The least number of complete years in which a sum of money put out at 20% compound interest will be more than doubled is

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 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 B

Explanation:

\begin{aligned}
C.I. = (4000 \times(1+\frac{10}{100})^2 - 4000) \\
= 4000 * \frac{11}{10} * \frac{11}{10} - 4000 \\
= 840 \\

\text{So S.I. = } \frac{840}{2} = 420\\

\text{So Sum = } \frac{S.I. * 100}{R*T} \\
= \frac{420 * 100}{3*8} \\
= Rs 1750
\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 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}