Find the compound interest on Rs.9000 in 2 years at rate of 10 p.c.p.astep by step explain pls About the author Brielle
Given : Principal = Rs. 9000 Time = 2 years Rate = 10 % To find : Compound Interest Concept Used : → Formula to calculate amount :- [tex]\dag \overline{\underline{\boxed{\sf\pmb{ \red{Amount = P\Bigg[1 + \dfrac{r}{100}\Bigg]^n}}}}}[/tex] where, P = Principal r = rate n = Time → Formula to calculate Compound Interest :- [tex]\dag \: \: \overline{ \underline {\boxed{ \sf{ \pmb{\red{Compound \: \: Interest = Amount – Principal}}}}}}[/tex] Solution : Amount :- [tex]\\ \longmapsto \quad \sf Amount = P\Bigg[1 + \dfrac{r}{100}\Bigg]^n[/tex] [tex]\\ \longmapsto \quad \sf Amount = 9000\Bigg[1 + \dfrac{10}{100}\Bigg]^2[/tex] [tex]\\ \longmapsto \quad \sf Amount = 9000\Bigg[1 + \dfrac{1 \not0}{10 \not0}\Bigg]^2[/tex] [tex]\\ \longmapsto \quad \sf Amount = 9000\Bigg[1 + \dfrac{1}{10}\Bigg]^2[/tex] [tex]\\ \longmapsto \quad \sf Amount = 9000\Bigg[ \dfrac{10 + 1}{10}\Bigg]^2[/tex] [tex]\\ \longmapsto \quad \sf Amount = 9000\Bigg[ \dfrac{11}{10}\Bigg]^2[/tex] [tex]\\ \longmapsto \quad \sf Amount = 9000 \times \dfrac{11}{10} \times \dfrac{11}{10} [/tex] [tex]\\ \longmapsto \quad \sf Amount = 90 \not0 \not0 \times \dfrac{11}{1 \not0} \times \dfrac{11}{1 \not0} [/tex] [tex]\\ \longmapsto \quad \sf Amount = 90 \times \dfrac{11}{1} \times \dfrac{11}{1} [/tex] [tex]\\ \longmapsto \quad \sf Amount = 90 \times 11 \times 11[/tex] [tex]\\ \longmapsto \quad \sf \green {Amount = \:Rs. \: 10,890} [/tex] Compound Interest :- [tex]\\ \longmapsto \quad \sf {Compound \: \: Interest = Amount – Principal}[/tex] [tex]\\ \longmapsto \quad \sf {Compound \: \: Interest = 10,890 – 9000}[/tex] [tex]\\ \longmapsto \quad \sf \green {Compound \: \: Interest = 1890}[/tex] [tex]\\ \longmapsto \quad \overline{\underline{\boxed{ \tt{ \pmb{\red {Compound \: \: Interest = 1890}}}}}}[/tex] Reply
Answer: 1,890 Step-by-step explanation: A=P(1+R/100)^n A=9000(1+10/100)^2 A=9000(1+1/10)^2 A=9000(10×1+1/10)^2 A=9000(11/10)^2 A=9000(11/10×11/10) A=90×11×11 A=90×121 A=10,890 C.I=A-P C.I=10890-9000=1890 Reply
Given :
To find :
Concept Used :
→ Formula to calculate amount :-
[tex]\dag \overline{\underline{\boxed{\sf\pmb{ \red{Amount = P\Bigg[1 + \dfrac{r}{100}\Bigg]^n}}}}}[/tex]
where,
→ Formula to calculate Compound Interest :-
[tex]\dag \: \: \overline{ \underline {\boxed{ \sf{ \pmb{\red{Compound \: \: Interest = Amount – Principal}}}}}}[/tex]
Solution :
Amount :-
[tex]\\ \longmapsto \quad \sf Amount = P\Bigg[1 + \dfrac{r}{100}\Bigg]^n[/tex]
[tex]\\ \longmapsto \quad \sf Amount = 9000\Bigg[1 + \dfrac{10}{100}\Bigg]^2[/tex]
[tex]\\ \longmapsto \quad \sf Amount = 9000\Bigg[1 + \dfrac{1 \not0}{10 \not0}\Bigg]^2[/tex]
[tex]\\ \longmapsto \quad \sf Amount = 9000\Bigg[1 + \dfrac{1}{10}\Bigg]^2[/tex]
[tex]\\ \longmapsto \quad \sf Amount = 9000\Bigg[ \dfrac{10 + 1}{10}\Bigg]^2[/tex]
[tex]\\ \longmapsto \quad \sf Amount = 9000\Bigg[ \dfrac{11}{10}\Bigg]^2[/tex]
[tex]\\ \longmapsto \quad \sf Amount = 9000 \times \dfrac{11}{10} \times \dfrac{11}{10} [/tex]
[tex]\\ \longmapsto \quad \sf Amount = 90 \not0 \not0 \times \dfrac{11}{1 \not0} \times \dfrac{11}{1 \not0} [/tex]
[tex]\\ \longmapsto \quad \sf Amount = 90 \times \dfrac{11}{1} \times \dfrac{11}{1} [/tex]
[tex]\\ \longmapsto \quad \sf Amount = 90 \times 11 \times 11[/tex]
[tex]\\ \longmapsto \quad \sf \green {Amount = \:Rs. \: 10,890} [/tex]
Compound Interest :-
[tex]\\ \longmapsto \quad \sf {Compound \: \: Interest = Amount – Principal}[/tex]
[tex]\\ \longmapsto \quad \sf {Compound \: \: Interest = 10,890 – 9000}[/tex]
[tex]\\ \longmapsto \quad \sf \green {Compound \: \: Interest = 1890}[/tex]
[tex]\\ \longmapsto \quad \overline{\underline{\boxed{ \tt{ \pmb{\red {Compound \: \: Interest = 1890}}}}}}[/tex]
Answer:
1,890
Step-by-step explanation:
A=P(1+R/100)^n
A=9000(1+10/100)^2
A=9000(1+1/10)^2
A=9000(10×1+1/10)^2
A=9000(11/10)^2
A=9000(11/10×11/10)
A=90×11×11
A=90×121
A=10,890
C.I=A-P
C.I=10890-9000=1890