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 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 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 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:

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 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}