Question

what sum will become ₹1331 after three years for 10% for compound interest ?​

Answer 1

$\large\underline{ \underline{ \sf \maltese{ \: Question:- }}}$

• What sum will become ₹1331 after three years for 10% for compound interest ?

━━━━━━━━━━━━━━━━━━━━━━━━━

$\large\underline{ \underline{ \sf \maltese{ \: Given:- }}}$

1. Amount = ₹1331
2. Time = 3years
3. Rate = 10%
4. Principal = ?

━━━━━━━━━━━━━━━━━━━━━━━━━

$\large\underline{ \underline{ \sf \maltese{ \: Solution:- }}}$

• There are four variables( Amount (A) , Principal (P) , Rate (r) , Time (n) .Out of these four if any three are known then we can calculate the fourth variable.

$\begin{gathered}\begin{gathered}\begin{gathered}: \implies \underline\blue{ \boxed{\displaystyle \sf \bold{\: \sf{A = p\:\: \bigg(1 + \dfrac{r}{100}\bigg)^{n}} }} }\\ \\\end{gathered}\end{gathered}\end{gathered}$

${\qquad \quad {:}\longrightarrow\sf{\sf{1331\: = \: p \bigg(1 + \dfrac{10}{100}\bigg)^{3}} }}$

${\qquad \quad {:}\longrightarrow\sf{\sf{1331\: = \: p \bigg( \dfrac{110}{100}\bigg)^{3}} }}$

${\qquad \quad {:}\longrightarrow\sf{\sf{1331\: = \: p \bigg( \dfrac{110}{100}\bigg) \bigg( \dfrac{110}{100}\bigg) \bigg( \dfrac{110}{100}\bigg) } }}$

${\qquad \quad {:}\longrightarrow\sf{\sf{1331\: = \: p \: (\:1.331\:)} }}$

${\qquad \quad {:}\longrightarrow\sf{\sf{\: \frac{1331}{1.331} = \: p \: } }}$

$\qquad\quad {:} \longrightarrow \underline \red{\boxed{\sf{1000 \: = \: Principal }}}$

━━━━━━━━━━━━━━━━━━━━━━━━━

$\begin{gathered}\begin{gathered}\qquad \therefore\: \sf{ Principal \: = \underline {\underline{ ₹1000 }}}\\\end{gathered}\end{gathered}$

━━━━━━━━━━━━━━━━━━━━━━━━━

Answer 2

A=P(1+ r/100)n

⟹13310=P(1+ 10/100 )3

Solving for P we get P=10,000

Hence sum=Rs.10,000.