What will be the compound interest on Rs. 25000 after 3 years at the rate of 12 % per annum

Answer: Option A

Explanation:

As per question we need something like following

\begin{aligned}
P(1+\frac{R}{100})^n > 2P \\
(1+\frac{20}{100})^n > 2 \\
(\frac{6}{5})^n > 2 \\
\frac{6}{5} \times \frac{6}{5} \times \frac{6}{5}\times\frac{6}{5} > 2

\end{aligned}

So answer is 4 years

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 B

Explanation:

Let the Sum be P
\begin{aligned}
S.I. = \frac{P*4*2}{100} = \frac{2P}{25}\\

C.I. = P(1+\frac{4}{100})^2 - P \\

= \frac{676P}{625} - P \\
= \frac{51P}{625} \\
\text{As, C.I. - S.I = 1}\\
=> \frac{51P}{625} - \frac{2P}{25} = 1 \\
=> \frac{51P - 50P}{625} = 1 \\
P = 625
\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 C

Explanation:

\begin{aligned}
S.I. = \frac{1000*10*4}{100} = 400 \\
C.I. = [1000(1+\frac{10}{100})^4 - 1000] \\
= 464.10
\end{aligned}

So difference between simple interest and compound interest will be 464.10 - 400 = 64.10