Question

Approximately how much principal would need to be placed into an account earning 3.575% interest compounded quarterly so that it has an accumulated value of $68,000 at the end of 30 years?
a.
$23,706
b.
$23,377
c.
$52,069
d.
$58,944

Answer 1

Answer-

$23377 must be deposited to get $68000 at the end of 30 years.

Solution-

We know that for compound interest,

$A=P(1+\dfrac{r}{n})^{nt}$

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,

$\Rightarrow 68000=P(1+\dfrac{0.03575}{4})^{4\times 30}$

$\Rightarrow 68000=P(1.0089375)^{120}$

$\Rightarrow P=\dfrac{68000}{(1.0089375)^{120}}$

$\Rightarrow P=23377.45$

Therefore, $23377 must be deposited to get $68000 at the end of 30 years.

Answer 2

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,

$A=P(1+\dfrac{r}{n})^{nt}$

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,

$68000=P(1+\dfrac{0.03575}{4})^{4\times 30}$

$\Rightarrow 68000=P(1.0089375)^{120}$

$\Rightarrow P=\dfrac{68000}{(1.0089375)^{120}}$

$\Rightarrow P=23377.45$

Therefore, $23,377 must be deposited to get $68000 at the end of 30 years.

Hence, Option b is correct.