Question Detail

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}

1. The present worth of Rs.169 due in 2 years at 4% per annum compound interest is

Rs 155.25
Rs 156.25
Rs 157.25
Rs 158.25

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}

2. 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}

3. A man saves Rs 200 at the end of each year and lends the money at 5% compound interest. How much will it become at the end of 3 years.

Rs 662
Rs 662.01
Rs 662.02
Rs 662.03

Answer: Option C

Explanation:

\begin{aligned}
[200(1+\frac{5}{100})^3 + 200(1+\frac{5}{100})^2+ \\ 200(1+\frac{5}{100})]

= [200(\frac{21}{20} \times \frac{21}{20} \times \frac{21}{20})\\
+ 200(\frac{21}{20}\times\frac{21}{20})+200(\frac{21}{20})] \\

= 662.02
\end{aligned}

4. 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}

5. 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