Question Detail

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

1. 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:

2. 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

3. 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}

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

Rs. 610
Rs. 612
Rs. 614
Rs. 616

Answer: Option B

Explanation:

\begin{aligned}
Amount = [7500 \times (1+ \frac{4}{100})^2] \\
= (7500 \times \frac{26}{25} \times \frac{26}{25}) \\
= 8112 \\

\end{aligned}

So compound interest = (8112 - 7500) = 612

5. 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%