3 }}}} \\ \\ \ {\longmapsto{\qquad{\sf{ A = 100000 \bigg\lgroup 1 + 0.13 \bigg\rgroup ^3 }}}} \\ \\ \ {\longmapsto{\qquad{\sf{ A = 100000 \bigg\lgroup 1.13 \bigg\rgroup ^3 }}}} \\ \\ \ {\longmapsto{\qquad{\sf{ A = 10000 \times 1.13 \times 1.13 \times 1.13}}}} \\ \\ \ {\longmapsto{\qquad{\sf{ A = 100000 \times 1.442897 }}}} \\ \\ \ {\qquad{\textsf{ Amount at the end of 2nd year = {\red{\sf{ ₹ \; 144289.7 }}}}}}$

~ Calculating the Compund Interest :

${\dashrightarrow{\qquad{\sf{ C.I = Amount - Principal }}}} \\ \\ \ {\dashrightarrow{\qquad{\sf{ C.I = 144289.7 - 100000 }}}} \\ \\ \ {\qquad{\textsf{ Compound Interest for the 3rd year = {\color{maroon}{\sf{ ₹ \; 44289.7 }}}}}}$

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❝ Amount at the end of 2nd year is ₹ 127690 and the Compound interst for 3rd year is ₹ 144289.7 . ❞

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Answer 2

$\begin{gathered} \\ {\underline{\rule{200pt}{3pt}}} \end{gathered}$

${\huge{\color{red}{\rm{Question}}}}$

Abhimanyu invested 100000 13% p.a compounded annually. Find the following . i.The amount standing to his credit the end of the second year . ii.The interest for the third year.

${\large{\underline{\underline{\rm{Answer}}}}}$

$\begin{gathered} \\ {\underline{\rule{200pt}{3pt}}} \end{gathered}$

${\large{\underline{\underline{\color{blue}{\rm{Given}}}}}}$

• Sum Invested=100,000
• Rate of Interest=13 percent
• Compounded Annually

$\begin{gathered} \\ {\underline{\rule{200pt}{3pt}}} \end{gathered}$

${\large{\underline{\underline{\color{orange}{\rm{To\:Find}}}}}}$

• Amount=?
• Interest?

$\begin{gathered} \\ {\underline{\rule{200pt}{3pt}}} \end{gathered}$

${\large{\underline{\underline{\rm{Formula\:Used}}}}}$

${\large{\underline{\underline{\rm{Amount}}}}}$

${\large{\underline{\boxed{\color{red}{\sf{A=P(1+\frac{R}{100})^n}}}}}}$

${\large{\underline{\underline{\rm{C.I}}}}}$

${\large{\underline{\boxed{\color{red}{\sf{C.I.=A-P}}}}}}$

${\large{\underline{\underline{\rm{Here}}}}}$

• A=Amount
• P=Principal
• R=Rate
• n=time

$\begin{gathered} \\ {\underline{\rule{200pt}{3pt}}} \end{gathered}$

$\begin{gathered} \\ \qquad{\rule{150pt}{1pt}} \end{gathered}$

${\qquad{\qquad{\qquad{\large{\underline{\underline{\rm{Solution}}}}}}}}$

${\qquad{\qquad{\qquad{\large{\sf{Amount\:at\:the\:end\:of\:2nd\:year:}}}}}}$

${\qquad{\qquad{\qquad{\large{\longmapsto{\sf{A=P[1+\frac{R}{100}]^n}}}}}}}$

${\qquad{\qquad{\qquad{\large{\longmapsto{\sf{A=1000001+\frac{13}{100}^2}}}}}}}$

${\qquad{\qquad{\qquad{\large{\longmapsto{\sf{A=1000001+\frac{13}{100}^2}}}}}}}$

${\qquad{\qquad{\qquad{\large{\longmapsto{\sf{A=1000001+0.13^2}}}}}}}$

${\qquad{\qquad{\qquad{\large{\longmapsto{\sf{A=1000001.13^2}}}}}}}$

${\qquad{\qquad{\qquad{\large{\longmapsto{\sf{A=100000\times 1.13\times 1.13}}}}}}}$

${\qquad{\qquad{\qquad{\large{\longmapsto{\sf{A=100000\times 1.13\times 1.13}}}}}}}}$

${\qquad{\qquad{\qquad{\large{\longmapsto{\sf{A=100000\times 1.2679}}}}}}}$

${\qquad{\qquad{\qquad{\large{\longmapsto{\sf{A=1267900}}}}}}}$

$\begin{gathered} \\ \qquad{\rule{150pt}{1pt}} \end{gathered}$

${\qquad{\qquad{\qquad{\large{\sf{Amount\:at\:the\:end\:of\:3rd\:year:}}}}}}$

$\begin{gathered} \\ \qquad{\rule{150pt}{1pt}} \end{gathered}$

${\qquad{\qquad{\qquad{\large{\longmapsto{\sf{A=P(1+\frac{R}{100})^n}}}}}}}$

${\qquad{\qquad{\qquad{\large{\longmapsto{\sf{A=P(1+\frac{R}{100})^n}}}}}}}$

${\qquad{\qquad{\qquad{\large{\longmapsto{\sf{A=100000(1+\frac{13}{100})^3}}}}}}}$

${\qquad{\qquad{\qquad{\large{\longmapsto{\sf{A=100000(1+\frac{13}{100})^3}}}}}}}$

${\qquad{\qquad{\qquad{\large{\longmapsto{\sf{A=100000(1+0.13)^3}}}}}}}$

${\qquad{\qquad{\qquad{\large{\longmapsto{\sf{A=100000(1.13)^3}}}}}}}$

${\qquad{\qquad{\qquad{\large{\longmapsto{\sf{A=100000(1.13\times1 .13 \times 1.13)}}}}}}}$

${\qquad{\qquad{\qquad{\large{\longmapsto{\sf{A=100000\times 1.442897}}}}}}}$

• C.I

A-P

${\sf{1.442897-100000}}$

${\large{\sf{C.I=144289.7}}}$

$\begin{gathered} \\ {\underline{\rule{200pt}{9pt}}} \end{gathered}$