Question Detail

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

1. Find the compound interest on Rs.16,000 at 20% per annum for 9 months, compounded quarterly

Rs 2520
Rs 2521
Rs 2522
Rs 2523

Answer: Option C

Explanation:

Please remember, when we have to calculate C.I. quarterly then we apply following formula if n is the number of years

\begin{aligned}
Amount = P(1+\frac{\frac{R}{4}}{100})^{4n}
\end{aligned}

Principal = Rs.16,000;
Time=9 months = 3 quarters;
Rate = 20%, it will be 20/4 = 5%

So lets solve this question now,

\begin{aligned}
Amount = 16000(1+\frac{5}{100})^3 \\
= 18522\\
C.I = 18522 - 16000 = 2522
\end{aligned}

2. 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}

3. 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

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