Given : Invested amount = $5000.
Rate of interet = r% compunded monthly.
Number of years = 4 years.
Total interest received = $1866.
Therefore, total amount after 4 years = 5000+1866 = $6866.
We know, formula for compound interest :
[tex]A=P(1+\frac{r}{n})^{t\times n}[/tex], where P is the invested amount, n is the number of monthly installments in an year.
Number of months in an year are 12.
Plugging n=12, P=5000, t=4 in the formula now, we get
[tex]6866=5000{(1+\frac{r}{12})^{12\times 4}[/tex]
[tex]6866=5000{(1+\frac{r}{12})^{48}[/tex]
Dividing both sides by 5000, we get
[tex]\frac{6866}{5000} =\frac{5000}{5000} {(1+\frac{r}{12})^{48}[/tex]
[tex]\frac{6866}{5000} = {(1+\frac{r}{12})^{48}[/tex]
[tex]1.3732= {(1+\frac{r}{12})^{48}[/tex]
Taking 48th root on both sides, we get
[tex]\sqrt[48]{1.3732} = \sqrt[48]{(1+\frac{r}{12})^{48}}[/tex]
[tex]\sqrt[48]{1.3732}=\left(1+\frac{r}{12}\right)[/tex]
[tex]1.00662903758=1+\frac{r}{12}[/tex]
Subtracting 1 from both sides, we get
0.00662903758 = [tex]\frac{r}{12}[/tex]
Multiplying by 12 on both sides, we get
r=0.07954845101
r≈0.0795
Or 7.95%.
Now, we need to find the interest after four years if the rate of interest is 3.6% compounded quarterly.
There are 4 quarters in an year.
So, n=4 and r=3.6%= 0.036.
Plugging values in compound interest formula now, we get
[tex]A=5000{(1+\frac{0.036}{4})^{4\times 4}[/tex]
[tex]\mathrm{Divide\:the\:numbers:}\:\frac{0.036}{4}=0.009[/tex]
[tex]A=5000\times \:1.009^{16}[/tex]
[tex]A=5000\times \:1.15414\dots[/tex]
[tex]=5770.70222\dots[/tex]
A≈5770.70
Subtracting 5770.70 -5000.00 = 770.70.
(A)
We are given
Mr tan invested 5000 Swiss francs(CHf) bank
so,
[tex]P=5000[/tex]
nominal annual interest rate of r% compounded monthly
so, n=12
t=4
total interest he received was 1866 CHF
I=1866
we can use formula
[tex]I=P(1+\frac{r}{n} )^{nt}-P[/tex]
now, we can plug values
[tex]1866=5000(1+\frac{r}{12} )^{12*4}-5000[/tex]
now, we can solve for r
[tex]r=0.0795[/tex]
so, interest rate is 7.95%..........Answer
(B)
we are given
P=5000
r=0.036
n=4
t=4
so, we can use interest formula
[tex]I=P(1+\frac{r}{n} )^{nt}-P[/tex]
now, we can plug values
[tex]I=5000(1+\frac{0.036}{4} )^{4*4}-5000[/tex]
now, we can simplify it
and we get
[tex]I=770.702[/tex]
The total interest is 771 CHF...........Answer
