Albert invested amount of 8000 in a fixed deposit for 2 years at compound interest rate of 5 % per annum. How much Albert will get on the maturity of the fixed deposit.

Answer: Option B

Explanation:

In this type of question we apply formula
\begin{aligned}
Amount = \frac{P}{(1+\frac{R}{100})^n} \\
Amount = \frac{169}{(1+\frac{4}{100})^2} \\
Amount = \frac{169 * 25 * 25}{26*26} \\
Amount = 156.25
\end{aligned}

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 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 D

Explanation:

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