Question

Surya borrowed some money from bank at 8% per annum simple interest After 3 years, he paid an interest of ₹ 480. Find the principal amount borrowed by him.

Answer 1

$\large \color{red} \bold{Given,} \\ \\ \sf \: \boxed{ \sf \: Rate \: of \: Intrest \: = \: 8\% \: \: \: | \: \: \: Time \: = \: 3 \: years \: \: \: | \: \: \: Intrest \: = \: ₹ \: 480} \\ \\ \color{orange} \bold{To \: find \: the \: Principal \: amount \: borrowed \: by \: him.} \\ \\ \large \color{lime} \bold{Solution \: ―} \\ \\ \boxed{ \sf \: Simple\:Intrest\:=\: \frac{P \: \times \: R \: \times \: T}{100}} \\ \\ \sf \: 480 \: = \: \frac{P \: \times \: 8 \: \times 3}{100} \\ \\ \sf \: P \: = \: \frac{480 \: \times \: 100}{8 \: \times \: 3} \\ \\ \sf \: \color{olive}P \: = \: ₹\:2000 \\ \\ \sf \color{skyblue} \: The \: Principal \: amount \: borrowed \: by \: him \: is \: ₹\:2000.$

$\color{red}\rule{200000000 pt}{2pt}$

$\huge \color{gold}\begin{gathered}\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\cal{\pmb{ \pink{More \: Formulae}}}} \\ \\ \bigstar \: Si \: = \: \frac{P \times R \times T}{100} \\ \\ \\ \color{gold}Amount \: = \: Principal\:+\:Intrest \\ \\ \\ \color{gold}Formula \: for \: Ci \: = \: A\:=\:P( {\frac{1 + r}{n} )}^{n \times t } \\ \\ \\ \color{gold}where \: \: A =\:final\: amount \\\color{aqua} \\ \\ P\:=\: Principal \\ \\ \\ \color{gold}r \: = \:rate \: of \: intrest \\ \color{aqua} \\ \\ n \: = \: number \: of \: times \: interest \: applied \: per \: time \: period \\ \color{aqua} \end{array} }}\end{gathered}\end{gathered}\end{gathered} \\ \end{gathered}$