*The difference between simple interest and compound interest at the rate of 10% for two years on an amount is Rs. 20. Find the principal value?* 1️⃣ ₹.1002️⃣ ₹ 2003️⃣ ₹. 200004️⃣ ₹ 2000 About the author Melody
Given : Difference between simple Interest and Compound Interest = Rs. 20 Rate = 10% Time = 2 years To find : Principal Concept : → Formula to calculate Simple Interest :- [tex] \star \quad\boxed{\pmb{ \sf{Simple \: \: Interest = \dfrac{P \times R \times T}{100}}}}[/tex] → Formula to calculate Compound Interest :- [tex]\\ \star \quad \boxed{\pmb{ \sf Compound \: \: Interest = P\Bigg[1 + \dfrac{r}{100}\Bigg]^n – P}}[/tex] where, P = Principal R = Rate T and n = Time Solution : Let the Sum [Principal] as P. Simple Interest :- [tex] \\ \dashrightarrow\sf \quad Simple \: \: Interest = \dfrac{P \times R \times T}{100}[/tex] [tex] \\ \dashrightarrow\sf \quad Simple \: \: Interest = \dfrac{P \times 10 \times 2}{100}[/tex] [tex] \\ \dashrightarrow\sf \quad Simple \: \: Interest = \dfrac{P \times 1 \not0 \times 2}{10 \not0}[/tex] [tex] \\ \dashrightarrow\sf \quad Simple \: \: Interest = \dfrac{P \times 2}{10}[/tex] [tex] \\ \dashrightarrow\sf \quad Simple \: \: Interest = \dfrac{2P}{10}[/tex] [tex] \\ \dashrightarrow\sf \quad \pmb{ \quad Simple \: \: Interest =Rs. \: \dfrac{2P}{10}}[/tex] Compound Interest :- [tex] \\ \dashrightarrow\sf \quad Compound \: \: Interest = P\Bigg[1 + \dfrac{r}{100}\Bigg]^n – P[/tex] [tex] \\ \dashrightarrow\sf \quad Compound \: \: Interest = P\Bigg[1 + \dfrac{10}{100}\Bigg]^2 – P[/tex] [tex] \\ \dashrightarrow\sf \quad Compound \: \: Interest = P\Bigg[1 + \dfrac{1 \not0}{10 \not0}\Bigg]^2 – P[/tex] [tex] \\ \dashrightarrow\sf \quad Compound \: \: Interest = P\Bigg[1 + \dfrac{1}{10}\Bigg]^2 – P[/tex] [tex] \\ \dashrightarrow\sf \quad Compound \: \: Interest = P\Bigg[\dfrac{10 + 1}{10}\Bigg]^2 – P[/tex] [tex] \\ \dashrightarrow\sf \quad Compound \: \: Interest = P\Bigg[\dfrac{11}{10}\Bigg]^2 – P[/tex] [tex] \\ \dashrightarrow\sf \quad Compound \: \: Interest = P\Bigg[\dfrac{11}{10} \times \dfrac{11}{10} \Bigg] – P[/tex] [tex] \\ \dashrightarrow\sf \quad Compound \: \: Interest = P\Bigg[\dfrac{121}{100} \Bigg] – P[/tex] [tex] \\ \dashrightarrow\sf \quad Compound \: \: Interest = \dfrac{121}{100}P – P[/tex] [tex] \\ \dashrightarrow\sf \quad Compound \: \: Interest = \dfrac{121P – 100P}{100}[/tex] [tex] \\ \dashrightarrow\sf \quad Compound \: \: Interest = \dfrac{21P}{100}[/tex] [tex] \\ \dashrightarrow \quad \sf \pmb{ Compound \: \: Interest = \dfrac{21P}{100}}[/tex] According to the question, ★ Compound Interest – Simple Interest= Rs. 20 [tex] \\ \dashrightarrow\sf \quad\dfrac{21P}{100} – \dfrac{2P}{10} = 20[/tex] [tex] \\ \dashrightarrow\sf \quad\dfrac{21P – 20P}{100} = 20[/tex] [tex] \\ \dashrightarrow\sf \quad P = 20 \times 100[/tex] [tex] \\ \dashrightarrow\sf \quad P = 2000[/tex] Answer → Option (4) Principal = ₹ 2000 Reply
Given : Difference between Compound Interest and Simple interest is Rs.20 , The rate is 10% p.a & the time is 2 yrs Need To Find : The Principal. ⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀ ❍ Let’s Consider the Principal be P . ⠀⠀⠀⠀⠀Finding Simple interest : [tex]\dag\:\:\it{ As,\:We\:know\:that\::}\\[/tex] [tex]\qquad \dag\:\:\bigg\lgroup \sf{ Simple \:Interest \:: \dfrac{P\times R \times T }{100} }\bigg\rgroup \\\\[/tex] ⠀⠀⠀⠀⠀Here P is the Principal , R is the Rate of Interest & T is the Time. ⠀⠀⠀⠀⠀⠀[tex]\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\[/tex] [tex]\qquad \longmapsto\sf Simple \:Interest \:= \dfrac{ P \times 10 \times 2 }{100} \\ [/tex] [tex]\qquad \longmapsto\sf Simple \:Interest \:= \dfrac{ P \times \cancel {10} \times 2 }{10\cancel {0}} \\ [/tex] [tex]\qquad \longmapsto\sf Simple \:Interest \:= \dfrac{P \times 2 }{10} \\ [/tex] [tex]\qquad \longmapsto\sf Simple \:Interest \:= \dfrac{2P }{10} \\ [/tex] [tex]\qquad \longmapsto\bf \bigg( 0.2P \bigg) \qquad\: \longrightarrow\:\: Simple \:Interest \\ [/tex] ⠀⠀⠀⠀⠀Finding Compound interest : [tex]\dag\:\:\it{ As,\:We\:know\:that\::}\\[/tex] [tex]\qquad \dag\:\:\bigg\lgroup \sf{ Compound \:Interest \:: P \bigg( 1 +\dfrac{R }{100}\bigg) ^T – P }\bigg\rgroup \\\\[/tex] ⠀⠀⠀⠀⠀Here P is the Principal , R is the Rate of Interest & T is the Time. ⠀⠀⠀⠀⠀⠀[tex]\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\[/tex] [tex]\qquad \longmapsto\sf Compound \:Interest \:= P\bigg( 1 + \dfrac{ 10 }{100}\bigg)^2 – P \\ [/tex] [tex]\qquad \longmapsto\sf Compound \:Interest \:= P\bigg( 1 + \dfrac{ \cancel{10} }{10\cancel{0}}\bigg)^2 -P \\ [/tex] [tex]\qquad \longmapsto\sf Compound \:Interest \:= P\bigg( 1 + \dfrac{ 1 }{10}\bigg)^2 – P \\ [/tex] [tex]\qquad \longmapsto\sf Compound \:Interest \:= P\bigg( \dfrac{ 10 +1 }{10}\bigg)^2- P \\ [/tex] [tex]\qquad \longmapsto\sf Compound \:Interest \:= P\bigg( \dfrac{ 11 }{10}\bigg)^2 – P \\ [/tex] [tex]\qquad \longmapsto\sf Compound \:Interest \:= P\bigg( \cancel {\dfrac{ 11 }{10}}\bigg)^2 – P \\ [/tex] [tex]\qquad \longmapsto\sf Compound \:Interest \:= P\bigg( 1.1\bigg)^2 \\ [/tex] [tex]\qquad \longmapsto\sf Compound \:Interest \:= P\bigg( 1.21\bigg) – P \\ [/tex] [tex]\qquad \longmapsto\sf Compound \:Interest \:= 1.21P – P \\ [/tex] [tex]\qquad \longmapsto\bf \bigg( Rs. \:0.21P\bigg) \qquad\: \longrightarrow\:\: Compound \:Interest \\ [/tex] ⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀ ⠀⠀⠀⠀⠀⠀[tex]\underline {\boldsymbol{\star\:According \: To \: Question \: \: \::}}\\[/tex] Difference between Compound Interest and Simple interest is Rs.20 . [tex]\qquad \longmapsto \sf 0.21P – 0.2P =20 \\ [/tex] [tex]\qquad \longmapsto \sf 0.01P =20 \\ [/tex] [tex]\qquad \longmapsto \sf P =\cancel {\dfrac{20}{0.01}} \\ [/tex] [tex]\qquad \longmapsto \frak{\underline{\purple{\:P = Rs.2,000 }} }\bigstar \\[/tex] Therefore, ⠀⠀⠀⠀⠀[tex]\therefore {\underline{ \mathrm {\:Principal \:is\:\bf{Rs.2,000}}}}\\[/tex] ⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀ Reply
Given :
To find :
Concept :
→ Formula to calculate Simple Interest :-
[tex] \star \quad\boxed{\pmb{ \sf{Simple \: \: Interest = \dfrac{P \times R \times T}{100}}}}[/tex]
→ Formula to calculate Compound Interest :-
[tex]\\ \star \quad \boxed{\pmb{ \sf Compound \: \: Interest = P\Bigg[1 + \dfrac{r}{100}\Bigg]^n – P}}[/tex]
where,
Solution :
Let the Sum [Principal] as P.
Simple Interest :-
[tex] \\ \dashrightarrow\sf \quad Simple \: \: Interest = \dfrac{P \times R \times T}{100}[/tex]
[tex] \\ \dashrightarrow\sf \quad Simple \: \: Interest = \dfrac{P \times 10 \times 2}{100}[/tex]
[tex] \\ \dashrightarrow\sf \quad Simple \: \: Interest = \dfrac{P \times 1 \not0 \times 2}{10 \not0}[/tex]
[tex] \\ \dashrightarrow\sf \quad Simple \: \: Interest = \dfrac{P \times 2}{10}[/tex]
[tex] \\ \dashrightarrow\sf \quad Simple \: \: Interest = \dfrac{2P}{10}[/tex]
[tex] \\ \dashrightarrow\sf \quad \pmb{ \quad Simple \: \: Interest =Rs. \: \dfrac{2P}{10}}[/tex]
Compound Interest :-
[tex] \\ \dashrightarrow\sf \quad Compound \: \: Interest = P\Bigg[1 + \dfrac{r}{100}\Bigg]^n – P[/tex]
[tex] \\ \dashrightarrow\sf \quad Compound \: \: Interest = P\Bigg[1 + \dfrac{10}{100}\Bigg]^2 – P[/tex]
[tex] \\ \dashrightarrow\sf \quad Compound \: \: Interest = P\Bigg[1 + \dfrac{1 \not0}{10 \not0}\Bigg]^2 – P[/tex]
[tex] \\ \dashrightarrow\sf \quad Compound \: \: Interest = P\Bigg[1 + \dfrac{1}{10}\Bigg]^2 – P[/tex]
[tex] \\ \dashrightarrow\sf \quad Compound \: \: Interest = P\Bigg[\dfrac{10 + 1}{10}\Bigg]^2 – P[/tex]
[tex] \\ \dashrightarrow\sf \quad Compound \: \: Interest = P\Bigg[\dfrac{11}{10}\Bigg]^2 – P[/tex]
[tex] \\ \dashrightarrow\sf \quad Compound \: \: Interest = P\Bigg[\dfrac{11}{10} \times \dfrac{11}{10} \Bigg] – P[/tex]
[tex] \\ \dashrightarrow\sf \quad Compound \: \: Interest = P\Bigg[\dfrac{121}{100} \Bigg] – P[/tex]
[tex] \\ \dashrightarrow\sf \quad Compound \: \: Interest = \dfrac{121}{100}P – P[/tex]
[tex] \\ \dashrightarrow\sf \quad Compound \: \: Interest = \dfrac{121P – 100P}{100}[/tex]
[tex] \\ \dashrightarrow\sf \quad Compound \: \: Interest = \dfrac{21P}{100}[/tex]
[tex] \\ \dashrightarrow \quad \sf \pmb{ Compound \: \: Interest = \dfrac{21P}{100}}[/tex]
According to the question,
★ Compound Interest – Simple Interest= Rs. 20
[tex] \\ \dashrightarrow\sf \quad\dfrac{21P}{100} – \dfrac{2P}{10} = 20[/tex]
[tex] \\ \dashrightarrow\sf \quad\dfrac{21P – 20P}{100} = 20[/tex]
[tex] \\ \dashrightarrow\sf \quad P = 20 \times 100[/tex]
[tex] \\ \dashrightarrow\sf \quad P = 2000[/tex]
Answer → Option (4)
Given : Difference between Compound Interest and Simple interest is Rs.20 , The rate is 10% p.a & the time is 2 yrs
Need To Find : The Principal.
⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀
❍ Let’s Consider the Principal be P .
⠀⠀⠀⠀⠀Finding Simple interest :
[tex]\dag\:\:\it{ As,\:We\:know\:that\::}\\[/tex]
[tex]\qquad \dag\:\:\bigg\lgroup \sf{ Simple \:Interest \:: \dfrac{P\times R \times T }{100} }\bigg\rgroup \\\\[/tex]
⠀⠀⠀⠀⠀Here P is the Principal , R is the Rate of Interest & T is the Time.
⠀⠀⠀⠀⠀⠀[tex]\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\[/tex]
[tex]\qquad \longmapsto\sf Simple \:Interest \:= \dfrac{ P \times 10 \times 2 }{100} \\ [/tex]
[tex]\qquad \longmapsto\sf Simple \:Interest \:= \dfrac{ P \times \cancel {10} \times 2 }{10\cancel {0}} \\ [/tex]
[tex]\qquad \longmapsto\sf Simple \:Interest \:= \dfrac{P \times 2 }{10} \\ [/tex]
[tex]\qquad \longmapsto\sf Simple \:Interest \:= \dfrac{2P }{10} \\ [/tex]
[tex]\qquad \longmapsto\bf \bigg( 0.2P \bigg) \qquad\: \longrightarrow\:\: Simple \:Interest \\ [/tex]
⠀⠀⠀⠀⠀Finding Compound interest :
[tex]\dag\:\:\it{ As,\:We\:know\:that\::}\\[/tex]
[tex]\qquad \dag\:\:\bigg\lgroup \sf{ Compound \:Interest \:: P \bigg( 1 +\dfrac{R }{100}\bigg) ^T – P }\bigg\rgroup \\\\[/tex]
⠀⠀⠀⠀⠀Here P is the Principal , R is the Rate of Interest & T is the Time.
⠀⠀⠀⠀⠀⠀[tex]\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\[/tex]
[tex]\qquad \longmapsto\sf Compound \:Interest \:= P\bigg( 1 + \dfrac{ 10 }{100}\bigg)^2 – P \\ [/tex]
[tex]\qquad \longmapsto\sf Compound \:Interest \:= P\bigg( 1 + \dfrac{ \cancel{10} }{10\cancel{0}}\bigg)^2 -P \\ [/tex]
[tex]\qquad \longmapsto\sf Compound \:Interest \:= P\bigg( 1 + \dfrac{ 1 }{10}\bigg)^2 – P \\ [/tex]
[tex]\qquad \longmapsto\sf Compound \:Interest \:= P\bigg( \dfrac{ 10 +1 }{10}\bigg)^2- P \\ [/tex]
[tex]\qquad \longmapsto\sf Compound \:Interest \:= P\bigg( \dfrac{ 11 }{10}\bigg)^2 – P \\ [/tex]
[tex]\qquad \longmapsto\sf Compound \:Interest \:= P\bigg( \cancel {\dfrac{ 11 }{10}}\bigg)^2 – P \\ [/tex]
[tex]\qquad \longmapsto\sf Compound \:Interest \:= P\bigg( 1.1\bigg)^2 \\ [/tex]
[tex]\qquad \longmapsto\sf Compound \:Interest \:= P\bigg( 1.21\bigg) – P \\ [/tex]
[tex]\qquad \longmapsto\sf Compound \:Interest \:= 1.21P – P \\ [/tex]
[tex]\qquad \longmapsto\bf \bigg( Rs. \:0.21P\bigg) \qquad\: \longrightarrow\:\: Compound \:Interest \\ [/tex]
⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀
⠀⠀⠀⠀⠀⠀[tex]\underline {\boldsymbol{\star\:According \: To \: Question \: \: \::}}\\[/tex]
[tex]\qquad \longmapsto \sf 0.21P – 0.2P =20 \\ [/tex]
[tex]\qquad \longmapsto \sf 0.01P =20 \\ [/tex]
[tex]\qquad \longmapsto \sf P =\cancel {\dfrac{20}{0.01}} \\ [/tex]
[tex]\qquad \longmapsto \frak{\underline{\purple{\:P = Rs.2,000 }} }\bigstar \\[/tex]
Therefore,
⠀⠀⠀⠀⠀[tex]\therefore {\underline{ \mathrm {\:Principal \:is\:\bf{Rs.2,000}}}}\\[/tex]
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