Answer-
$23377 must be deposited to get $68000 at the end of 30 years.
Solution-
We know that for compound interest,
[tex]A=P(1+\dfrac{r}{n})^{nt}[/tex]
Where,
A = Future amount = $68,000
P = ??
r = 3.575% annual = 0.03575
n = 4 as interest is compounded quarterly
t = time in year = 30 years
Putting the values,
[tex]\Rightarrow 68000=P(1+\dfrac{0.03575}{4})^{4\times 30}[/tex]
[tex]\Rightarrow 68000=P(1.0089375)^{120}[/tex]
[tex]\Rightarrow P=\dfrac{68000}{(1.0089375)^{120}}[/tex]
[tex]\Rightarrow P=23377.45[/tex]
Therefore, $23377 must be deposited to get $68000 at the end of 30 years.
Answer:
Option b- $23,377
Step-by-step explanation:
Given : An account earning 3.575% interest compounded quarterly so that it has an accumulated value of $68,000 at the end of 30 years.
To find : How much principal would need to be placed into an account?
Solution :
Using compound interest formula,
[tex]A=P(1+\dfrac{r}{n})^{nt}[/tex]
Where,
A = Future amount = $68,000
P = Principal value =?
r = 3.575% annual = 0.03575
n = 4 (interest is compounded quarterly)
t = time in year = 30 years
Putting the values,
[tex]68000=P(1+\dfrac{0.03575}{4})^{4\times 30}[/tex]
[tex]\Rightarrow 68000=P(1.0089375)^{120}[/tex]
[tex]\Rightarrow P=\dfrac{68000}{(1.0089375)^{120}}[/tex]
[tex]\Rightarrow P=23377.45[/tex]
Therefore, $23,377 must be deposited to get $68000 at the end of 30 years.
Hence, Option b is correct.
