Find the compound interest on Rs. 7500 at 4% per annum for 2 years, compounded annually.

Answer: Option D

Explanation:

Let Rate will be R%

\begin{aligned}
1200(1+\frac{R}{100})^2 = \frac{134832}{100} \\

(1+\frac{R}{100})^2 = \frac{134832}{120000} \\

(1+\frac{R}{100})^2 = \frac{11236}{10000} \\

(1+\frac{R}{100}) = \frac{106}{100} \\
=> R = 6\%
\end{aligned}

Answer: Option C

Explanation:

\begin{aligned}
S.I. = \frac{P*R*T}{100} \\
P = \frac{50*100}{5*2} = 500\\

Amount = 500(1+\frac{5}{100})^2 \\
500(\frac{21}{20} * \frac{21}{20}) \\
= 551.25 \\

C.I. = 551.25 - 500 = 51.25


\end{aligned}

Answer: Option A

Explanation:

\begin{aligned}
(25000 \times (1 + \frac{12}{100})^3) \\
=> 25000\times\frac{28}{25}\times\frac{28}{25}\times\frac{28}{25} \\
=> 35123.20 \\
\end{aligned}

So Compound interest will be 35123.20 - 25000
= Rs 10123.20

Answer: Option B

Explanation:

\begin{aligned}
Amount = [7500 \times (1+ \frac{4}{100})^2] \\
= (7500 \times \frac{26}{25} \times \frac{26}{25}) \\
= 8112 \\

\end{aligned}

So compound interest = (8112 - 7500) = 612

Answer: Option A

Explanation:

Let the amount Rs 100 for 1 year when compounded half yearly, n = 2, Rate = 6/2 = 3%

\begin{aligned}
Amount = 100(1+\frac{3}{100})^2 = 106.09
\end{aligned}

Effective rate = (106.09 - 100)% = 6.09%