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

Answer: Option B

Explanation:

Principal = Rs.1000;
Amount = Rs.1331;
Rate = Rs.10%p.a.

Let the time be n years then,

\begin{aligned}
1000(1+\frac{10}{100})^n = 1331 \\

(\frac{11}{10})^n = \frac{1331}{1000} \\

(\frac{11}{10})^3 = \frac{1331}{1000} \\

\end{aligned}
So answer is 3 years

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 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 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 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