Question

The amount to be paid at the end of 2 years when principal is Rs: 5,000 and rate of interest is

6% is​

Answer 1

Answer:

Detailed Solution. ∴ The amount is Rs. 5750.

Answer 2

$\large {\dag \; {\underline{\underline{\blue{\pmb{\textbf{\textsf{ \; Solution :- }}}}}}}}$

Amount is found by a Formula is =

$\begin{gathered}\boxed{\tt{ \: \: Amount \: = \: p \: {\bigg[1 + \dfrac{r}{200} \bigg]}^{n} \: \: }} \\ \end{gathered}$

Where P is principal, R is rate of interest ,n is number of Times/years .

P = ₹5,000

R = 6%(per annum)

N = 2 years .

Amount :-

$\rm \: \: Amount \: = \: 5000\: {\bigg[1 + \dfrac{6 }{100} \bigg]} {}^{2}$

$\rm \: \: Amount \: = \: 5000 \: {\bigg[\dfrac{106}{100} \bigg]} {}^{2}$

$\rm \: \: Amount \: = \: 5000 \: {\bigg[\dfrac{10 6}{100} \bigg]} \times {\bigg[\dfrac{10 6}{100} \bigg]}$

$A \sf \: mount \: = ₹561.8$

Hence , the Amount is RS.561.8

More Formulae`s :-

$\begin{gathered}\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{P = \dfrac{SI \times 100}{r \times n} }\\ \\ \bigstar \: \bf{r = \dfrac{SI \times 100}{P \times n} }\\ \\ \bigstar \: \bf{n = \dfrac{SI \times 100}{P \times r} }\\ \\ \bigstar \: \bf{Amount = P\bigg(\dfrac{100 + rn}{100} \bigg) }\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$