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. What will be the difference between simple and compound interest @ 10% per annum on the sum of Rs 1000 after 4 years

Rs 62.10
Rs 63.10
Rs 64.10
Rs 65.10

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

2. 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%

3. 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}

4. If the simple interest on a sum of money for 2 years at 5% per annum is Rs.50, what will be the compound interest on same values

Rs.51.75
Rs 51.50
Rs 51.25
Rs 51

Answer: Option C

Explanation:

\begin{aligned}
S.I. = \frac{P*R*T}{100} \\
P = \frac{50*100}{5*2} = 500\\

Amount = 500(1+\frac{5}{100})^2 \\
500(\frac{21}{20} * \frac{21}{20}) \\
= 551.25 \\

C.I. = 551.25 - 500 = 51.25

\end{aligned}

5. 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}