Question

what sm will amount to rupees 40,00 in 3 tears at 6% c.a. compond intest ???¿??¿

Answer with full explanation !​

Answer 1

Corrected Question :

• What sum will amount to rupees 4000 in 3 years at 6% p.a. compound interest ?

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Given : sum will amount to rupees 4000 in 3 years 6% c.a.

To Find : Find Amount ?

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$\underline{\frak{As ~we~ know~ that~:}}$

• $\boxed{\sf\pink{Amount~=~P\bigg(1~+~\dfrac{R}{100}\bigg)^{T}}}$

• $\boxed{\sf\purple{Compound~-~Interest~=~Amount~-~Principle}}$

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Solution : Firstly, Finding the amount

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• ${\sf\longmapsto{Amount~=~P\bigg(1~+~\dfrac{R}{100}\bigg)^{T}}}$

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

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$~~~~~~~~~~{\sf:\implies{Amount~=~4000\bigg(1~+~\dfrac{6}{100}\bigg)^{3}}}$

$~~~~~~~~~{\sf:\implies{Amount~=~4000\bigg(1~×~100~+~\dfrac{6}{100}\bigg)^{3}}}$

$~~~~~~~~~{\sf:\implies{Amount~=~4000\bigg(\dfrac{100~+~6}{100}\bigg)^{3}}}$

$~~~~~~~~~{\sf:\implies{Amount~=~4000\bigg(\dfrac{106}{100}\bigg)^{3}}}$

$~~~~~~~~~{\sf:\implies{Amount~=~4000\bigg(\cancel\dfrac{106}{100}\bigg)^{3}}}$

$~~~~~~~~~~{\sf:\implies{Amount~=~4000\bigg(\dfrac{53}{50}\bigg)^{3}}}$

$~~~~~~~~~~{\sf:\implies{Amount~=~4000\bigg(\dfrac{53}{50}~×~\dfrac{53}{50}~×~\dfrac{53}{50}\bigg)}}$

$~~~~~~~~~~{\sf:\implies{Amount~=~4000\bigg(\dfrac{148877}{125000}\bigg)}}$

$~~~~~~~~~~{\sf:\implies{Amount~=~4000~×~\dfrac{148877}{125000}}}$

$~~~~~~~~~~{\sf:\implies{Amount~=~4\cancel{000}~×~\dfrac{148877}{125\cancel{000}}}}$

$~~~~~~~~~~{\sf:\implies{Amount~=~\dfrac{148877~×~4}{125}}}$

$~~~~~~~~~~{\sf:\implies{Amount~=~\dfrac{595508}{125}}}$

$~~~~~~~~~~{\sf:\implies{Amount~=~\cancel\dfrac{595508}{125}}}$

$~~~~~~~~~~{\sf:\implies{Amount~=~4764.064}}$

$~~~~~~~~~~:\implies{\underline{\boxed{\frak{\purple{Amount~=~Rs.~4764.064}}}}}$★

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Therefore,

• The Amount is Rs. 4764.064.

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$\underline{\frak{Now ~Finding ~the ~Compound ~Interest~-}}$

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• ${\sf\longmapsto{Compound ~Interest~=~Amount~-~Principle}}$

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$\underline{\bf{Now~ By ~Substituting~ the ~Given ~and~ Found~ Values~:}}$

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$~~~~~~~~~~{\sf:\implies{Compound ~Interest~=~47.64.064~-~4000}}$

$~~~~~~~~~~{\sf:\implies{Compound ~Interest~=~764.046}}$

• $\underset{\blue{\rm Required\ Answer}}{\underbrace{\boxed{\frak{\pink{Compound ~Interest~=~Rs.~764.046}}}}}$

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Hence,

$\therefore\underline{\sf{The~ Compound ~Interest~is~\bf{Rs.~764.046}}}$

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More Information :

• ${\rm\leadsto{Amount~=~ Principal~+~Interest}}$

• ${\rm\leadsto{Principal~=~Amount~-~Interest}}$

• ${\rm\leadsto{S.I~=~\dfrac{P~×~R~×~T}{100}}}$

• ${\rm\leadsto{Principal~=~\dfrac{Interest~×~100}{Time~×~Rate}}}$

• ${\rm\leadsto{Principal~=~\dfrac{Amount~×~100}{100~+~(Time~×~Rate)}}}$

Answer 2

Appropriate Question :

• Principal is Rs. 4000 , Time is 3 years and Rate of interest is 6 % p.a . Find it's Compound Interest and Amount .

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Given : Principal is Rs. 4000 , Time is 3 years and Rate of interest is 6 % p.a .

Need To Find : Amount & Compound Interest ?

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          Finding Amount to find Compound Interest :

As , We know that ,

$\qquad \dag \:\:\:\boxed { \pink{\pmb{\:\: \: Amount \:=\: P \:\bigg( 1 + \dfrac{R}{100} \bigg)^n }}}$

          Here , P is the Principal , R is the Rate of interest & n is the Time .

$\qquad\dashrightarrow \sf Amount \:=\: P \:\bigg( 1 + \dfrac{R}{100} \bigg)^n$

$\qquad \bigstar \:\underline {\boldsymbol { Now \:,\:By\:Substituting \:known \:values \: }} :$

$\qquad\dashrightarrow \sf Amount \:=\: P \:\bigg( 1 + \dfrac{R}{100} \bigg)^n$

$\qquad\dashrightarrow \sf Amount \:=\: 4000 \:\bigg( 1 + \dfrac{6}{100} \bigg)^3$

$\qquad\dashrightarrow \sf Amount \:=\: 4000 \:\bigg( \dfrac{106}{100} \bigg)^3$

$\qquad\dashrightarrow \sf Amount \:=\: 4000 \:\bigg( \cancel {\dfrac{106}{100}} \bigg)^3$

$\qquad\dashrightarrow \sf Amount \:=\: 4000 \:\bigg( 1.06 \bigg)^3$

$\qquad\dashrightarrow \sf Amount \:=\: 4000 \:\times 1.191016$

$\qquad\dashrightarrow \sf Amount \:=\: 4,764.064$

$\qquad \therefore \: \underline {\purple{ \pmb {\bf Amount \:\:=\:\: Rs.\: 4,764.064}}}\:\: \bigstar$

$\therefore \underline {\sf The \:Amount \:\:is \bf\:Rs. \:\: 4764.064 \:\: }$

                    Now , Finding Compound Interest :

As , We know that ,

$\qquad \dag \:\:\:\boxed { \pink{\pmb{\:\:Compound \:Interest \: \:=\: Amount\:\: - \:\:Principal \:\:}}}$

              Here , Amount is Rs. 4764.064 & Principal is Rs. 4000

$\qquad\dashrightarrow \sf \:Compound \:Interest \: \:=\: Amount\:\: - \:\:Principal \:$

$\qquad \bigstar \:\underline {\boldsymbol { Now \:,\:By\:Substituting \:known \:values \: }} :$

$\qquad\dashrightarrow \sf \:Compound \:Interest \: \:=\: Amount\:\: - \:\:Principal \:$

$\qquad\dashrightarrow \sf \:Compound \:Interest \: \:=\: 4764.064\:\: - \:\:4000 \:$

$\qquad\dashrightarrow \sf \:Compound \:Interest \: \:=\: 764.064\:\: \:$

$\qquad \therefore \: \underline {\purple{ \pmb {\bf Compound \:Interest \:\:=\:\: Rs.\: 764.064}}}\:\: \bigstar$

$\therefore \underline {\sf Hence , \:The \:Compound \:Interest \:\:is \bf\:Rs. \:\: 764.064 \:\: }$