Question

At what rate percent per annum, compound interest will Rs. 10000 amount to Rs. 13310 in three years ?

Answer 1

Given :-

• Principal  is  Rs. 10000  

• Amount is Rs. 13310  

• Time is three years

To Find :-

• The rate percent

Solution :-

~Here, we’re given the principal , amount and time for which the money is invested and we need to find the rate of interest on which it was compounded for the given time. We can easily find the rate of interest by putting the values in the formula of finding amount.

_____________

Here,  

• The principal  ( P ) is Rs. 10,000

• The rate percent is ( r ) which we need to find  

• The time ( n )  is 3 years  

• The amount ( A )  is Rs. 13310  

_____________

As we know that ,  

 $\boxed{\sf{ \maltese \;\; A = P \bigg\{ 1 + \dfrac{r}{100} \bigg\} }}$

Finding the rate percent :-  

$\sf \implies 13310 = 10000 \bigg\{ 1 + \dfrac{r}{100} \bigg\}^{3}$ 

$\sf \implies \dfrac{13310}{10000} = \bigg\{ 1 + \dfrac{r}{100} \bigg\}$  

$\sf \implies \bigg\{ \dfrac{11^{3}}{10^{3}} \bigg\} = \bigg\{ 1 + \dfrac{r}{100} \bigg\}$

$\sf \implies 1 + \dfrac{r}{100} = \dfrac{11}{10}$  

$\sf \implies \dfrac{r}{100} = \dfrac{11}{10} -1$ 

$\sf \implies \dfrac{r}{100} = \dfrac{1}{10}$  

$\sf \implies r = \dfrac{100}{10}$ 

$\boxed{\sf{ \star \;\; r = 10 \% }}$ 

Hence,  

• The rate percent is 10 % per annum.  

Answer 2

Given : Principal  is  Rs. 10000   , Amount is Rs. 13310  & Time is 3 years

Need To Find : The Rate of Interest .

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⠀⠀⠀⠀Formula for Finding Amount is given by :

$\dag\:\:\boxed{ \sf{ Amount = \bigg[ P \bigg( 1 + \dfrac {R}{100} \bigg) ^T \bigg] }}$

Where,

• P is the Principal , R is the Rate of Interest & T is the Time .

⠀⠀⠀⠀⠀⠀$\underline {\bf{\star\:Now \: By \: Substituting \: the \: Given \: Values \::}}$

$\qquad \:\::\implies \sf{ 13310 = \bigg[ 10000 \bigg( 1 + \dfrac {R}{100} \bigg) ^3 \bigg] }$

$\qquad \:\::\implies \sf{ \dfrac{13310}{10000} = \bigg( 1 + \dfrac {R}{100} \bigg) ^3 }$

$\qquad \:\::\implies \sf{ \dfrac{(11)^3}{(10)^3} = \bigg( 1 + \dfrac {R}{100} \bigg) ^3 }$

$\qquad \:\::\implies \sf{ \bigg(\dfrac{11}{10}\bigg)^3 = \bigg( 1 + \dfrac {R}{100} \bigg) ^3 }$

Canceling Exponent 3 from Both side :

$\qquad \:\::\implies \sf{ \bigg(\dfrac{11}{10}\bigg)^{\cancel {3}} = \bigg( 1 + \dfrac {R}{100} \bigg) ^{\cancel {3}} }$

$\qquad \:\::\implies \sf{ \dfrac{11}{10} = 1 + \dfrac {R}{100} }$

$\qquad \:\::\implies \sf{ \dfrac{11}{10} - 1= \dfrac {R}{100} }$

$\qquad \:\::\implies \sf{ \dfrac{1}{10} = \dfrac {R}{100} }$

$\qquad \:\::\implies \sf{R = \cancel {\dfrac {100}{10} } }$

⠀⠀⠀⠀⠀$\underline {\boxed{\pink{ \mathrm {  R = 10\%\: }}}}\:\bf{\bigstar}$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {  Rate \:of\:Interest \:is\:\bf{10\: \%}}}}$

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