In what time will Rs.1000 become Rs.1331 at 10% per annum compounded annually

Answer: Option B

Explanation:

\begin{aligned}
Amount = [7500 \times (1+ \frac{4}{100})^2] \\
= (7500 \times \frac{26}{25} \times \frac{26}{25}) \\
= 8112 \\

\end{aligned}

So compound interest = (8112 - 7500) = 612

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 C

Explanation:

Please remember, when we have to calculate C.I. quarterly then we apply following formula if n is the number of years

\begin{aligned}
Amount = P(1+\frac{\frac{R}{4}}{100})^{4n}
\end{aligned}

Principal = Rs.16,000;
Time=9 months = 3 quarters;
Rate = 20%, it will be 20/4 = 5%

So lets solve this question now,

\begin{aligned}
Amount = 16000(1+\frac{5}{100})^3 \\
= 18522\\
C.I = 18522 - 16000 = 2522
\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}