At what rate of compound interest per annum will a sum of Rs. 1200 become Rs. 1348.32 in 2 years

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 D

Explanation:

Please apply the formula

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

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:

It is clear from question that Ram's share after five years = Sham's share after seven years
Hence we can conclude following :

$$\begin{aligned} \text{(Rams's present share)}\left(1 + \dfrac{5}{100}\right)^5 = \text{(Sham's present share)}\left(1 + \dfrac{5}{100}\right)^7\\ => \dfrac{\text{(Ram's present share)}}{\text{(Sham's present share)}}= \dfrac{\left(1 + \dfrac{5}{100}\right)^7}{\left(1 + \dfrac{5}{100}\right)^5} \\ = \left(1 + \dfrac{5}{100}\right)^{(7-5)} = \left(1 + \dfrac{5}{100}\right)^2 \\ = \left(\dfrac{21}{20}\right)^2 = \dfrac{441}{400} \end{aligned}$$
Ram's present share : B's present share = 441 : 400

$$\begin{aligned} \text{As amount is Rs.3364, Ram's share = }3364 \times \dfrac{441}{(441+400)} \\\\ = 3364 \times \dfrac{441}{841} = 4 \times 441 = \text{ Rs. 1764} \end{aligned}$$

So Sham's share is = 3364-1764 = 1600

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