Question

Ajay deposited rupees 20000 in a bank where compound interest at the rate of 6% annually. How much amount he will get at the end of third year. ​

Answer 1

Answer:

Solution :

$\sf{\longrightarrow{A=P(1 + \dfrac{R}{100} )^{n}}}$

$\sf{\longrightarrow{A=20,000 \: (1 \: + \: \dfrac{6}{100} ) ^{3}}}$

$\sf{\longrightarrow{A=20,000 \: (1 \: + \: 0.06)^{3}}}$

$\sf{\longrightarrow{A=20,000 \: (1.06) ^{3}}}$

$\sf{\longrightarrow{A=20,000 \: \times \: 1.191016}}$

A = 23,820.32

• Therefore, he gets Rs. 23,820.32 at the end of 3 years.

Answer 2

Answer:

$\begin{gathered}{\underline{\underline{\maltese{\large{\textsf{\textbf{\purple{ Given:}}}}}}}}\end{gathered}$

• $\dashrightarrow$ Principle = Rs.20000
• $\dashrightarrow$ Rate of Interest = 6%
• $\dashrightarrow$ Time = 3 years

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

$\begin{gathered}{\underline{\underline{\maltese{\large{\textsf{\textbf{\purple{ To Find:}}}}}}}}\end{gathered}$

• $\dashrightarrow$ Amount

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

$\begin{gathered}{\underline{\underline{\maltese{\large{\textsf{\textbf{\purple{ Using Formula:}}}}}}}}\end{gathered}$

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

Where

• $\Rightarrow$ A = Amount
• $\Rightarrow$ P = Principle
• $\Rightarrow$ R = Rate of Interest
• $\Rightarrow$ T = Time Period

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

$\begin{gathered}{\underline{\underline{\maltese{\large{\textsf{\textbf{\purple{ Solution:}}}}}}}}\end{gathered}$

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

• Substituting the values

$\quad{: \implies{\sf{Amount ={20000{\bigg(1 + \dfrac{6}{100}{\bigg)}^{3}}}}}}$

$\quad{: \implies{\sf{Amount ={20000{\bigg(\dfrac{(1 \times 100) + 6}{100}{\bigg)}^{3}}}}}}$

$\quad{: \implies{\sf{Amount ={20000{\bigg(\dfrac{ 100 + 6}{100}{\bigg)}^{3}}}}}}$

$\quad{: \implies{\sf{Amount ={20000{\bigg(\dfrac{ 106}{100}{\bigg)}^{3}}}}}}$

$\quad{: \implies{\sf{Amount ={20000{\bigg( \cancel{\dfrac{106}{100}}{\bigg)}^{3}}}}}}$

$\quad{: \implies{\sf{Amount ={20000{\bigg(\dfrac{53}{50}{\bigg)}^{3}}}}}}$

$\quad{: \implies{\sf{Amount ={20000{\bigg(\dfrac{53}{50} \times \dfrac{53}{50} \times \dfrac{53}{50}{\bigg)}}}}}}$

$\quad{: \implies{\sf{Amount ={20000{\bigg(\dfrac{148877}{125000}{\bigg)}^{3}}}}}}$

$\quad{: \implies{\sf{Amount ={20000 \times \dfrac{148877}{125000}}}}}$

$\quad{: \implies{\sf{Amount = {\cancel{20000} \times \dfrac{148877}{\cancel{125000}}}}}}$

$\quad{: \implies{\sf{Amount ={0.16 \times 148877}}}}$

$\quad{: \implies{\sf{Amount ={Rs.23820.32}}}}$

$\quad\dag\underline{\boxed{\sf{\red{Amount ={Rs.23820.32}}}}}$

• $\dashrightarrow$ Henceforth,The Amount is Rs.23820.32.

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

$\begin{gathered}{\underline{\underline{\maltese{\large{\textsf{\textbf{\purple{ Learn More :}}}}}}}}\end{gathered}$

$\quad\circ{\underline{\boxed{\sf{\red{Amount = Principle + Interest}}}}}$

$\quad\circ{\underline{\boxed{\sf{\red{ P=Amount - Interest }}}}}$

$\quad\circ{\underline{\boxed{\sf{\red{ S.I = \dfrac{P \times R \times T}{100}}}}}}$

$\quad\circ{\underline{\boxed{\sf{\red{P = \dfrac{Amount\times 100 }{100 + (Time \times Rate)}}}}}}$

$\quad\circ{\underline{\boxed{\sf{\red{P = \dfrac{Interest \times 100 }{Time \times Rate}}}}}}$