At what rate of compound interest per annum will a sum of Rs. 1200 become Rs. 1348.32 in 2 years

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 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}
[200(1+\frac{5}{100})^3 + 200(1+\frac{5}{100})^2+ \\ 200(1+\frac{5}{100})]

= [200(\frac{21}{20} \times \frac{21}{20} \times \frac{21}{20})\\
+ 200(\frac{21}{20}\times\frac{21}{20})+200(\frac{21}{20})] \\

= 662.02
\end{aligned}

Answer: Option D

Explanation:

\begin{aligned}
=> (8000 \times(1+\frac{5}{100})^2) \\
=> 8000 \times \frac{21}{20}\times \frac{21}{20} \\
=> 8820
\end{aligned}

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%