Question

find the compound interest on rupees 40000 for 2 year at 5% per annum compounded annually.

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Answer 1

Step-by-step explanation:

Given:

• P = ₹40,000
• n = 2 years
• R = 5% p.a
• CI = ?

To find Amount:

$\implies$ $\sf{ A=P\left(1+\Large\frac{r}{100}\right)^{n}}$

$\implies$ $\sf{ A=40000\left(1+\Large\frac{5}{100}\right)^{2}}$

$\implies$ $\sf{40000×\left({\Large\frac{ 105}{ 100}}\right)^2 }$

$\implies$ $\sf{ 40000 × {\Large\frac{ 21}{ 20 }}×{\Large\frac{ 21}{ 20 }} }$

$\implies$ $\sf{{\Large\frac{400 × 21×21 }{ 2 × 2 }} }$

$\therefore$ $\sf{ A = ₹44100}$

To find Compound interest :

$\implies$ $\sf{CI = A - P }$

$\implies$ $\sf{ 44100 - 40000}$

$\therefore$ $\boxed{\sf{ CI = ₹4100 }}$

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Answer 2

Answer:

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