Given,
Principal = Rs.4000
C.I = Rs.410
Time = 2 years
Let rate of interest = R %
So,
= compound interest \(={P}[({1}+\frac{R}{100})^T-1]\)
\(=4000[({1}+\frac{R}{100})^2-1]\) = 410
\(4000({1}+\frac{R}{100})^2-4000\) = 410
\(=4000({1}+\frac{R}{100})^2\) = 4000 + 410 = 4410
\(=({1}+\frac{R}{100})^2\) \(=\frac{4410}{4000}\) \(=\frac{441}{400}\) \(=(\frac{21}{20})^2\)
\(={1}+\frac{R}{100}\) \(=\frac{21}{20}\)
\(=\frac{R}{100}\) \(=\frac{21}{20}-1\) \(=\frac{1}{20}\)
= R =\(\frac{1\times100}{20}\) = 5%
Hence,
Rate of interest = 5% per annum
