Question Detail

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 B

Explanation:

S.I. on Rs 800 for 1 year = 40

Rate = (100*40)/(800*1) = 5%

1. Find compound interest on Rs. 7500 at 4% per annum for 2 years, compounded annually

Rs 312
Rs 412
Rs 512
Rs 612

Answer: Option D

Explanation:

Please apply the formula

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

2. The least number of complete years in which a sum of money put out at 20% compound interest will be more than doubled is

4 years
5 years
6 years
7 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

3. 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

Rs 1650
Rs 1750
Rs 1850
Rs 1950

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}

4. 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.

Rs. 8510
Rs. 8620
Rs. 8730
Rs. 8820

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}

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

2 Years
3 Years
4 Years
5 Years

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