Question

Find the compound interest on Rs.9000 in 2 years at rate of 10 p.c.p.a
step by step explain pls​

Answer 1

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

Answer 2

Given :

• Principal = Rs. 9000
• Time = 2 years
• Rate = 10 %

To find :

• Compound Interest

Concept Used :

→ Formula to calculate amount :-

$\dag \overline{\underline{\boxed{\sf\pmb{ \red{Amount = P\Bigg[1 + \dfrac{r}{100}\Bigg]^n}}}}}$

where,

• P = Principal
• r = rate
• n = Time

→ Formula to calculate Compound Interest :-

$\dag \: \: \overline{ \underline {\boxed{ \sf{ \pmb{\red{Compound \: \: Interest = Amount - Principal}}}}}}$

Solution :

Amount :-

$\\ \longmapsto \quad \sf Amount = P\Bigg[1 + \dfrac{r}{100}\Bigg]^n$

$\\ \longmapsto \quad \sf Amount = 9000\Bigg[1 + \dfrac{10}{100}\Bigg]^2$

$\\ \longmapsto \quad \sf Amount = 9000\Bigg[1 + \dfrac{1 \not0}{10 \not0}\Bigg]^2$

$\\ \longmapsto \quad \sf Amount = 9000\Bigg[1 + \dfrac{1}{10}\Bigg]^2$

$\\ \longmapsto \quad \sf Amount = 9000\Bigg[ \dfrac{10 + 1}{10}\Bigg]^2$

$\\ \longmapsto \quad \sf Amount = 9000\Bigg[ \dfrac{11}{10}\Bigg]^2$

$\\ \longmapsto \quad \sf Amount = 9000 \times \dfrac{11}{10} \times \dfrac{11}{10}$

$\\ \longmapsto \quad \sf Amount = 90 \not0 \not0 \times \dfrac{11}{1 \not0} \times \dfrac{11}{1 \not0}$

$\\ \longmapsto \quad \sf Amount = 90 \times \dfrac{11}{1} \times \dfrac{11}{1}$

$\\ \longmapsto \quad \sf Amount = 90 \times 11 \times 11$

$\\ \longmapsto \quad \sf \green {Amount = \:Rs. \: 10,890}$

Compound Interest :-

$\\ \longmapsto \quad \sf {Compound \: \: Interest = Amount - Principal}$

$\\ \longmapsto \quad \sf {Compound \: \: Interest = 10,890 - 9000}$

$\\ \longmapsto \quad \sf \green {Compound \: \: Interest = 1890}$

$\\ \longmapsto \quad \overline{\underline{\boxed{ \tt{ \pmb{\red {Compound \: \: Interest = 1890}}}}}}$