Question

Karen opens a savings account with $1500. She deposits $100 every month into the account that has a 0.85% interest rate, compounded annually. If she doesn’t withdraw any money, what will the account balance be in 10 years?

$14,102.05

$14,208.04

$12,575.55

$12,469.56

Answer 1

Have you tried solving this yourself? It's quite a big problem to just copy and paste on here. I'd like to know what you did, where you think you went wrong, etc.

Answer 2

I used the future value formula and the compound interest formula

Answer 3

im not getting any of the answers

Answer 4

what would be the present value?

Answer 5

so what would be the formula?

Answer 6

this is the same problem as yesterday

Answer 7

the 2nd problem in this thread
http://openstudy.com/users/chelseachels#/updates/51487480e4b05e69bfaaef33

Answer 8

the PMT(((1+I)^n)-1)?

Answer 9

yes

Answer 10

would the payment be $1500 or $100?

Answer 11

however, because it is compounded annually but payments are made monthly it is tricky

Answer 12

so then what do i do?

Answer 13

good question. I have not seen this problem before, and would have to figure it out.
However, I would expect that your book tells you how to do it. Can you find a similar problem in your book and post how they solve it ?

Answer 14

im in flvs so there is no book and the examples dont show how to do this problem

Answer 15

I am sure you have been taught how to do this... But I'll post something later on this problem.

Answer 16

ok thanks

Answer 17

In the meantime, you could grind it out one year at a time as I did and at the end of 10 years, I got the second choice.

Answer 18

can you write on here what steps you took?

Answer 19

There is no interest paid until the end of the first year at which time there will be 1500+1200 in the account. So multiply 2700 by 1.0085 to get the value of the account at the end of the first year.

Answer 20

ok

Answer 21

So now there is 2722.95 in the account. Interest will not be paid again until the end of the second year and at that time there will be 3922.95 in the account. Multiply that by 1.0085 to get 3956.30 at the end of the second year.

Answer 22

how did you get 3956.30?

Answer 23

Add 1200...Multiply by 1.0085
Until you have reached the end of the 10th year.

Answer 24

at that time there will be 3922.95 in the account. Multiply that by 1.0085

Answer 25

oh ok

Answer 26

3922.95 times 1.0085

Answer 27

3956.30 i understand that

Answer 28

and then you keep going until you get to the tenth year?

Answer 29

Add 1200...Multiply by 1.0085 Until you have reached the end of the 10th year.

Answer 30

thanks so much

Answer 31

yw

Answer 32

Here is what I came up with:

opens a savings account with $1500.
after 10 years, at interest rate 0.0085/year compounded yearly
P1= 1500* (1+0.0085)^10 = 1632.49

for the 2nd part, 100 per month, for 10 years
The trick (for me) is that you do not get interest until the end of the year, so after 12 months, she has put in 1200. now we find the interest.

This is like the problem of making a payment once a year of 1200,
That means we can use the formula
\[ PMT\frac{ (1+i)^n -1}{i} \]

PMT is 1200
I is 0.0085
n is 10 (years)

we get : 1200* ( (1.0085)^10 -1)/0.0085 = 12469.56

(notice that is one of the choices)
but we have to add in the dollars from the initial 1500: 1632.49
total is
1632.49 + 12469.56 = 14102.05