A sum of money invested at compound interest to Rs. 800 in 3 years and to Rs 840 in 4 years. The rate on interest per annum is.

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 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:

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:

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 B

Explanation:

Let the Sum be P
$$\begin{aligned} S.I. = \frac{P*4*2}{100} = \frac{2P}{25}\\ C.I. = P(1+\frac{4}{100})^2 - P \\ = \frac{676P}{625} - P \\ = \frac{51P}{625} \\ \text{As, C.I. - S.I = 1}\\ => \frac{51P}{625} - \frac{2P}{25} = 1 \\ => \frac{51P - 50P}{625} = 1 \\ P = 625 \end{aligned}$$