The present worth of Rs.169 due in 2 years at 4% per annum compound interest is

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 D

Explanation:

Please apply the formula

\begin{aligned}
Amount = P(1+\frac{R}{100})^n \\
\text{C.I. = Amount - P}
\end{aligned}

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:

\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 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}