Question Detail

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

1. 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}

2. 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%

3. On a sum of money, simple interest for 2 years is Rs 660 and compound interest is Rs 696.30, the rate of interest being the same in both cases.

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%

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

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}

5. The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 4% per annum is Rs 1. Find the sum

Rs 600
Rs 625
Rs 650
Rs 675

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}