Question

Find compound interest and amount when principal is Rs 5,000, rate is 10 paise per rupee per annum and time is 2 years.​

Answer 1

Answer:

Rs 1050

Step-by-step explanation:

a=p(1+r/100)n

Answer 2

Answer:-

$\boxed{\sf{\large { Amount =\:Rs.\:6050 }}}$

$\boxed{\sf{\large { Compound \:Interest\:=\:Rs.1050 \: }}}$

$\green\dag{\sf{\large { Eҳ℘Ɩąŋątเօղ:\:}}}$

$\begin{gathered} \frak{Given \:\: -:} \begin{cases} \sf{Principal= \frak{Rs.5000}} & \\\\ \sf{Rate \:=\:\frak{10\%}}& \\\\ \sf{Time \:=\:\frak{2 \ year}}\end{cases} \\\\ \end{gathered}$

$\begin{gathered} \frak{To \:Find\: -:} \begin{cases} \sf{The\:Amount, \: and, } & \\\\ \sf{The\:Compound\:Interest. }\end{cases} \\\\ \end{gathered}$

$\red\dag{\sf{\large { SօӀմtเօղ-:\:}}}$

$\implies{\sf{\large { Amount \: =P\left(1 \: +\dfrac{R}{100} \right)^{n}}}}$

$\begin{gathered} \frak{Here \:\: -:} \begin{cases} \sf{P \:= \:Principal\:= \frak{Rs.5000}} & \\\\ \sf{R\:=Rate \:=\:\frak{10\%}}& \\\\ \sf{n \: = Time \:=\:\frak{2 \ year}}& \\\\ \sf{A\:=Amount \:=\:\frak{??}}\end{cases} \\\\ \end{gathered}$

$\pink\dag{\sf{\large { Now,\:Substitute \:the\:Known\:value\:in\:Formula.-: }}}$

$\implies{\sf{\large { Amount \: =5000 \left(1 \: +\dfrac{10}{100} \right)^{2}}}}$

$\bigstar{\sf{\large { Oɾ\:}}}$

$\implies{\sf{\large { Amount \: =5000\left(\dfrac{1}{1} \: +\dfrac{10}{100} \right)^{2}}}}$

$\implies{\sf{\large { Amount \: =5000\left(\dfrac{1}{1} \: +\dfrac{1}{10} \right)^{2}}}}$

$\implies{\sf{\large { LCM \:of\:10\:and\:1\:is = \:10}}}$

$\implies{\sf{\large { Amount \: =5000\left( \: \dfrac{10+1}{10} \right)^{2}}}}$

$\implies{\sf{\large { Amount \: =5000\left( \: \dfrac{11}{10} \right)^{2}}}}$

$\implies{\sf{\large { 11^{2} = \:121\:and\:,\:10^{2}= 100}}}$

$\implies{\sf{\large { Amount \: =5000 \times \dfrac{121}{100} }}}$

$\implies{\sf{\large { Amount \: =50 \times 121 }}}$

$\implies{\sf{\large { Amount =\:Rs.\:6050 }}}$