CoCubes Simple and Compound Interest Question and Answers Quiz-1

\text{ Amount }=  P \left ( 1 + \frac{R}{100} \right )^{t}

\text{ Amount }= 8000\left ( 1 + \frac{15}{100} \right )^{2\frac{1}{3}}

\text{ Amount }= 8000\left ( 1 + \frac{3}{20} \right )^{2}\left ( 1 + \frac{3}{20\times 3} \right )

\text{ Amount }= 8000\times \frac{23}{20}\times \frac{23}{20}\times \frac{21}{20}

Amount = Rs 11109

Compound Interest = 11109 - 8000 = Rs 3109

\text{ Amount }=p\left ( 1 + \frac{R}{100} \right )^{t}

\text{ Amount }= 7500\left ( 1 + \frac{4}{100} \right )^{2}

Amount=7500\times \frac{26}{25}\times \frac{26}{25}=8112

Compound Interest = 8112 - 7500 = Rs 612

Given SI = $6750 at 6\frac{2}{3}\% for the time period of 3 years

as we know

SI = \frac{P\times R\times T}{100}

6750 = \frac{P\times \frac{20}{3}\times 3}{100}

P = $33,750

Compound Interest = (Amount - P)

\text{ Amount }=P\left ( 1 + \frac{R}{100} \right )^{t}

\text{ Amount } = 33750\left ( 1 + \frac{\frac{20}{3}}{100} \right )^{3}

\text{ Amount } = 33750\left ( 1 + \frac{1}{15} \right )^{3}

\text{ Amount } = 33750\left ( \frac{16}{15}\right )^{3}

Amount = $ 40,960

Compound Interest = 40960 - 33750

Compound Interest= $ 7210

Let P = Rs.100
Simple Interest = Rs. 80 ( 80% increase is due to the simple interest)
Rate of interest=\frac{100×SI}{P\times T}

=\frac{100×80}{100×8} =10% per annum
Now let's find out the compound interest of Rs. 14,000 after 3 years at 10%
P = Rs.14000
T = 3 years
R = 10%
Amount after 3 years =P\left ( 1 + \frac{R}{100} \right )^{t}
=14000\left ( 1 + \frac{10}{100} \right )^{3}
=14000\times \frac{110}{100}^{3}
=14000 \left( \frac{11}{10}^{3}\right)
=14×(11)³=18634.
Compound Interest = Rs.18634 - Rs.14000 = Rs.4634

According to the question

SI = Rs 90, T = 2 yrs, R = 10%

P = \frac{SI\times 100}{RT}

P = \frac{100\times 90}{10\times 2}

P = Rs 450

Compound Interest = Amount - P

\text{ Amount }=P\left ( 1 + \frac{R}{100} \right )^{t}

\text{ Amount }= 450\left ( 1 + \frac{10}{100} \right )^{2}

Amount = Rs 544.5

Therefore C.I. = Rs 544.5 - Rs 450 = Rs. 94.50

Let Amount be P

Amount with compound interest = P(1 +\frac{r}{100})^{t}

3P = P ( 1 + \frac{r}{100})^{10}

3 = ( 1 +\frac{r}{100})^{10} - eq 1

Let say in T years it become 9 times

9P = P(1 + \frac{r}{100})^{T}

=> 9 = (1 +\frac{ r}{100})^{T}

=> 3^{2} = (1 + \frac{r}{100})^{T}

from eq 1

=> (( 1 + \frac{r}{100})^{10})^{2} = (1 +\frac{ r}{100})^{T}

=> ( 1 + \frac{r}{100})^{20} = (1 + \frac{r}{100})^{T}

Comparing both

T = 20

Time = 20 Years

Principal = Rs.1000;
Amount = Rs.1331;
Rate = Rs. 10% p.a.
Let the time be n years then,
1000(1+\frac{10}{100})^{n}=1331
(\frac{11}{10})^{n}=\frac{1331}{1000}
(\frac{11}{10})^{3}=\frac{1331}{1000}
So answer is 3 years

Let the required rate be R% per annum
Here, A = $ 2205, P = $ 2000 and n = 2 years
Using the formula \text{ Amount }=p\left ( 1 + \frac{R}{100} \right )^{t}
2205 = 2000\left ( 1 + \frac{R}{100} \right )^{2}
( 1 + \frac{R}{100})^{2} = \frac{2205}{2000} = \frac{441}{400} = (\frac{21}{20})^{2}
1 + \frac{R}{100} = \frac{21}{20}
\frac{R}{100} = \left( \frac{21}{20} – 1\right) = \frac{1}{20}
R = (100 \times \frac{1}{20}) = 5
Hence, the required rate of interest is 5% per annum

Let the sum be $ 100. Then,
SI = (100 \times 6 \times \frac{2}{100}) = 12
and compound interest =   {100 \times (1+\frac{6}{100})^{2} - 100}
= {(100 \times \frac{53}{50} \times \frac{53}{50}) - 100}

=   (\frac{2809}{25} - 100)

= \frac{309}{25}
Therefore, (CI) - (SI) =  (\frac{309}{25} – 12) = \frac{9}{25}
If the difference between the CI and SI is  \frac{9}{25} , then the sum = 100
If the difference between the CI and SI is 18, then the sum =  (100 × \frac{25}{9} × 18 ) =  5000
Hence, the required sum is  5000.

For question these type of the question we use this particular formula

x = \frac{100A}{100t + \frac{r(t - 1)(t)}{2}}

Now putting the values in the above formula:

80 = \frac{A\times 100}{5\times 100 + \frac{5(4)(5)}{2}}

A = \frac{80\times 550}{100} Rs. 440