Question

Find the compound interest on Rs.2000 at the rate of 10%p.a.for 2 years.

Answer 1

Given:-

• Principal = 2000rupees
• Rate = 10%
• Time = 2 years

To Find:-

• Compound Interest

Solution:-

we have,

• P = 2000rupees
• R = 10%
• T = 2 years

Using Formula:

$\: \: \sf \: c.i = p \times { {(1 + \frac{r}{100} )}^{t} - 1}$

Now substitute the values,

$\: \: \sf \: c.i = 2000 \times {(1 + \frac{10}{100}) }^{2} - 1 \\ \\ \: \: \sf \: c.i = 2000 \times ( \frac{121 - 100}{100} ) \\ \\ \: \: \sf \: c.i = 2000 \times \frac{21}{100} \\ \\ \: \: \sf \: c.i = 420rupees$

Hence, Compound Interest is 420 rupees.

Answer 2

Answer:

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

• ➤ Principle = Rs.2000
• ➤ Rate = 10%
• ➤ Time = 2 Years

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

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{green}{To Find:}}}}}}}\end{gathered}$

• ➤ Compound Interest

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

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{green}{Using Formula:}}}}}}}\end{gathered}$

${\dag{\underline{\boxed{\sf{Amount ={P{\bigg(1 + \dfrac{R}{100}{\bigg)}^{T}}}}}}}}$

$\dag{\underline{\boxed{\sf{Compound \: Interest = Amount- Principle }}}}$

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

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{green}{Full Solution:}}}}}}}\end{gathered}$

${\bigstar \:{\underline{\pmb{\frak{\red{Firstly,Finding \: the \: Amount }}}}}}$

$\quad {:\implies{\sf{Amount = \bf{P{\bigg(1 + \dfrac{R}{100}{\bigg)}^{T}}}}}}$

• Substituting the values

$\quad {:\implies{\sf{Amount = \bf{2000{\bigg(1 + \dfrac{10}{100}{\bigg)}^{2}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{2000{\bigg(\dfrac{1 \times 100 + 10}{100}{\bigg)}^{2}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{2000{\bigg(\dfrac{100 + 10}{100}{\bigg)}^{2}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{2000{\bigg( \dfrac{110}{100}{\bigg)}^{2}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{2000{\bigg({\cancel\dfrac{110}{100}}{\bigg)}^{2}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{2000{\bigg( \dfrac{11}{10}{\bigg)}^{2}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{2000{\bigg( \dfrac{11}{10} \times \dfrac{11}{10}{\bigg)}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{2000{\bigg( \dfrac{121}{100}{\bigg)}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{2000 \times {\dfrac{121}{100}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{\cancel{2000} \times {\dfrac{121}{\cancel{100}}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{20\times{121}}}}}$

$\quad {:\implies{\sf{Amount = \bf{{2420}}}}}$

$\begin{gathered} \dag{\boxed{\textsf{\textbf{\underline{\color{purple}{Amount = {Rs.2420}}}}}}}\end{gathered}$

• ➤ Hence, The Amount is Rs.2420.

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

${\bigstar \:{\underline{\pmb{\frak{\red{ Now,Finding \: The \: Compound \: Interest }}}}}}$

$\quad{: \implies{\sf{Compound \: Interest = \bf{Amount- Principle }}}}$

• Substituting the values

$\quad{: \implies{\sf{Compound \: Interest = \bf{2420-2000}}}}$

$\quad{: \implies{\sf{Compound \: Interest = \bf{420}}}}$

$\begin{gathered} \dag{\boxed{\textsf{\textbf{\underline{\color{purple}{Compound Interest = Rs.420}}}}}}\end{gathered}$

• ➤ Henceforth,The Compound Interest is Rs.420.

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

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{green}{Learn More:}}}}}}}\end{gathered}$

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