what sm will amount to rupees 40,00 in 3 tears at 6% c.a. compond intest ???¿??¿ Answer with full explanation ! About the author Alexandra
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 . ________________________ Given : Principal is Rs. 4000 , Time is 3 years and Rate of interest is 6 % p.a . Need To Find : Amount & Compound Interest ? ________________________ Finding Amount to find Compound Interest : As , We know that , [tex]\qquad \dag \:\:\:\boxed { \pink{\pmb{\:\: \: Amount \:=\: P \:\bigg( 1 + \dfrac{R}{100} \bigg)^n }}}\\\\[/tex] Here , P is the Principal , R is the Rate of interest & n is the Time . [tex]\qquad\dashrightarrow \sf Amount \:=\: P \:\bigg( 1 + \dfrac{R}{100} \bigg)^n \\\\[/tex] [tex]\qquad \bigstar \:\underline {\boldsymbol { Now \:,\:By\:Substituting \:known \:values \: }} : \\\\[/tex] [tex]\qquad\dashrightarrow \sf Amount \:=\: P \:\bigg( 1 + \dfrac{R}{100} \bigg)^n \\\\[/tex] [tex]\qquad\dashrightarrow \sf Amount \:=\: 4000 \:\bigg( 1 + \dfrac{6}{100} \bigg)^3 \\\\[/tex] [tex]\qquad\dashrightarrow \sf Amount \:=\: 4000 \:\bigg( \dfrac{106}{100} \bigg)^3 \\\\[/tex] [tex]\qquad\dashrightarrow \sf Amount \:=\: 4000 \:\bigg( \cancel {\dfrac{106}{100}} \bigg)^3 \\\\[/tex] [tex]\qquad\dashrightarrow \sf Amount \:=\: 4000 \:\bigg( 1.06 \bigg)^3 \\\\[/tex] [tex]\qquad\dashrightarrow \sf Amount \:=\: 4000 \:\times 1.191016 \\\\[/tex] [tex]\qquad\dashrightarrow \sf Amount \:=\: 4,764.064 \\\\[/tex] [tex]\qquad \therefore \: \underline {\purple{ \pmb {\bf Amount \:\:=\:\: Rs.\: 4,764.064}}}\:\: \bigstar \\\\[/tex] [tex]\therefore \underline {\sf The \:Amount \:\:is \bf\:Rs. \:\: 4764.064 \:\: }\\\\[/tex] Now , Finding Compound Interest : As , We know that , [tex]\qquad \dag \:\:\:\boxed { \pink{\pmb{\:\:Compound \:Interest \: \:=\: Amount\:\: – \:\:Principal \:\:}}}\\\\[/tex] Here , Amount is Rs. 4764.064 & Principal is Rs. 4000 [tex]\qquad\dashrightarrow \sf \:Compound \:Interest \: \:=\: Amount\:\: – \:\:Principal \: \\\\[/tex] [tex]\qquad \bigstar \:\underline {\boldsymbol { Now \:,\:By\:Substituting \:known \:values \: }} : \\\\[/tex] [tex]\qquad\dashrightarrow \sf \:Compound \:Interest \: \:=\: Amount\:\: – \:\:Principal \: \\\\[/tex] [tex]\qquad\dashrightarrow \sf \:Compound \:Interest \: \:=\: 4764.064\:\: – \:\:4000 \: \\\\[/tex] [tex]\qquad\dashrightarrow \sf \:Compound \:Interest \: \:=\: 764.064\:\: \: \\\\[/tex] [tex]\qquad \therefore \: \underline {\purple{ \pmb {\bf Compound \:Interest \:\:=\:\: Rs.\: 764.064}}}\:\: \bigstar \\\\[/tex] [tex]\therefore \underline {\sf Hence , \:The \:Compound \:Interest \:\:is \bf\:Rs. \:\: 764.064 \:\: }\\\\[/tex] Reply
Corrected Question : What sum will amount to rupees 4000 in 3 years at 6% p.a. compound interest ? [tex]~[/tex] Given : sum will amount to rupees 4000 in 3 years 6% c.a. To Find : Find Amount ? ________________________ [tex]\underline{\frak{As ~we~ know~ that~:}}[/tex] [tex]\boxed{\sf\pink{Amount~=~P\bigg(1~+~\dfrac{R}{100}\bigg)^{T}}}[/tex] [tex]\boxed{\sf\purple{Compound~-~Interest~=~Amount~-~Principle}}[/tex] [tex]~[/tex] Solution : Firstly, Finding the amount [tex]~[/tex] [tex]{\sf\longmapsto{Amount~=~P\bigg(1~+~\dfrac{R}{100}\bigg)^{T}}}[/tex] [tex]~[/tex] [tex]\underline{\bf{Now ~By ~Substituting ~the ~Given~ Values~:}}[/tex] [tex]~[/tex] [tex]~~~~~~~~~~{\sf:\implies{Amount~=~4000\bigg(1~+~\dfrac{6}{100}\bigg)^{3}}}[/tex] [tex]~~~~~~~~~{\sf:\implies{Amount~=~4000\bigg(1~×~100~+~\dfrac{6}{100}\bigg)^{3}}}[/tex] [tex]~~~~~~~~~{\sf:\implies{Amount~=~4000\bigg(\dfrac{100~+~6}{100}\bigg)^{3}}}[/tex] [tex]~~~~~~~~~{\sf:\implies{Amount~=~4000\bigg(\dfrac{106}{100}\bigg)^{3}}}[/tex] [tex]~~~~~~~~~{\sf:\implies{Amount~=~4000\bigg(\cancel\dfrac{106}{100}\bigg)^{3}}}[/tex] [tex]~~~~~~~~~~{\sf:\implies{Amount~=~4000\bigg(\dfrac{53}{50}\bigg)^{3}}}[/tex] [tex]~~~~~~~~~~{\sf:\implies{Amount~=~4000\bigg(\dfrac{53}{50}~×~\dfrac{53}{50}~×~\dfrac{53}{50}\bigg)}}[/tex] [tex]~~~~~~~~~~{\sf:\implies{Amount~=~4000\bigg(\dfrac{148877}{125000}\bigg)}}[/tex] [tex]~~~~~~~~~~{\sf:\implies{Amount~=~4000~×~\dfrac{148877}{125000}}}[/tex] [tex]~~~~~~~~~~{\sf:\implies{Amount~=~4\cancel{000}~×~\dfrac{148877}{125\cancel{000}}}}[/tex] [tex]~~~~~~~~~~{\sf:\implies{Amount~=~\dfrac{148877~×~4}{125}}}[/tex] [tex]~~~~~~~~~~{\sf:\implies{Amount~=~\dfrac{595508}{125}}}[/tex] [tex]~~~~~~~~~~{\sf:\implies{Amount~=~\cancel\dfrac{595508}{125}}}[/tex] [tex]~~~~~~~~~~{\sf:\implies{Amount~=~4764.064}}[/tex] [tex]~~~~~~~~~~:\implies{\underline{\boxed{\frak{\purple{Amount~=~Rs.~4764.064}}}}}[/tex]★ [tex]~[/tex] Therefore, The Amount is Rs. 4764.064. [tex]~[/tex] [tex]\underline{\frak{Now ~Finding ~the ~Compound ~Interest~-}}[/tex] [tex]~[/tex] [tex]{\sf\longmapsto{Compound ~Interest~=~Amount~-~Principle}}[/tex] [tex]~[/tex] [tex]\underline{\bf{Now~ By ~Substituting~ the ~Given ~and~ Found~ Values~:}}[/tex] [tex]~[/tex] [tex]~~~~~~~~~~{\sf:\implies{Compound ~Interest~=~47.64.064~-~4000}}[/tex] [tex]~~~~~~~~~~{\sf:\implies{Compound ~Interest~=~764.046}}[/tex] [tex]\underset{\blue{\rm Required\ Answer}}{\underbrace{\boxed{\frak{\pink{Compound ~Interest~=~Rs.~764.046}}}}}[/tex] [tex]~[/tex] Hence, [tex]\therefore\underline{\sf{The~ Compound ~Interest~is~\bf{Rs.~764.046}}}[/tex] [tex]~[/tex] _____________________________________ More Information : [tex]{\rm\leadsto{Amount~=~ Principal~+~Interest}}[/tex] [tex]{\rm\leadsto{Principal~=~Amount~-~Interest}}[/tex] [tex]{\rm\leadsto{S.I~=~\dfrac{P~×~R~×~T}{100}}}[/tex] [tex]{\rm\leadsto{Principal~=~\dfrac{Interest~×~100}{Time~×~Rate}}}[/tex] [tex]{\rm\leadsto{Principal~=~\dfrac{Amount~×~100}{100~+~(Time~×~Rate)}}}[/tex] Reply
Appropriate Question :
________________________
Given : Principal is Rs. 4000 , Time is 3 years and Rate of interest is 6 % p.a .
Need To Find : Amount & Compound Interest ?
________________________
Finding Amount to find Compound Interest :
As , We know that ,
[tex]\qquad \dag \:\:\:\boxed { \pink{\pmb{\:\: \: Amount \:=\: P \:\bigg( 1 + \dfrac{R}{100} \bigg)^n }}}\\\\[/tex]
Here , P is the Principal , R is the Rate of interest & n is the Time .
[tex]\qquad\dashrightarrow \sf Amount \:=\: P \:\bigg( 1 + \dfrac{R}{100} \bigg)^n \\\\[/tex]
[tex]\qquad \bigstar \:\underline {\boldsymbol { Now \:,\:By\:Substituting \:known \:values \: }} : \\\\[/tex]
[tex]\qquad\dashrightarrow \sf Amount \:=\: P \:\bigg( 1 + \dfrac{R}{100} \bigg)^n \\\\[/tex]
[tex]\qquad\dashrightarrow \sf Amount \:=\: 4000 \:\bigg( 1 + \dfrac{6}{100} \bigg)^3 \\\\[/tex]
[tex]\qquad\dashrightarrow \sf Amount \:=\: 4000 \:\bigg( \dfrac{106}{100} \bigg)^3 \\\\[/tex]
[tex]\qquad\dashrightarrow \sf Amount \:=\: 4000 \:\bigg( \cancel {\dfrac{106}{100}} \bigg)^3 \\\\[/tex]
[tex]\qquad\dashrightarrow \sf Amount \:=\: 4000 \:\bigg( 1.06 \bigg)^3 \\\\[/tex]
[tex]\qquad\dashrightarrow \sf Amount \:=\: 4000 \:\times 1.191016 \\\\[/tex]
[tex]\qquad\dashrightarrow \sf Amount \:=\: 4,764.064 \\\\[/tex]
[tex]\qquad \therefore \: \underline {\purple{ \pmb {\bf Amount \:\:=\:\: Rs.\: 4,764.064}}}\:\: \bigstar \\\\[/tex]
[tex]\therefore \underline {\sf The \:Amount \:\:is \bf\:Rs. \:\: 4764.064 \:\: }\\\\[/tex]
Now , Finding Compound Interest :
As , We know that ,
[tex]\qquad \dag \:\:\:\boxed { \pink{\pmb{\:\:Compound \:Interest \: \:=\: Amount\:\: – \:\:Principal \:\:}}}\\\\[/tex]
Here , Amount is Rs. 4764.064 & Principal is Rs. 4000
[tex]\qquad\dashrightarrow \sf \:Compound \:Interest \: \:=\: Amount\:\: – \:\:Principal \: \\\\[/tex]
[tex]\qquad \bigstar \:\underline {\boldsymbol { Now \:,\:By\:Substituting \:known \:values \: }} : \\\\[/tex]
[tex]\qquad\dashrightarrow \sf \:Compound \:Interest \: \:=\: Amount\:\: – \:\:Principal \: \\\\[/tex]
[tex]\qquad\dashrightarrow \sf \:Compound \:Interest \: \:=\: 4764.064\:\: – \:\:4000 \: \\\\[/tex]
[tex]\qquad\dashrightarrow \sf \:Compound \:Interest \: \:=\: 764.064\:\: \: \\\\[/tex]
[tex]\qquad \therefore \: \underline {\purple{ \pmb {\bf Compound \:Interest \:\:=\:\: Rs.\: 764.064}}}\:\: \bigstar \\\\[/tex]
[tex]\therefore \underline {\sf Hence , \:The \:Compound \:Interest \:\:is \bf\:Rs. \:\: 764.064 \:\: }\\\\[/tex]
Corrected Question :
[tex]~[/tex]
Given : sum will amount to rupees 4000 in 3 years 6% c.a.
To Find : Find Amount ?
________________________
[tex]\underline{\frak{As ~we~ know~ that~:}}[/tex]
[tex]~[/tex]
Solution : Firstly, Finding the amount
[tex]~[/tex]
[tex]~[/tex]
[tex]\underline{\bf{Now ~By ~Substituting ~the ~Given~ Values~:}}[/tex]
[tex]~[/tex]
[tex]~~~~~~~~~~{\sf:\implies{Amount~=~4000\bigg(1~+~\dfrac{6}{100}\bigg)^{3}}}[/tex]
[tex]~~~~~~~~~{\sf:\implies{Amount~=~4000\bigg(1~×~100~+~\dfrac{6}{100}\bigg)^{3}}}[/tex]
[tex]~~~~~~~~~{\sf:\implies{Amount~=~4000\bigg(\dfrac{100~+~6}{100}\bigg)^{3}}}[/tex]
[tex]~~~~~~~~~{\sf:\implies{Amount~=~4000\bigg(\dfrac{106}{100}\bigg)^{3}}}[/tex]
[tex]~~~~~~~~~{\sf:\implies{Amount~=~4000\bigg(\cancel\dfrac{106}{100}\bigg)^{3}}}[/tex]
[tex]~~~~~~~~~~{\sf:\implies{Amount~=~4000\bigg(\dfrac{53}{50}\bigg)^{3}}}[/tex]
[tex]~~~~~~~~~~{\sf:\implies{Amount~=~4000\bigg(\dfrac{53}{50}~×~\dfrac{53}{50}~×~\dfrac{53}{50}\bigg)}}[/tex]
[tex]~~~~~~~~~~{\sf:\implies{Amount~=~4000\bigg(\dfrac{148877}{125000}\bigg)}}[/tex]
[tex]~~~~~~~~~~{\sf:\implies{Amount~=~4000~×~\dfrac{148877}{125000}}}[/tex]
[tex]~~~~~~~~~~{\sf:\implies{Amount~=~4\cancel{000}~×~\dfrac{148877}{125\cancel{000}}}}[/tex]
[tex]~~~~~~~~~~{\sf:\implies{Amount~=~\dfrac{148877~×~4}{125}}}[/tex]
[tex]~~~~~~~~~~{\sf:\implies{Amount~=~\dfrac{595508}{125}}}[/tex]
[tex]~~~~~~~~~~{\sf:\implies{Amount~=~\cancel\dfrac{595508}{125}}}[/tex]
[tex]~~~~~~~~~~{\sf:\implies{Amount~=~4764.064}}[/tex]
[tex]~~~~~~~~~~:\implies{\underline{\boxed{\frak{\purple{Amount~=~Rs.~4764.064}}}}}[/tex]★
[tex]~[/tex]
Therefore,
[tex]~[/tex]
[tex]\underline{\frak{Now ~Finding ~the ~Compound ~Interest~-}}[/tex]
[tex]~[/tex]
[tex]~[/tex]
[tex]\underline{\bf{Now~ By ~Substituting~ the ~Given ~and~ Found~ Values~:}}[/tex]
[tex]~[/tex]
[tex]~~~~~~~~~~{\sf:\implies{Compound ~Interest~=~47.64.064~-~4000}}[/tex]
[tex]~~~~~~~~~~{\sf:\implies{Compound ~Interest~=~764.046}}[/tex]
[tex]~[/tex]
Hence,
[tex]\therefore\underline{\sf{The~ Compound ~Interest~is~\bf{Rs.~764.046}}}[/tex]
[tex]~[/tex]
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