Question

Calculate the amount and the compound interest on 12,000 in 3 years when the rates
of interest for successive years are 8%, 10% and 15% respectively.
please explain on paper​

Answer 1

Answer:

${\large{\pmb{\sf{\underline{Given...}}}}}$

• $\red\bigstar$ Principle = 12000
• $\red\bigstar$ Time = 3 years
• $\red\bigstar$ Rates of Interest = 8%, 10%, 15%

$\begin{gathered}\end{gathered}$

${\large{\pmb{\sf{\underline{To \: Find ...}}}}}$

• $\red\bigstar$ Amount
• $\red\bigstar$ Compound Interest

$\begin{gathered}\end{gathered}$

${\large{\pmb{\sf{\underline{Using \: Formulae...}}}}}$

$\begin{gathered}\bigstar{\underline{\boxed{\sf {\purple{A =P\bigg\lgroup{1 +\dfrac{R_1}{100} }\bigg\rgroup\bigg\lgroup{1 + \dfrac{R_2}{100}}\bigg\rgroup\bigg\lgroup{1 + {\dfrac{R_3}{100}}}\bigg\rgroup}}}}} \end{gathered}$

Where

• $\green\star$ A = Amount
• $\green\star$ P = Principle
• $\green\star$ $\sf{R_1}$ = Rate Interest of first year
• $\green\star$ $\sf{R_2}$ = Rate Interest of second year
• $\green\star$ $\sf{R_1}$ = Rate Interest of third year

$\bigstar{\underline{\boxed{\sf{\purple{C.I = \big\lgroup{A - P \big\rgroup}}}}}}$

Where

• $\green\star$ C.I = Compound Interest
• $\green\star$ A = Amount
• $\green\star$ P = Principle

$\begin{gathered}\end{gathered}$

${\large{\pmb{\sf{\underline{Solution...}}}}}$

$\pink{\dag \: {\underline{\frak{Finding \: the \: Amount : }}}}$

$\quad{: \implies{\sf{A = \bf{P\bigg\lgroup{1 +\dfrac{R_1}{100} }\bigg\rgroup\bigg\lgroup{1 + \dfrac{R_2}{100}}\bigg\rgroup\bigg\lgroup{1 + {\dfrac{R_3}{100}}}\bigg\rgroup}}}}$

• Substituting the values

$\quad{: \implies{\sf{A = \bf{12000\bigg\lgroup{1 +\dfrac{8}{100} }\bigg\rgroup\bigg\lgroup{1 + \dfrac{10}{100}}\bigg\rgroup\bigg\lgroup{1 + {\dfrac{15}{100}}}\bigg\rgroup}}}}$

$\quad{: \implies{\sf{A = \bf{12000\bigg\lgroup{\dfrac{(1 \times 100) + 8}{100}}\bigg\rgroup\bigg\lgroup{\dfrac{(1 \times 100) + 10}{100}}\bigg\rgroup\bigg\lgroup{\dfrac{(1 \times 100) + 15}{100}}\bigg\rgroup}}}}$

$\quad{: \implies{\sf{A = \bf{12000\bigg\lgroup{\dfrac{100+ 8}{100}}\bigg\rgroup\bigg\lgroup{\dfrac{100 + 10}{100}}\bigg\rgroup\bigg\lgroup{\dfrac{100 + 15}{100}}\bigg\rgroup}}}}$

$\quad{: \implies{\sf{A = \bf{12000\bigg\lgroup{\dfrac{108}{100}}\bigg\rgroup\bigg\lgroup{\dfrac{110}{100}}\bigg\rgroup\bigg\lgroup{\dfrac{115}{100}}\bigg\rgroup}}}}$

$\quad{: \implies{\sf{A = \bf{12000\ \times {\dfrac{108}{100}} \times {\dfrac{110}{100}} \times {\dfrac{115}{100}}}}}}$

$\quad{: \implies{\sf{A = \bf{12\cancel{000}\ \times {\dfrac{1366200}{1000\cancel{000}}}}}}}$

$\quad{: \implies{\sf{A = \bf{\dfrac{12 \times 1366200}{1000}}}}}$

$\quad{: \implies{\sf{A = \bf{\dfrac{16394400}{1000}}}}}$

$\quad{:\implies{\sf{A = \bf{\cancel{\dfrac{16394400}{1000}}}}}}$

$\quad{: \implies{\sf{\purple{A = \bf{16394.4}}}}}$

${\bigstar{\underline{\boxed{\sf{Amount = Rs.16394.40}}}}}$

$\begin{gathered}\end{gathered}$

$\pink{\dag \: {\underline{\frak{Finding \: Compound \: Interest : }}}}$

$\quad{ : \implies{\sf{C.I = \bf\big\lgroup{A - P \big\rgroup}}}}$

• Substituting the values

$\quad{ : \implies{\sf{C.I = \bf\big\lgroup{ 16394.40 - 12000\big\rgroup}}}}$

$\quad{ : \implies{\sf{\purple{C.I = \bf{Rs.4394.40}}}}}$

${\bigstar{\underline{\boxed{\sf{Compound \: Interest = Rs.4394.40}}}}}$

$\begin{gathered}\end{gathered}$

$\pink{\dag \: {\underline{\frak{Hence : }}}}$

• $\red\bigstar$ The Amount is Rs.16394.40..
• $\red\bigstar$ The Compound Interest is Rs.4384.40..

$\begin{gathered}\end{gathered}$

${\large{\pmb{\sf{\underline{Learn \: More...}}}}}$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \dag \: \underline{\bf{More \: Useful \: Formula}}\\ {\boxed{\begin{array}{cc}\dashrightarrow {\sf{Amount = Principle + Interest}} \\ \\ \dashrightarrow \sf{ P=Amount - Interest }\\ \\ \dashrightarrow \sf{ S.I = \dfrac{P \times R \times T}{100}} \\ \\ \dashrightarrow \sf{P = \dfrac{Interest \times 100 }{Time \times Rate}} \\ \\ \dashrightarrow \sf{P = \dfrac{Amount\times 100 }{100 + (Time \times Rate)}} \\ \end{array}}}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$