Simple interest on a certain sum of money for 3 years at 8% per annum is half the compound interest on Rs. 4000 for 2 years at 10% per annum. The sum placed on simple interest is

Answer: Option C

Explanation:

\begin{aligned}
S.I. = \frac{1000*10*4}{100} = 400 \\
C.I. = [1000(1+\frac{10}{100})^4 - 1000] \\
= 464.10
\end{aligned}

So difference between simple interest and compound interest will be 464.10 - 400 = 64.10

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 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 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}