Calculate Interest & Payments - John's $4000 Lump Sum Investment

[gillgill]

John brought a stamp for his collection. He agreed to pay a lump sum of $4000 after 5 years. Until then, he pays 6% simple interest semiannually on the $4000.
a) Find the amount of each semiannual interest payment.
b) John sets up a sinking fund so that enought money will be present to pay off the $4000. He will make annual payments into the fund. the accound pays 8% compounded annually. Find the amound of each payment.

a)
is it m=2
n=5x2=10
i=0.06/2=0.03
P=4000

and sub in A=P(1+i)^n
and then solve for A
interest=A-P
therefore A-4000
so the interest is $1374.67?

b) i found that the amount of each payment is $681.83?

I am not really sure about my answers.

 

[HallsofIvy]

John brought a stamp for his collection. He agreed to pay a lump sum of $4000 after 5 years. Until then, he pays 6% simple interest semiannually on the $4000.
a) Find the amount of each semiannual interest payment.
b) John sets up a sinking fund so that enought money will be present to pay off the $4000. He will make annual payments into the fund. the accound pays 8% compounded annually. Find the amound of each payment.
a)
is it m=2

Is WHAT m= 2??

n=5x2=10
i=0.06/2=0.03
P=4000
and sub in A=P(1+i)^n
and then solve for A
interest=A-P
therefore A-4000
so the interest is $1374.67?

No, the problem specifically said "simple interest".
$4000 at 6% interest would be (0.06)(4000)= $240 for a year and so $120 for 6 months. The problem asked you to find "the amount of each semiannual interest payment", not the total interest.

b) i found that the amount of each payment is $681.83?
I am not really sure about my answers.

Assuming he makes the same payment each year, let P be that amount. The amount deposited at the beginning earns 8% compounded annually for 5 years. At the end of the 5 years, it will have increased to
P(1.08)⁵.
The amount he puts in at the beginning of the second year will earn 8% compounded annually for 4 years. At the end, it will have increased to
P(1.08)⁴.
The amount he puts in at the beginning of the third year will have increased to P(1.08)³.
The amount he puts in at the beginning of the 4th year will have increased to P(1.08)<sup>2.
The amount he puts in at the beginning of the 5th year will have increased to P(1.08).
At the end of the 5th year, he will have P(1.08+ 1.082+ 1.083+ 1.084+ 1.085). That short enough to do on directly but you might recognize it as a geometric series with
initial term a<sub>1= P(1.08), common ratio r= 1.08, and summed from n= 0 to n= 4.
The formula for such a sum is $a_1\frac{1- r^5}{1-r}$.
With these values that will be [itex]P(1.08)\frac{1- 1.469}{1-1.08}= p(1.08)\frac{-0.469}{-0.08}= 6.34P[/tex].

Set that equal to $4000 and solve for P. I get $631.32.</sub></sup>

 

[Architeuthis Dux]

Can you please verify them?

a) Your calculations for the semiannual interest payment are correct. The semiannual interest rate is 3%, so the amount of each semiannual payment would be $120 ($4000 x 0.03). After 5 years, John would have paid a total of $120 x 10 = $1200 in interest.

b) Your calculations for the annual payments into the sinking fund are not correct. The annual interest rate for the sinking fund is 8%, and John needs to save enough money to pay off the $4000 after 5 years. This means that he needs to save $4000 in total, and the annual interest earned on the savings will help him reach that goal. To find the amount of each annual payment, we can use the formula for the future value of an annuity:

A = P * ((1+i)^n - 1) / i

Where:
A = annual payment
P = present value (in this case, $4000)
i = annual interest rate (8%)
n = number of periods (5 years)

Plugging in the values, we get:

A = $4000 * ((1+0.08)^5 - 1) / 0.08
A = $4000 * (1.08^5 - 1) / 0.08
A = $4000 * (1.469328 - 1) / 0.08
A = $4000 * 0.469328 / 0.08
A = $4000 * 5.8666
A = $23466.40

So, John would need to make annual payments of $23466.40 in order to have enough money in the sinking fund to pay off the $4000 after 5 years. This amount may seem high, but keep in mind that the annual interest earned on the savings will also contribute to reaching the goal of $4000.