The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 4% per annum is Rs 1. Find the sum

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%

Answer: Option D

Explanation:

Difference between C.I and S.I for 2 years = 36.30
S.I. for one year = 330.
S.I. on Rs 330 for one year = 36.30

So R% = \frac{100*36.30}{330*1} = 11%

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 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 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}